Essential Rules for Finding Derivatives

In this video, we're going to talk about how to find derivatives. So the first thing we're going to talk about is the derivative of a constant. What you need to know is that the derivative of any constant is always 0. So for example, the derivative of 7 is 0. The derivative of a constant like negative 4, that's also 0. Pi. is just a number. There's no variable to it. So the derivative of pi is zero. The derivative of pi to the e, that's also zero. Both pi and e, they're just a number. So if you don't have a variable, if you just have a constant, the derivative of any constant is always zero. Now the next thing we need to talk about is something known as the power rule. Let's say if we have a variable raised to a constant, such as x raised to the nth power. This is equal to n times x raised to the n minus 1. So let me give you some examples. Let's say we want to find the derivative of x to the second power. In this case, n is 2. This is going to be 2x raised to the 2 minus 1, which is 2x to the first power. In short, you need to move the exponent to the front. and then subtract this by 1. So let me give you some more examples. Let's say we want to find the derivative of x cubed. So what we're going to do is move the 3 to the front. So it's going to be 3x, and then we're going to take away 1 from 3. 3 minus 1 is 2, so it's 3x squared. The derivative of x to the fourth, that's going to be 4x cubed. And the derivative of x to the fifth power, that's 5x to the fourth power. So hopefully you see the pattern here, but that's the power rule. That's how you can use it. So with that in mind, what do you think the derivative of 4x to the fifth power is equal to? So this is the same as multiplying 4. by the derivative of x to the fifth power. And we could use the power rule on that part. The derivative of x to the fifth power is five x to the fourth power. And then we could simply multiply four and five, and this is gonna be 20 x to the fourth power. So this is known as the constant multiple rule. Go ahead and try these two examples. the derivative of 6x to the 8th power, and also the derivative of 5x to the 3rd power. So let's start with the first one. We're going to rewrite the constant 6, and then multiply it by the derivative of x to the 8th power. So moving the 8 to the front, it's going to be 6 times 8 times x, and then we're going to subtract 8 by 1, which is 7. So it's 6 times 8x to the 7th power. And then we can multiply 6 by 8. That's going to be 48 times x to the 7th power. So that's the answer for the first example. For the second one, it's going to be 5 times the derivative of x cubed, which is 3x squared. And then 5 times 3 is 15. So the answer is 15x squared. Now let's say we have... some polynomial function f of x is equal to, let's say, 4x cubed plus 7x squared minus 9x plus 5. So what is the derivative of this function? By the way, the notation for the derivative of a function is f prime of x. So if we take the derivative of f of x, this is going to equal f prime of x. for those of you who might be wondering. So for this problem, we just got to take it one step at a time. First, let's find the derivative of 4x cubed. It's going to be 4 times the derivative of x cubed, which is 3x squared. And then next, we need to find the derivative of 7x squared. So it's going to be 7 times the derivative of x squared, which is 2x. Now, what about the derivative of negative 9x? what is that equal to well first we need to know what the derivative of x is equal to x is the same as x to the first power. So using the power rule, we can move the exponent to the front. It's going to be 1x, and then if we subtract that number by 1, we get 0. Anything raised to the 0 power is 1. So the derivative of x, for instance, is 1. The derivative of 5x is going to be 5 times 1, or 5. The derivative of negative 7x is simply negative 7. So the derivative of negative 9x is simply going to be negative 9. And we know that the derivative of a constant that is a number without a variable, it's always 0. So the answer for this example is going to be 4 times 3, which is 12. 7 times 2 is 14. And so it's going to be 12x squared plus 14x minus 9. So that is the derivative of the original function. Now let's work on finding the derivative of a rational function. What is the derivative of 1 over x squared? How would you find the answer for this one? Well, it turns out that you could use the power rule, but you need to rewrite the expression first. The first thing you need to do is move the x variable to the top. As you do so, the sine of the exponent changes from positive 2 to negative 2. So you can write it like this, or you can simply just write it as x to the minus 2. It's the same. So now using the power rule, we can move the exponent to the front. This is going to be negative 2x, and then we need to subtract the exponent by 1. Negative 2 minus 1 is negative 3. So this is our answer, but we can simplify it or rewrite it a bit. What you want to do is get rid of the negative exponent. So you want to move the x variable back to the bottom. Thus, we're going to have a negative 2 on top, and x raised to the positive 3 on the bottom. This is the final answer in its simplified form. So that's the derivative of 1 over x squared. Let's try another example. What is the derivative of 1 over x cubed? Feel free to pause the video if you want to work on it. Step one, we need to rewrite the expression. Let's move the x variable to the top. So this is equal to the derivative of x to the negative 3. And now let's use the power rule. Let's move the exponent to the front. And so it's going to be negative 3x, and then let's subtract that number by 1. So negative 3 minus 1 is negative 4. Finally, let's move the x variable back to the bottom. And so this is going to be negative 3 over... x raised to the fifth power actually I need to make a slight correction that's negative 3 over X to the fourth power that should be a 4 so this is the answer to the problem now let's work on the next problem let's find the derivative of negative 6 over X to the fifth power go ahead and work on that So like before we're going to move the x variable to the top. So the first step is to rewrite the problem. So this is going to be negative 6x to the negative 5th power over 1. Now let's move the exponent to the front. So we're going to have negative 6 times negative 5, and then it's going to be x. And then let's subtract negative 5 by 1. Negative 5 minus 1 is negative 6. So what we're going to do next is we're going to move the x variable back to the bottom, and we're going to multiply those two numbers. Negative 6 times negative 5, that's positive 30. And we have x to the 6 on the bottom. So this is the derivative of negative 6 over x to the 5th power. So now you know how to find the derivative of a rational function, let's move on to finding the derivative of a radical function. What is the derivative of the square root of x? How can we find the answer to that? Well, it's important to know that the exponent is 1 and the index number is 2. So like before, we need to rewrite this problem. We can rewrite the square root of x as x to the 1 half. Now in this form, we could use the square root of x to the 1 half. use the power rule. So we're going to move the exponent to the front. This is going to be one half x, and then we're going to subtract the exponent by one. So one half minus one, we could change one to two over two. So it becomes one minus two, which is negative one over the same denominator of two. So one half minus one is negative a half. So now at this point we can rewrite the expression. So this is equivalent to saying x raised to the negative 1 half divided by 2. And now let's move the exponent, I mean the variable, back to the bottom. So it's going to be 1 over 2 x to the 1 half. And now at this point, we can replace x to the 1 half with what we had at the beginning, the square root of x. So the final answer is 1 over 2 times the square root of x. Now let's work on this example. Let's find the derivative of the cube root of x to the fifth power. Go ahead and take a minute to work on it. Feel free to pause the video, by the way, if you want to try any of these examples. So let's begin by rewriting the expression. The cube root of x to the fifth power is the same as x raised to the 5 over 3. And now let's take the derivative by moving the exponent to the front. So it's going to be 5 over 3x, and now we need to subtract that by 1. So 5 over 3 minus 1. Well, we need to get common denominators in order to subtract these two numbers. So we're going to change 1 into 3 over 3, because 3 divided by 3 is 1. And now we can subtract the numerators of the two fractions. 5 minus 3 is 2, and the denominator is going to stay the same. So 5 over 3 minus 3 over 3 is 2 over 3. That's going to be our new exponent. And so you can move this x to the top and write it as 5x raised to the 2 thirds divided by 3. Notice that we have a positive exponent. So we're not going to move the x variable to the bottom and get a negative exponent. There's no need to do that in this problem. The only thing that we do need to do is convert this expression from an exponential fraction into a radical expression. So x raised to the 2 thirds can be rewritten as the cube root of x squared. So make sure you understand this. The b root of x to the a is equal to x raised to the a over b. The number inside the radical goes on top of the fraction. The index number goes on the bottom of the exponential fraction. But for this problem, this is the final answer. Now let's work on one more example of differentiating radical functions. So let's say we have the eighth root of x to the fifth power. Go ahead and find the derivative of that. So let's begin by rewriting the expression. So the eighth root of x to the fifth is the same as x raised to the 5 over 8 power. So now we're going to move the exponent to the front, just like before. This is going to be 5 over 8x. Now let's subtract the exponent by 1. We're following the same steps. So let's get common denominators. Let's replace negative 1 with negative 8 over 8. 5 minus 8 is minus 3. So this is going to be negative 3 over 8. So we can rewrite this as 5x to the negative 3 over 8 over 8. Now this time we do have a negative exponent, so we're going to move the x variable to the bottom of the fraction. So it's 5 over 8x raised to the 3 over 8. Now the next thing we're going to do is convert the exponential fraction back into its radical form. So x to the 3 eighths is the same as the eighth root of x cubed. And we're going to leave the answer like this. Now, if you want to, you could rationalize the denominator, but that's a whole other topic. I'm going to leave the answer in this form. Now, let's move on to finding the derivatives of trigonometric functions. So for now, hopefully, you could take out a sheet of paper and a pen or pencil and jot down some notes. The derivative of sine of the u variable, where u is a function of x, is going to be cosine of that u. variable times the derivative of u or times u prime. The derivative of cosine u is going to be negative sine of the u variable times u prime. The derivative of tangent u is going to be positive secant squared u times u prime. The derivative of cotangent u is going to be negative cosecant squared of u times u prime. The derivative of secant is secant tangent. times u prime. So hopefully you see a pattern here. And finally, the derivative of cosecant is going to be negative cosecant cotangent times u prime. So note that every time you differentiate a trig function that starts with a c, you're going to have a negative sign. That's one thing you want to keep in mind. The derivative of tangent and cotangent, they're similar to each other. This is secant squared, that's cosecant squared. And the derivative of secant and cosecant are also similar to each other. This is secant tangent, this is cosecant cotangent. So hopefully that will help you to remember these trigonometric derivatives. Now let's try some example problems. Let's start with this one. What is the derivative of sine x? Well, we know that the derivative of sine to the u is going to be cosine u times u prime. In this example, we can clearly see that u is equal to x. If u is equal to x, what's u prime? Well, we know that the derivative of x is simply 1. So this is going to be cosine x times 1, or just simply cosine x. Here's another example. What is the derivative of sine x to the third power? Try that one. So let's rewrite the formula. The derivative of sine to the u is going to be cosine. of u times u prime. So the derivative of sine x cubed, well first we know that u is x cubed, so u prime is going to be 3x squared. And then if we just replace u with x cubed and u prime with 3x squared, we're going to get the answer. So notice the process that we're taking here. First we differentiate sine, it turns into cosine. Whatever is inside of the sine function, that is the angle, it remains the same in the answer. So the angle of sine is x cubed, the angle of cosine remains x cubed. Next we take the derivative of the inside part of the function, x cubed, and that gives us 3x squared. So now let's move on to some other examples. The derivative of cosine x squared. Try that one. And also try the derivative of tangent raised to the x to the 5th power. I mean tangent of x to the 5th power. So first, we're going to find the derivative of cosine. The derivative of cosine is negative sine. And we're going to keep the angle the same. So the angle is x squared. Next, we're going to find the derivative of x squared, which is 2x. So the final answer is negative 2x sine x squared. Now, moving on to the next example, the derivative of tangent is secant squared. Now, the angle of tangent is x to the fifth, so the angle of secant squared will be the same thing. Next, we're going to find the derivative of x to the fifth power, which is 5x to the fourth power. So the final answer is 5x to the fourth power times secant squared x to the fifth power. Now what about the derivative of secant for x? What's the answer to that one? The derivative of secant is secant tangent, so it's going to be secant 4x tangent 4x the angle is going to remain the same after that we could find the derivative of 4x which is just 4 and I'm going to move the 4 to the front so the derivative of secant 4x is going to be 4 secant 4x tangent 4x now what about the derivative of let's say you cotangent x cubed plus x raised to the fifth power. The derivative of cotangent is negative cosecant squared. So we're going to have negative cosecant squared and then whatever we have inside of the angle of cotangent that's going to be the same for the angle of cosecant squared. which is x cubed plus x to the fifth power. Next, we need to find the derivative of the inside angle. The derivative of x cubed is 3x squared, and the derivative of x to the fifth is 5x to the fourth. And so I'm going to leave the answer like that, but if you want to rewrite it, you could rewrite it like this. You could say negative 3x squared plus 5x to the fourth power times cosecant squared. x cubed plus x to the fifth power. So that's the answer for this example. Now let's move on to the derivatives of natural logs. The derivative of ln u, this is equal to u prime divided by u. So let's say if we want to find the derivative of the natural log of x. In this case, u is equal to x. u prime, the derivative of x is going to be 1. So it's u prime over u, or just 1 over x. So that's the derivative of the natural log of x. Now let's try some more examples. So what if we want to find the derivative of, let's say, the natural log of x cubed. What's the answer? Well, we can see that u is equal to x cubed, u prime is 3x squared. So it's u prime over u, that's 3x squared over x cubed. x squared is basically x times x. x cubed is x times x times x. So we can cross out two x variables. And this is going to give us 3 over x as a final answer. That's one way in which you can get the answer. Here's another way. A property of logs and natural logs allows you to move the exponent to the front. This is without differentiation, by the way. So we can rewrite the expression as 3 times ln x. Now from the previous example, we know that the derivative of ln x is 1 over x. So the derivative of 3 ln x is 3 times 1 over x, which is the same as 3 over x. So that's another way in which you can find the derivative of that particular natural log function. Go ahead and try this one. Find the derivative of ln x to the fourth minus x to the fifth. So we can see that u is x to the fourth minus x to the fifth. u prime is the derivative of that. It's 4x cubed minus 5x to the fourth. So the answer is going to be the derivative of what we see inside the natural log function divided by whatever is inside of it. And that's really an easier way to think about it in order to find the answer. So let's say if we want to find the derivative of the natural log of tangent x. So first, take the derivative of the inside part of ln. The derivative of tangent is secant squared. And then simply divide it by whatever is inside of the natural log function, which is tangent. So the answer is secant squared over tangent, but chances are we can reduce this expression. Secant is 1 over cosine, so secant squared is going to be 1 over cosine squared. Tangent is sine over cosine. What we could do is multiply the top and the bottom by, I'm going to multiply by cosine squared. These two will cancel, and I'm going to get 1 on top. One of the cosines will cancel, leaving 1 cosine here, so I'm going to get sine times cosine. which we can write that as 1 over sine times 1 over cosine. 1 over sine is cosecant x, and 1 over cosine is secant x. So the final answer is cosecant x times secant x. So now that we've talked about how to find the derivative of a natural log function, let's talk about how we could find the derivative of a regular logarithmic function, such as log base a of u. This is going to be equal to u prime over u ln a. Now, I'm going to compare that to the derivative of the natural log of u. which we said is u prime over u. Now, notice that here we have a base a, and it turns into l and a. The base of a natural log is e, and what we should have here is l and e. So these two are essentially the same thing. Instead of a, we have e. But the natural log of e, what you need to know is equal to 1. So because of that, we really don't need this expression here. But I want you to compare and see the similarities between the two equations. So let's find the derivative of, let's say, log base 2 of x to the fifth power. Go ahead and try that. u in this example is what we see here, x to the fifth power, and a is 2. u prime, the derivative of x to the fifth power, that's going to be 5x to the fourth power. So all we got to do is plug in the information into that formula, and we're going to get the answer. So it's u prime divided by u times the natural log of a, or ln2. Now, we can simplify our answer. x to the 5th power, I can rewrite that as x to the 4th power times x, because 1 plus 4 is 5. And so we can cancel x to the 4th power. Thus, the final answer is going to be 5 over x ln 2. Now remember, you could move this 5 to the front, and thus we see that here. So that's the final simplified answer for that problem. Let's try another problem with regular logs. So let's say we have log base 4 of x cubed plus 4x squared. Let's try that one. So u is going to be whatever we see inside of the log expression. So it's x cubed plus 4x squared. a is the base of the log expression. a is 4. And u prime is just the derivative of what we see inside of here. The derivative of x cubed is 3x squared. The derivative of 4x squared is 4 times 2x, or 8x. And so plugging everything into that formula, it's going to be u prime, which is 3x squared plus 8x, divided by u, which is x cubed plus 4x squared, and then times the natural log of a, or ln4. Now, you could leave your answer like this, but understand that you could simplify this expression. In the numerator, we can factor out an x, and we'll be left with 3x plus 8. In the denominator, we can take out an x squared, which I'm going to write as x times x. If we take out x squared, we're going to be left with x plus 4, and then times ln 4. So now we can cross out an x variable. So we can leave our answer as 3x plus 8 divided by x times x plus 4 times the natural log of 4. Or you can just rewrite that part as x squared plus 4x. There's many ways in which you can write your final answer. Hopefully your teacher won't be too picky about exactly how she wants you to leave your final answer. But that's it. So that's how you could find the derivative of this logarithmic function. Now let's move on to our next topic, and that is the derivative of exponential functions. So let's say we have a constant raised to a variable. Before, we had a variable raised to a constant, which we would use the power rule. But now we have a constant raised to a variable. And this is going to equal that constant raised to that variable times the derivative of that variable. So let's put this information to practice. Let's say if we want to find the derivative of e to the x. So it's going to be e to the x times the derivative of x, which is 1. So the derivative of e to the x is simply e raised to the x power. Now what about the derivative of e to the 2x? It's going to be the same thing, e to the 2x, times the derivative of 2x, which is 2. So you can write it as 2e to the 2x. Now let's work on some other examples. What is the derivative of e to the 5x? It's going to be e to the 5x times 5, which is 5e to the 5x. And then finally, if we have the derivative of e to the x squared, it's going to be the same thing, e to the x squared times the derivative of x squared, which is 2x. Now let's do one more. Let's say we want to find the derivative of e to the sine x. This is going to be e raised to the sine x times the derivative of sine x, which is cosine x. So that's how you can differentiate an exponential function with a base e. Now what if the base is not e? Let's say it's some number other than e, because e is a special number. It's equal to 2.7182818, and it just continues. But what if we have, let's say, some other constant, race the u variable? What's the derivative of that? It's going to be the same thing a to the u times u prime, but times ln a. If we contrast that with e to the u, which is e to the u times u prime, but technically times ln e. But L and E is 1, so we don't need that part when dealing with E. It's just E to U times U prime. But if the constant is a number other than E, then you need to add the L and A part to the answer. So let's put this to practice. Let's say we want to find the derivative of 5 raised to the x power. This is going to be the same thing, 5 raised to the x, times the derivative of x, which is 1, times the natural log of 5. So based on that, go ahead and try these examples. Find the derivative of 4 raised to the x squared, and also... the derivative of 7 raised to the x to the fourth. So this is going to be whatever we see here. So we're just going to rewrite it. That's 4 raised to the x squared, and then times u prime. u in this example is x squared, so the derivative of x squared is 2x, and then times the natural log of whatever the base is. The base in this example is 4. For the next one, it's going to be exactly what we started with. 7 raised to the x to the 4th times the derivative of the exponent. The derivative of x to the 4th is 4x cubed times the natural log of the base. And the base is 7. So that's the derivative of 7x to the 4th. You simply need to just follow the equation. As you can see, this is the a to the u part. U prime, that's the derivative. of the exponent and then L and A we can see a is 7 now the next topic of discussion is the product rule so let's say you have a function multiplied to another function u times V What is the derivative of u times v? Well, first you need to differentiate the first part, leave the second part the same, plus leave the first part the same, differentiate the second part. So let's say we have the derivative of x cubed times sine x. In this example, you could say that u is x cubed and v is sine x. So we're going to differentiate the first part, that is x cubed. The derivative of x cubed is 3x squared. And then we're going to leave the second part the same. So we're just going to rewrite sine x. Plus, now we're going to rewrite the first part, x cubed, and then take the derivative of the second part. The derivative of sine x is cosine x. So this... the answer. Now for those of you who want to simplify it you could factor out the GCF which is x squared. If you decide to do that here's where you're gonna get. Taking out x squared in the first term you'll be left with 3 sine x. Taking out x squared in the second term, you'll be left with x cosine x. So you could leave your answer in this format if you want to. Now let's try another example. What is the derivative of x squared ln x? So feel free to pause the video if you want to try that. So this is u and this is v. So let's begin by differentiating the first part, x squared. So we're going to get 2x, and then we're going to rewrite the second part, ln x. We're not going to change that. Plus, let's rewrite the first part, x squared, and then let's take the derivative of the second part. The derivative of ln x is 1 over x. So we're going to get 2x ln x, and then x squared times 1 over x, that's like x squared divided by x, which is x. Now like before, we can factor on x. If we do so, we're going to get 2 ln x plus 1. So that's the final answer, if we wish to simplify it that far. So now you know how to use the product rule when finding derivatives. Here's another example. So let's say we have 7x plus 4 times x squared plus 8. Now we could use the product rule to find the derivative of this expression, or we can FOIL and then use the power rule to find the derivative of the resultant polynomial function. But let's use the product rule. So first, let's find the derivative of the first part, 7x plus 4. The derivative of 7x is 7. The derivative of 4 is 0, so we just get 7. Next, we're going to rewrite the second part. And then we're going to rewrite the first part, and then take the derivative of the second part. The derivative of x squared is 2x, the derivative of the constant 8 is 0. So this is the answer that we get. Now what I'm going to do is I'm going to distribute and combine like terms. So this is going to be 7x squared plus 56. And this is going to be 14x squared plus 8x. So I can combine 7x squared and 14x squared. So I'm going to get 21x squared plus 8x plus 56. And so this is the final answer. Now, note the other way in which we can get the answer. We could have initially distributed before we take the derivative. 7x times x squared is 7x cubed. And then this is 7x times 8, which is 56x. This is 4x squared. And this 4 times 8 is 32. Now, if we were to take the derivative of 7x cubed, that's going to be 7 times 3x squared, which is 21x squared. The derivative of 4x squared is 8x. And the derivative of... 56 x is 56. The derivative of the constant is 0. So we would still get the same answer in the end. Now let's talk about the quotient rule. Let's say if we have a function divided by another function, or u over v. To find the derivative of that expression, we could use this formula. It's going to be v u prime minus u v prime over v squared. So let's say we want to find the derivative of 3x minus 5 over 7x plus 4. Now if you want to, it helps to write down a few things. So u is whatever is on the numerator of the fraction. So we can say that u is 3x minus 5. v is the stuff on the bottom or on the denominator of the fraction. So that's 7x plus 4. u prime, the derivative of 3x minus 5 is just 3. v prime in this example is simply 7. So now we just got to plug in everything into that formula. So it's v, that's 7x plus 4, times u prime, which is 3. And then minus u, that's 3x minus 5 times v prime, which is 7, divided by v squared. So that's 7x plus 4 squared. Now let's do some algebra. So let's begin by distributing the 3 to 7x and 4. 7x times 3 is 21x. 4 times 3 is 12. Next, let's distribute 7 to 3x, and keep in mind it's a negative sign, so that's going to be negative 21x. And then 7 times negative 5, combined with that negative sign, that's going to be positive 35. So immediately we could see that we could cancel the 21x and then we can combine 12 and 35 which will be 47. So the final answer is 47 divided by 7x plus 4 squared for this problem. So that's how you can use the quotient rule whenever you have a division of two functions. Now let's talk about the chain rule. We've covered some examples of it early in this video. What is the derivative of sine x to the 5th power? From what we talked about, we would differentiate the outer part of the function. The derivative of sine is cosine. And then we would keep the inside the same. And then we would differentiate the inside part of the function. The derivative of x to the fifth power is 5x to the fourth power. That process can help you to solve derivatives associated with the chain rule. So when dealing with the chain rule, you're dealing with composite functions. So if you want to find the derivative of a function inside of another function, like f of g of x, It's equal to the derivative of the outside function, so f prime, and then the inside function stays the same, that is g of x, and then times the derivative of the inside function, g prime of x. So that's how you could find the derivative of a composite function, when you have a function inside of another function. So let's say we want to find the derivative of 4x cubed plus 7x raised to the fifth power. In this case, we need to start with the out of function, that is the power rule. So we're going to move the 5 to the front. We're going to keep the inside function the same. And then we're going to subtract the exponent by 1. Next, we're going to take the derivative of the inside function. The derivative of 4x cubed is 12x squared. The derivative of 7x is 7. So that's the derivative of this function. So you don't have to follow this five times. You could just simply use the chain rule to get the answer. now let's move on to our next example what is the derivative of tangent of sine x cubed. Go ahead and work on that example. So first we're going to find the derivative of the outer function, tangent, and that's going to be secant squared. Now, inside of tangent, we have sine x cubed. So we're going to have the same thing inside of secant squared. So we're starting with the outer function, and we're working our way in. So now let's work our way in towards sine. The derivative of sine is cosine, and the angle of sine is x cubed. So the angle of cosine will be x cubed. And then we're going to work even further towards the inside part of the functions. The derivative of sine is x cubed. of x cubed is 3x squared. Now, that's as far as we can go. So this is the answer of this expression. Secant squared sine x cubed times cosine x cubed times 3x squared. Here's another example you can work on. Find the derivative of sine to the fourth power. cosine of tangent x squared now what I would recommend doing since we have an exponent here is rewriting the expression So basically what this means is that this whole thing is raised to the fourth power. And it affects the way you're going to apply the chain rule. Because before you can differentiate the sine function, you need to use the power rule on that exponent. So let's move the 4 to the front. So it's going to be 4. And then we're going to rewrite everything we see inside of the brackets highlighted in red. Now we're going to subtract the exponent by 1. And then we're going to differentiate the sine function. So that's going to be cosine. And then the angle of sine is everything we see here. That's cosine of tangent x squared. So now we're going to move inward towards cosine. By the way, there's a multiplication for all of these functions, so it's times. The derivative of cosine is negative sine, and the angle of cosine is tangent x squared. So we're going to rewrite that. Now let's work our way in towards tangent. The derivative of tangent is secant squared, and the angle is x squared. Finally, we can find the derivative of x squared, which is 2x. And so this whole thing is our final answer. Now, of course, you can combine a few things. You can combine the 4 and the 2x and the negative sign. And you can put negative 8x in front. But besides that, there's nothing else you can do here. So that's it for this problem. Now the next topic we need to talk about is implicit differentiation. Sometimes you'll have equations with two different variables, like this one, x and y. And you want to find the derivative of y with respect to x. So basically you want to determine the value of dy dx. How do you do that? Well, what you do is you're going to differentiate both sides with respect to x. Now before we do that, I want to talk about a few things. When you differentiate x cubed with respect to x, you're going to get 3x squared. And technically you get times dx over dx, which cancels to 1. But if you want to differentiate y cubed with respect to x, you're going to get 3y squared, but times dy dx. Note the difference here, since these variables don't match up. Here, these, they do match up, so this would be dx over dx, which is pointless in writing. So implicit, when dealing with implicit differentiation problems, every time you differentiate y with respect to x, you need to add dy dx. Now sometimes you'll be dealing with what is known as related rates. Instead of differentiating with respect to x, you'll be differentiating with respect to t, or time. So if you differentiate y with respect to x, you'll be differentiating with respect to t, or time. If you differentiate x cubed with respect to time, it's going to be 3x squared, but times dx dt. Likewise, if you differentiate y cubed with respect to time, it's going to be 3y squared times dy dt. So just keep that in mind. But when dealing with implicit differentiation, I mean, let me say that again. When dealing with implicit differentiation, we are differentiating with respect to x. So the derivative of x to the fourth is simply going to be 4x cubed. But the derivative of y to the fourth is going to be 4y cubed, but times dy over dx. And this is what we need to solve for. The derivative of 12 is 0. So I'm going to take this term and move it to the other side. So I'm going to have 4y cubed times dy dx, and that's going to be equal to negative 4x cubed. And then all we need to do is divide both sides by 4y cubed. and then the force will cancel. And so we get dy over dx is equal to negative x cubed over y cubed. So that's a quick and simple example of how to perform. implicit differentiation. For the sake of practice, let's work on one more example. Let's say we have x cubed plus 4xy plus y squared is equal to 9. Go ahead and calculate the value of dy over dx. So we're going to differentiate this entire equation with respect to x. The derivative of x cubed is 3x squared. Now, notice that we have x and y. So we need to use the product rule. I'm going to say the first part, the u part, is 4x, the v part is going to be y. So the derivative of the first part, 4x, is just 4, times the second part, which is just y, plus the first part, 4x, times the derivative of the second part. The derivative of y is going to be 1 times dy over dx, because we're differentiating y with respect to x. Now the derivative of y squared is going to be 2y times dy over dx. And the derivative of a constant is always 0. So everything that does not have a dy over dx attached to it, I'm going to move to the other side. So I'm going to move these two terms across the equal sign. So we're going to have 4x dy over dx plus... 2y dy over dx and that's going to equal negative 3x squared minus 4y. So because we have a common factor we can now factor out the dy over dx on the left side. And that's going to leave us with 4x plus 2y, which equals negative 3x minus 4y. So now we can divide both sides by this, 4x plus 2y. So these two will cancel. And thus we have our final answer. dy over dx is equal to negative 3x squared minus 4y. And we could factor out a 2 on the bottom, but I'm not going to worry about that in this example. So that's going to be divided by 4x plus 2y. So now you know how to find the derivative in or using implicit differentiation. So far, we talked about how we could find the derivative of a variable raised to a constant. And that is by using the power rule. X to the n, the derivative of that is going to be n times x raised to the n minus 1. And we also talk about having or finding the derivative of a constant raised to a variable. And that was a to the u times u prime times ln a. So we have a variable raised to a constant and a constant raised to a variable. But what about finding the derivative? Of a variable raised to a variable. How do we do that? How do we find the derivative of x raised to the x? Well, first, let's make y equal to x raised to the x. So if we could find the derivative of y, then we know what the derivative of x is. x to the x's. That's step one. Step two is to take the natural log of both sides. So ln y is equal to ln x raised to the x. Now a property of natural logs allows us to move the variable x to the front. And so what we're going to have is L and Y is equal to X, L and X. And now at this point, we are going to differentiate both sides with respect to X. So here's how you can express that in mathematical words. Now we know that the derivative of ln x is 1 over x, so the derivative of ln y is going to be 1 over y. Now notice that we're differentiating y with respect to x. So when you see that, you need to add dy over dx. This is basically implicit differentiation. On the right side, we need to use the product rule. So we're going to differentiate the first part, x, which is 1, and leave the second part by itself. And then we're going to leave the first part by itself, and then differentiate the second part. The derivative of ln x is 1 over x. So what we have now is 1 over y times dy over dx, and that's equal to ln x, and then x times 1 over x, you can cross out the x variables, so it becomes plus 1. And now what we're going to do is we're going to multiply both sides by y. So we can cross out the y variables here. So we're going to have dy dx is equal to y times, you can say ln x plus 1, or you can say 1 plus ln x. Now recall at the beginning, we said x raised to the x equal to y. At this point, we're going to substitute y with x raised to the x. So the final answer is going to be x raised to the x. times 1 plus ln x. So that's how you could perform what is known as logarithmic differentiation. That's how you could find the derivative of a variable raised to a variable. So this is the answer. So you could say that the derivative of x raised to the x in this example is x raised to the x times 1 plus ln x. Well, that's basically it for this video. We've covered a lot of topics, and hopefully you found this to be helpful. Thanks for watching.

So the derivative of pi is zero. The derivative of pi to the e, that's also zero. Both pi and e, they're just a number. So if you don't have a variable, if you just have a constant, the derivative of any constant is always zero. Now the next thing we need to talk about is something known as the power rule.

Let's say if we have a variable raised to a constant, such as x raised to the nth power. This is equal to n times x raised to the n minus 1. So let me give you some examples. Let's say we want to find the derivative of x to the second power. In this case, n is 2. This is going to be 2x raised to the 2 minus 1, which is 2x to the first power. In short, you need to move the exponent to the front.

and then subtract this by 1. So let me give you some more examples. Let's say we want to find the derivative of x cubed. So what we're going to do is move the 3 to the front.

So it's going to be 3x, and then we're going to take away 1 from 3. 3 minus 1 is 2, so it's 3x squared. The derivative of x to the fourth, that's going to be 4x cubed. And the derivative of x to the fifth power, that's 5x to the fourth power.

So hopefully you see the pattern here, but that's the power rule. That's how you can use it. So with that in mind, what do you think the derivative of 4x to the fifth power is equal to? So this is the same as multiplying 4. by the derivative of x to the fifth power. And we could use the power rule on that part.

The derivative of x to the fifth power is five x to the fourth power. And then we could simply multiply four and five, and this is gonna be 20 x to the fourth power. So this is known as the constant multiple rule. Go ahead and try these two examples.

the derivative of 6x to the 8th power, and also the derivative of 5x to the 3rd power. So let's start with the first one. We're going to rewrite the constant 6, and then multiply it by the derivative of x to the 8th power. So moving the 8 to the front, it's going to be 6 times 8 times x, and then we're going to subtract 8 by 1, which is 7. So it's 6 times 8x to the 7th power.

And then we can multiply 6 by 8. That's going to be 48 times x to the 7th power. So that's the answer for the first example. For the second one, it's going to be 5 times the derivative of x cubed, which is 3x squared.

And then 5 times 3 is 15. So the answer is 15x squared. Now let's say we have... some polynomial function f of x is equal to, let's say, 4x cubed plus 7x squared minus 9x plus 5. So what is the derivative of this function? By the way, the notation for the derivative of a function is f prime of x. So if we take the derivative of f of x, this is going to equal f prime of x.

for those of you who might be wondering. So for this problem, we just got to take it one step at a time. First, let's find the derivative of 4x cubed.

It's going to be 4 times the derivative of x cubed, which is 3x squared. And then next, we need to find the derivative of 7x squared. So it's going to be 7 times the derivative of x squared, which is 2x. Now, what about the derivative of negative 9x? what is that equal to well first we need to know what the derivative of x is equal to x is the same as x to the first power.

So using the power rule, we can move the exponent to the front. It's going to be 1x, and then if we subtract that number by 1, we get 0. Anything raised to the 0 power is 1. So the derivative of x, for instance, is 1. The derivative of 5x is going to be 5 times 1, or 5. The derivative of negative 7x is simply negative 7. So the derivative of negative 9x is simply going to be negative 9. And we know that the derivative of a constant that is a number without a variable, it's always 0. So the answer for this example is going to be 4 times 3, which is 12. 7 times 2 is 14. And so it's going to be 12x squared plus 14x minus 9. So that is the derivative of the original function. Now let's work on finding the derivative of a rational function.

What is the derivative of 1 over x squared? How would you find the answer for this one? Well, it turns out that you could use the power rule, but you need to rewrite the expression first. The first thing you need to do is move the x variable to the top. As you do so, the sine of the exponent changes from positive 2 to negative 2. So you can write it like this, or you can simply just write it as x to the minus 2. It's the same.

So now using the power rule, we can move the exponent to the front. This is going to be negative 2x, and then we need to subtract the exponent by 1. Negative 2 minus 1 is negative 3. So this is our answer, but we can simplify it or rewrite it a bit. What you want to do is get rid of the negative exponent.

So you want to move the x variable back to the bottom. Thus, we're going to have a negative 2 on top, and x raised to the positive 3 on the bottom. This is the final answer in its simplified form. So that's the derivative of 1 over x squared. Let's try another example.

What is the derivative of 1 over x cubed? Feel free to pause the video if you want to work on it. Step one, we need to rewrite the expression. Let's move the x variable to the top.

So this is equal to the derivative of x to the negative 3. And now let's use the power rule. Let's move the exponent to the front. And so it's going to be negative 3x, and then let's subtract that number by 1. So negative 3 minus 1 is negative 4. Finally, let's move the x variable back to the bottom.

And so this is going to be negative 3 over... x raised to the fifth power actually I need to make a slight correction that's negative 3 over X to the fourth power that should be a 4 so this is the answer to the problem now let's work on the next problem let's find the derivative of negative 6 over X to the fifth power go ahead and work on that So like before we're going to move the x variable to the top. So the first step is to rewrite the problem.

So this is going to be negative 6x to the negative 5th power over 1. Now let's move the exponent to the front. So we're going to have negative 6 times negative 5, and then it's going to be x. And then let's subtract negative 5 by 1. Negative 5 minus 1 is negative 6. So what we're going to do next is we're going to move the x variable back to the bottom, and we're going to multiply those two numbers.

Negative 6 times negative 5, that's positive 30. And we have x to the 6 on the bottom. So this is the derivative of negative 6 over x to the 5th power. So now you know how to find the derivative of a rational function, let's move on to finding the derivative of a radical function. What is the derivative of the square root of x?

How can we find the answer to that? Well, it's important to know that the exponent is 1 and the index number is 2. So like before, we need to rewrite this problem. We can rewrite the square root of x as x to the 1 half. Now in this form, we could use the square root of x to the 1 half. use the power rule.

So we're going to move the exponent to the front. This is going to be one half x, and then we're going to subtract the exponent by one. So one half minus one, we could change one to two over two. So it becomes one minus two, which is negative one over the same denominator of two. So one half minus one is negative a half.

So now at this point we can rewrite the expression. So this is equivalent to saying x raised to the negative 1 half divided by 2. And now let's move the exponent, I mean the variable, back to the bottom. So it's going to be 1 over 2 x to the 1 half. And now at this point, we can replace x to the 1 half with what we had at the beginning, the square root of x. So the final answer is 1 over 2 times the square root of x.

Now let's work on this example. Let's find the derivative of the cube root of x to the fifth power. Go ahead and take a minute to work on it.

Feel free to pause the video, by the way, if you want to try any of these examples. So let's begin by rewriting the expression. The cube root of x to the fifth power is the same as x raised to the 5 over 3. And now let's take the derivative by moving the exponent to the front. So it's going to be 5 over 3x, and now we need to subtract that by 1. So 5 over 3 minus 1. Well, we need to get common denominators in order to subtract these two numbers.

So we're going to change 1 into 3 over 3, because 3 divided by 3 is 1. And now we can subtract the numerators of the two fractions. 5 minus 3 is 2, and the denominator is going to stay the same. So 5 over 3 minus 3 over 3 is 2 over 3. That's going to be our new exponent. And so you can move this x to the top and write it as 5x raised to the 2 thirds divided by 3. Notice that we have a positive exponent. So we're not going to move the x variable to the bottom and get a negative exponent.

There's no need to do that in this problem. The only thing that we do need to do is convert this expression from an exponential fraction into a radical expression. So x raised to the 2 thirds can be rewritten as the cube root of x squared.

So make sure you understand this. The b root of x to the a is equal to x raised to the a over b. The number inside the radical goes on top of the fraction. The index number goes on the bottom of the exponential fraction.

But for this problem, this is the final answer. Now let's work on one more example of differentiating radical functions. So let's say we have the eighth root of x to the fifth power.

Go ahead and find the derivative of that. So let's begin by rewriting the expression. So the eighth root of x to the fifth is the same as x raised to the 5 over 8 power. So now we're going to move the exponent to the front, just like before.

This is going to be 5 over 8x. Now let's subtract the exponent by 1. We're following the same steps. So let's get common denominators. Let's replace negative 1 with negative 8 over 8. 5 minus 8 is minus 3. So this is going to be negative 3 over 8. So we can rewrite this as 5x to the negative 3 over 8 over 8. Now this time we do have a negative exponent, so we're going to move the x variable to the bottom of the fraction.

So it's 5 over 8x raised to the 3 over 8. Now the next thing we're going to do is convert the exponential fraction back into its radical form. So x to the 3 eighths is the same as the eighth root of x cubed. And we're going to leave the answer like this. Now, if you want to, you could rationalize the denominator, but that's a whole other topic.

I'm going to leave the answer in this form. Now, let's move on to finding the derivatives of trigonometric functions. So for now, hopefully, you could take out a sheet of paper and a pen or pencil and jot down some notes. The derivative of sine of the u variable, where u is a function of x, is going to be cosine of that u. variable times the derivative of u or times u prime.

The derivative of cosine u is going to be negative sine of the u variable times u prime. The derivative of tangent u is going to be positive secant squared u times u prime. The derivative of cotangent u is going to be negative cosecant squared of u times u prime.

The derivative of secant is secant tangent. times u prime. So hopefully you see a pattern here.

And finally, the derivative of cosecant is going to be negative cosecant cotangent times u prime. So note that every time you differentiate a trig function that starts with a c, you're going to have a negative sign. That's one thing you want to keep in mind.

The derivative of tangent and cotangent, they're similar to each other. This is secant squared, that's cosecant squared. And the derivative of secant and cosecant are also similar to each other.

This is secant tangent, this is cosecant cotangent. So hopefully that will help you to remember these trigonometric derivatives. Now let's try some example problems. Let's start with this one. What is the derivative of sine x?

Well, we know that the derivative of sine to the u is going to be cosine u times u prime. In this example, we can clearly see that u is equal to x. If u is equal to x, what's u prime? Well, we know that the derivative of x is simply 1. So this is going to be cosine x times 1, or just simply cosine x. Here's another example.

What is the derivative of sine x to the third power? Try that one. So let's rewrite the formula. The derivative of sine to the u is going to be cosine.

of u times u prime. So the derivative of sine x cubed, well first we know that u is x cubed, so u prime is going to be 3x squared. And then if we just replace u with x cubed and u prime with 3x squared, we're going to get the answer. So notice the process that we're taking here.

First we differentiate sine, it turns into cosine. Whatever is inside of the sine function, that is the angle, it remains the same in the answer. So the angle of sine is x cubed, the angle of cosine remains x cubed.

Next we take the derivative of the inside part of the function, x cubed, and that gives us 3x squared. So now let's move on to some other examples. The derivative of cosine x squared. Try that one.

And also try the derivative of tangent raised to the x to the 5th power. I mean tangent of x to the 5th power. So first, we're going to find the derivative of cosine.

The derivative of cosine is negative sine. And we're going to keep the angle the same. So the angle is x squared.

Next, we're going to find the derivative of x squared, which is 2x. So the final answer is negative 2x sine x squared. Now, moving on to the next example, the derivative of tangent is secant squared. Now, the angle of tangent is x to the fifth, so the angle of secant squared will be the same thing. Next, we're going to find the derivative of x to the fifth power, which is 5x to the fourth power.

So the final answer is 5x to the fourth power times secant squared x to the fifth power. Now what about the derivative of secant for x? What's the answer to that one?

The derivative of secant is secant tangent, so it's going to be secant 4x tangent 4x the angle is going to remain the same after that we could find the derivative of 4x which is just 4 and I'm going to move the 4 to the front so the derivative of secant 4x is going to be 4 secant 4x tangent 4x now what about the derivative of let's say you cotangent x cubed plus x raised to the fifth power. The derivative of cotangent is negative cosecant squared. So we're going to have negative cosecant squared and then whatever we have inside of the angle of cotangent that's going to be the same for the angle of cosecant squared.

which is x cubed plus x to the fifth power. Next, we need to find the derivative of the inside angle. The derivative of x cubed is 3x squared, and the derivative of x to the fifth is 5x to the fourth. And so I'm going to leave the answer like that, but if you want to rewrite it, you could rewrite it like this.

You could say negative 3x squared plus 5x to the fourth power times cosecant squared. x cubed plus x to the fifth power. So that's the answer for this example.

Now let's move on to the derivatives of natural logs. The derivative of ln u, this is equal to u prime divided by u. So let's say if we want to find the derivative of the natural log of x.

In this case, u is equal to x. u prime, the derivative of x is going to be 1. So it's u prime over u, or just 1 over x. So that's the derivative of the natural log of x. Now let's try some more examples.

So what if we want to find the derivative of, let's say, the natural log of x cubed. What's the answer? Well, we can see that u is equal to x cubed, u prime is 3x squared. So it's u prime over u, that's 3x squared over x cubed. x squared is basically x times x.

x cubed is x times x times x. So we can cross out two x variables. And this is going to give us 3 over x as a final answer. That's one way in which you can get the answer. Here's another way.

A property of logs and natural logs allows you to move the exponent to the front. This is without differentiation, by the way. So we can rewrite the expression as 3 times ln x. Now from the previous example, we know that the derivative of ln x is 1 over x. So the derivative of 3 ln x is 3 times 1 over x, which is the same as 3 over x.

So that's another way in which you can find the derivative of that particular natural log function. Go ahead and try this one. Find the derivative of ln x to the fourth minus x to the fifth.

So we can see that u is x to the fourth minus x to the fifth. u prime is the derivative of that. It's 4x cubed minus 5x to the fourth.

So the answer is going to be the derivative of what we see inside the natural log function divided by whatever is inside of it. And that's really an easier way to think about it in order to find the answer. So let's say if we want to find the derivative of the natural log of tangent x.

So first, take the derivative of the inside part of ln. The derivative of tangent is secant squared. And then simply divide it by whatever is inside of the natural log function, which is tangent. So the answer is secant squared over tangent, but chances are we can reduce this expression.

Secant is 1 over cosine, so secant squared is going to be 1 over cosine squared. Tangent is sine over cosine. What we could do is multiply the top and the bottom by, I'm going to multiply by cosine squared. These two will cancel, and I'm going to get 1 on top.

One of the cosines will cancel, leaving 1 cosine here, so I'm going to get sine times cosine. which we can write that as 1 over sine times 1 over cosine. 1 over sine is cosecant x, and 1 over cosine is secant x. So the final answer is cosecant x times secant x. So now that we've talked about how to find the derivative of a natural log function, let's talk about how we could find the derivative of a regular logarithmic function, such as log base a of u.

This is going to be equal to u prime over u ln a. Now, I'm going to compare that to the derivative of the natural log of u. which we said is u prime over u.

Now, notice that here we have a base a, and it turns into l and a. The base of a natural log is e, and what we should have here is l and e. So these two are essentially the same thing. Instead of a, we have e. But the natural log of e, what you need to know is equal to 1. So because of that, we really don't need this expression here.

But I want you to compare and see the similarities between the two equations. So let's find the derivative of, let's say, log base 2 of x to the fifth power. Go ahead and try that.

u in this example is what we see here, x to the fifth power, and a is 2. u prime, the derivative of x to the fifth power, that's going to be 5x to the fourth power. So all we got to do is plug in the information into that formula, and we're going to get the answer. So it's u prime divided by u times the natural log of a, or ln2. Now, we can simplify our answer. x to the 5th power, I can rewrite that as x to the 4th power times x, because 1 plus 4 is 5. And so we can cancel x to the 4th power.

Thus, the final answer is going to be 5 over x ln 2. Now remember, you could move this 5 to the front, and thus we see that here. So that's the final simplified answer for that problem. Let's try another problem with regular logs. So let's say we have log base 4 of x cubed plus 4x squared. Let's try that one.

So u is going to be whatever we see inside of the log expression. So it's x cubed plus 4x squared. a is the base of the log expression. a is 4. And u prime is just the derivative of what we see inside of here.

The derivative of x cubed is 3x squared. The derivative of 4x squared is 4 times 2x, or 8x. And so plugging everything into that formula, it's going to be u prime, which is 3x squared plus 8x, divided by u, which is x cubed plus 4x squared, and then times the natural log of a, or ln4. Now, you could leave your answer like this, but understand that you could simplify this expression. In the numerator, we can factor out an x, and we'll be left with 3x plus 8. In the denominator, we can take out an x squared, which I'm going to write as x times x.

If we take out x squared, we're going to be left with x plus 4, and then times ln 4. So now we can cross out an x variable. So we can leave our answer as 3x plus 8 divided by x times x plus 4 times the natural log of 4. Or you can just rewrite that part as x squared plus 4x. There's many ways in which you can write your final answer. Hopefully your teacher won't be too picky about exactly how she wants you to leave your final answer.

But that's it. So that's how you could find the derivative of this logarithmic function. Now let's move on to our next topic, and that is the derivative of exponential functions.

So let's say we have a constant raised to a variable. Before, we had a variable raised to a constant, which we would use the power rule. But now we have a constant raised to a variable. And this is going to equal that constant raised to that variable times the derivative of that variable.

So let's put this information to practice. Let's say if we want to find the derivative of e to the x. So it's going to be e to the x times the derivative of x, which is 1. So the derivative of e to the x is simply e raised to the x power. Now what about the derivative of e to the 2x? It's going to be the same thing, e to the 2x, times the derivative of 2x, which is 2. So you can write it as 2e to the 2x.

Now let's work on some other examples. What is the derivative of e to the 5x? It's going to be e to the 5x times 5, which is 5e to the 5x.

And then finally, if we have the derivative of e to the x squared, it's going to be the same thing, e to the x squared times the derivative of x squared, which is 2x. Now let's do one more. Let's say we want to find the derivative of e to the sine x.

This is going to be e raised to the sine x times the derivative of sine x, which is cosine x. So that's how you can differentiate an exponential function with a base e. Now what if the base is not e? Let's say it's some number other than e, because e is a special number.

It's equal to 2.7182818, and it just continues. But what if we have, let's say, some other constant, race the u variable? What's the derivative of that?

It's going to be the same thing a to the u times u prime, but times ln a. If we contrast that with e to the u, which is e to the u times u prime, but technically times ln e. But L and E is 1, so we don't need that part when dealing with E. It's just E to U times U prime. But if the constant is a number other than E, then you need to add the L and A part to the answer.

So let's put this to practice. Let's say we want to find the derivative of 5 raised to the x power. This is going to be the same thing, 5 raised to the x, times the derivative of x, which is 1, times the natural log of 5. So based on that, go ahead and try these examples.

Find the derivative of 4 raised to the x squared, and also... the derivative of 7 raised to the x to the fourth. So this is going to be whatever we see here. So we're just going to rewrite it.

That's 4 raised to the x squared, and then times u prime. u in this example is x squared, so the derivative of x squared is 2x, and then times the natural log of whatever the base is. The base in this example is 4. For the next one, it's going to be exactly what we started with. 7 raised to the x to the 4th times the derivative of the exponent. The derivative of x to the 4th is 4x cubed times the natural log of the base.

And the base is 7. So that's the derivative of 7x to the 4th. You simply need to just follow the equation. As you can see, this is the a to the u part.

U prime, that's the derivative. of the exponent and then L and A we can see a is 7 now the next topic of discussion is the product rule so let's say you have a function multiplied to another function u times V What is the derivative of u times v? Well, first you need to differentiate the first part, leave the second part the same, plus leave the first part the same, differentiate the second part. So let's say we have the derivative of x cubed times sine x. In this example, you could say that u is x cubed and v is sine x.

So we're going to differentiate the first part, that is x cubed. The derivative of x cubed is 3x squared. And then we're going to leave the second part the same. So we're just going to rewrite sine x.

Plus, now we're going to rewrite the first part, x cubed, and then take the derivative of the second part. The derivative of sine x is cosine x. So this...

the answer. Now for those of you who want to simplify it you could factor out the GCF which is x squared. If you decide to do that here's where you're gonna get. Taking out x squared in the first term you'll be left with 3 sine x. Taking out x squared in the second term, you'll be left with x cosine x.

So you could leave your answer in this format if you want to. Now let's try another example. What is the derivative of x squared ln x? So feel free to pause the video if you want to try that.

So this is u and this is v. So let's begin by differentiating the first part, x squared. So we're going to get 2x, and then we're going to rewrite the second part, ln x. We're not going to change that. Plus, let's rewrite the first part, x squared, and then let's take the derivative of the second part. The derivative of ln x is 1 over x.

So we're going to get 2x ln x, and then x squared times 1 over x, that's like x squared divided by x, which is x. Now like before, we can factor on x. If we do so, we're going to get 2 ln x plus 1. So that's the final answer, if we wish to simplify it that far.

So now you know how to use the product rule when finding derivatives. Here's another example. So let's say we have 7x plus 4 times x squared plus 8. Now we could use the product rule to find the derivative of this expression, or we can FOIL and then use the power rule to find the derivative of the resultant polynomial function. But let's use the product rule. So first, let's find the derivative of the first part, 7x plus 4. The derivative of 7x is 7. The derivative of 4 is 0, so we just get 7. Next, we're going to rewrite the second part.

And then we're going to rewrite the first part, and then take the derivative of the second part. The derivative of x squared is 2x, the derivative of the constant 8 is 0. So this is the answer that we get. Now what I'm going to do is I'm going to distribute and combine like terms. So this is going to be 7x squared plus 56. And this is going to be 14x squared plus 8x.

So I can combine 7x squared and 14x squared. So I'm going to get 21x squared plus 8x plus 56. And so this is the final answer. Now, note the other way in which we can get the answer. We could have initially distributed before we take the derivative.

7x times x squared is 7x cubed. And then this is 7x times 8, which is 56x. This is 4x squared. And this 4 times 8 is 32. Now, if we were to take the derivative of 7x cubed, that's going to be 7 times 3x squared, which is 21x squared.

The derivative of 4x squared is 8x. And the derivative of... 56 x is 56. The derivative of the constant is 0. So we would still get the same answer in the end.

Now let's talk about the quotient rule. Let's say if we have a function divided by another function, or u over v. To find the derivative of that expression, we could use this formula. It's going to be v u prime minus u v prime over v squared.

So let's say we want to find the derivative of 3x minus 5 over 7x plus 4. Now if you want to, it helps to write down a few things. So u is whatever is on the numerator of the fraction. So we can say that u is 3x minus 5. v is the stuff on the bottom or on the denominator of the fraction.

So that's 7x plus 4. u prime, the derivative of 3x minus 5 is just 3. v prime in this example is simply 7. So now we just got to plug in everything into that formula. So it's v, that's 7x plus 4, times u prime, which is 3. And then minus u, that's 3x minus 5 times v prime, which is 7, divided by v squared. So that's 7x plus 4 squared.

Now let's do some algebra. So let's begin by distributing the 3 to 7x and 4. 7x times 3 is 21x. 4 times 3 is 12. Next, let's distribute 7 to 3x, and keep in mind it's a negative sign, so that's going to be negative 21x. And then 7 times negative 5, combined with that negative sign, that's going to be positive 35. So immediately we could see that we could cancel the 21x and then we can combine 12 and 35 which will be 47. So the final answer is 47 divided by 7x plus 4 squared for this problem.

So that's how you can use the quotient rule whenever you have a division of two functions. Now let's talk about the chain rule. We've covered some examples of it early in this video.

What is the derivative of sine x to the 5th power? From what we talked about, we would differentiate the outer part of the function. The derivative of sine is cosine. And then we would keep the inside the same.

And then we would differentiate the inside part of the function. The derivative of x to the fifth power is 5x to the fourth power. That process can help you to solve derivatives associated with the chain rule. So when dealing with the chain rule, you're dealing with composite functions. So if you want to find the derivative of a function inside of another function, like f of g of x, It's equal to the derivative of the outside function, so f prime, and then the inside function stays the same, that is g of x, and then times the derivative of the inside function, g prime of x.

So that's how you could find the derivative of a composite function, when you have a function inside of another function. So let's say we want to find the derivative of 4x cubed plus 7x raised to the fifth power. In this case, we need to start with the out of function, that is the power rule. So we're going to move the 5 to the front.

We're going to keep the inside function the same. And then we're going to subtract the exponent by 1. Next, we're going to take the derivative of the inside function. The derivative of 4x cubed is 12x squared. The derivative of 7x is 7. So that's the derivative of this function. So you don't have to follow this five times.

You could just simply use the chain rule to get the answer. now let's move on to our next example what is the derivative of tangent of sine x cubed. Go ahead and work on that example. So first we're going to find the derivative of the outer function, tangent, and that's going to be secant squared. Now, inside of tangent, we have sine x cubed.

So we're going to have the same thing inside of secant squared. So we're starting with the outer function, and we're working our way in. So now let's work our way in towards sine. The derivative of sine is cosine, and the angle of sine is x cubed.

So the angle of cosine will be x cubed. And then we're going to work even further towards the inside part of the functions. The derivative of sine is x cubed. of x cubed is 3x squared.

Now, that's as far as we can go. So this is the answer of this expression. Secant squared sine x cubed times cosine x cubed times 3x squared. Here's another example you can work on.

Find the derivative of sine to the fourth power. cosine of tangent x squared now what I would recommend doing since we have an exponent here is rewriting the expression So basically what this means is that this whole thing is raised to the fourth power. And it affects the way you're going to apply the chain rule.

Because before you can differentiate the sine function, you need to use the power rule on that exponent. So let's move the 4 to the front. So it's going to be 4. And then we're going to rewrite everything we see inside of the brackets highlighted in red.

Now we're going to subtract the exponent by 1. And then we're going to differentiate the sine function. So that's going to be cosine. And then the angle of sine is everything we see here. That's cosine of tangent x squared.

So now we're going to move inward towards cosine. By the way, there's a multiplication for all of these functions, so it's times. The derivative of cosine is negative sine, and the angle of cosine is tangent x squared. So we're going to rewrite that.

Now let's work our way in towards tangent. The derivative of tangent is secant squared, and the angle is x squared. Finally, we can find the derivative of x squared, which is 2x.

And so this whole thing is our final answer. Now, of course, you can combine a few things. You can combine the 4 and the 2x and the negative sign.

And you can put negative 8x in front. But besides that, there's nothing else you can do here. So that's it for this problem.

Now the next topic we need to talk about is implicit differentiation. Sometimes you'll have equations with two different variables, like this one, x and y. And you want to find the derivative of y with respect to x.

So basically you want to determine the value of dy dx. How do you do that? Well, what you do is you're going to differentiate both sides with respect to x.

Now before we do that, I want to talk about a few things. When you differentiate x cubed with respect to x, you're going to get 3x squared. And technically you get times dx over dx, which cancels to 1. But if you want to differentiate y cubed with respect to x, you're going to get 3y squared, but times dy dx. Note the difference here, since these variables don't match up.

Here, these, they do match up, so this would be dx over dx, which is pointless in writing. So implicit, when dealing with implicit differentiation problems, every time you differentiate y with respect to x, you need to add dy dx. Now sometimes you'll be dealing with what is known as related rates. Instead of differentiating with respect to x, you'll be differentiating with respect to t, or time.

So if you differentiate y with respect to x, you'll be differentiating with respect to t, or time. If you differentiate x cubed with respect to time, it's going to be 3x squared, but times dx dt. Likewise, if you differentiate y cubed with respect to time, it's going to be 3y squared times dy dt. So just keep that in mind.

But when dealing with implicit differentiation, I mean, let me say that again. When dealing with implicit differentiation, we are differentiating with respect to x. So the derivative of x to the fourth is simply going to be 4x cubed. But the derivative of y to the fourth is going to be 4y cubed, but times dy over dx.

And this is what we need to solve for. The derivative of 12 is 0. So I'm going to take this term and move it to the other side. So I'm going to have 4y cubed times dy dx, and that's going to be equal to negative 4x cubed.

And then all we need to do is divide both sides by 4y cubed. and then the force will cancel. And so we get dy over dx is equal to negative x cubed over y cubed.

So that's a quick and simple example of how to perform. implicit differentiation. For the sake of practice, let's work on one more example. Let's say we have x cubed plus 4xy plus y squared is equal to 9. Go ahead and calculate the value of dy over dx.

So we're going to differentiate this entire equation with respect to x. The derivative of x cubed is 3x squared. Now, notice that we have x and y.

So we need to use the product rule. I'm going to say the first part, the u part, is 4x, the v part is going to be y. So the derivative of the first part, 4x, is just 4, times the second part, which is just y, plus the first part, 4x, times the derivative of the second part. The derivative of y is going to be 1 times dy over dx, because we're differentiating y with respect to x. Now the derivative of y squared is going to be 2y times dy over dx.

And the derivative of a constant is always 0. So everything that does not have a dy over dx attached to it, I'm going to move to the other side. So I'm going to move these two terms across the equal sign. So we're going to have 4x dy over dx plus...

2y dy over dx and that's going to equal negative 3x squared minus 4y. So because we have a common factor we can now factor out the dy over dx on the left side. And that's going to leave us with 4x plus 2y, which equals negative 3x minus 4y.

So now we can divide both sides by this, 4x plus 2y. So these two will cancel. And thus we have our final answer.

dy over dx is equal to negative 3x squared minus 4y. And we could factor out a 2 on the bottom, but I'm not going to worry about that in this example. So that's going to be divided by 4x plus 2y.

So now you know how to find the derivative in or using implicit differentiation. So far, we talked about how we could find the derivative of a variable raised to a constant. And that is by using the power rule. X to the n, the derivative of that is going to be n times x raised to the n minus 1. And we also talk about having or finding the derivative of a constant raised to a variable. And that was a to the u times u prime times ln a.

So we have a variable raised to a constant and a constant raised to a variable. But what about finding the derivative? Of a variable raised to a variable.

How do we do that? How do we find the derivative of x raised to the x? Well, first, let's make y equal to x raised to the x.

So if we could find the derivative of y, then we know what the derivative of x is. x to the x's. That's step one. Step two is to take the natural log of both sides. So ln y is equal to ln x raised to the x. Now a property of natural logs allows us to move the variable x to the front.

And so what we're going to have is L and Y is equal to X, L and X. And now at this point, we are going to differentiate both sides with respect to X. So here's how you can express that in mathematical words.

Now we know that the derivative of ln x is 1 over x, so the derivative of ln y is going to be 1 over y. Now notice that we're differentiating y with respect to x. So when you see that, you need to add dy over dx. This is basically implicit differentiation. On the right side, we need to use the product rule.

So we're going to differentiate the first part, x, which is 1, and leave the second part by itself. And then we're going to leave the first part by itself, and then differentiate the second part. The derivative of ln x is 1 over x.

So what we have now is 1 over y times dy over dx, and that's equal to ln x, and then x times 1 over x, you can cross out the x variables, so it becomes plus 1. And now what we're going to do is we're going to multiply both sides by y. So we can cross out the y variables here. So we're going to have dy dx is equal to y times, you can say ln x plus 1, or you can say 1 plus ln x. Now recall at the beginning, we said x raised to the x equal to y. At this point, we're going to substitute y with x raised to the x.

So the final answer is going to be x raised to the x. times 1 plus ln x. So that's how you could perform what is known as logarithmic differentiation. That's how you could find the derivative of a variable raised to a variable. So this is the answer.

So you could say that the derivative of x raised to the x in this example is x raised to the x times 1 plus ln x. Well, that's basically it for this video. We've covered a lot of topics, and hopefully you found this to be helpful.

Thanks for watching.

Transcript for:Essential Rules for Finding Derivatives

Transcript for:
Essential Rules for Finding Derivatives