Essential Rules for Calculating Derivatives

Hi, so for today we're going to talk about rules for finding the derivative. So we have rules for finding the derivative. So for our number one rule is the derivative of constant. So always keep in mind that the derivative derivative of constant is always equal to zero. So it means if you have a y that is equal to three, if you get the derivative of that dy over dx, the derivative of y with respect to x is always equal to zero. So here is the formula d over dx of any constant c is actually equal to zero. Okay, so if you're asked to find the derivative, for example, let's say again, y is equal to 1000. The derivative of y with respect to x is always equal to 0. Okay, so for our number 2 rule is what we call the power rule. So the power rule has this. formula dy oh okay d over dx of x raised to any number n then it is actually equal to n times x raised to n minus 1 okay that is we're going to get the power then multiply it to our original x but we're going to subtract 1 from its original exponent. So let's say for example we have y is equals to x cubed. If we're going to get the derivative of y with respect to x we get the exponent 3 then multiply it by the function x cubed but we have to apply minus 1 because the formula says n times x raised to n minus 1 so that we have this 3x raised to 3 minus 1 is actually 3x squared and we now get the derivative of this function y is equals to x cubed so another example is for example we have a function that has x raised to the fifth okay so we have dy over dx is equal to 5 times x raised to 5 minus 1 or in other words simplifying this 5x times x raised to the fourth okay so very simple formula okay the power rule for our third is the derivative constant times a function times a function so it says here the formula dy over dx is actually equal to d over dx of the function constant times a certain f of X So we have here, since the derivative of constant is, we have to, since this is constant c, we can actually isolate that and get the derivative of that f of x. Okay? So for example, we have here an example. So for example, find the derivative of y is equals to 3 times x raised to 6. So this is our constant 3. So all we have to do to find the derivative of that is to apply that formula. So what are we going to do is 3 times dy over dx of x raised to 6. So in order to get that, we have 3 apply power formula 6 times x raised to 6 minus 1. so we have applied the power formula So to simplify, we have 3 times 6, that is 18. x raised to 6 minus 1 is 18 times x raised to 5. So if we have a constant, simply we're going to get the derivative of this function and isolate this constant. Okay, so another example is this. For example, we have here y is equal to 2x squared. Okay. In order to get the derivative of this, we have to leave the constant itself and get the derivative of the function x squared. So we have here 2 times the derivative of x squared by power formula. We get the exponent. We multiply it. Then the original function, x raised to 2 minus 1, that will be equal to 2 times 2, that is 4. times x 2 minus 1 is x raised to 1 or in other words x so that we have the answer for x okay so how about the fourth rule the derivative of sum and difference So in this case, we shall be applying the formula dy over dx is equal to d over dx times u plus or minus v. And that is equal to the du over dx plus or minus dv over dx, wherein u and v are two terms. So for example, we have here this. y is equal to 3x squared plus 2x minus 1 and find the dy over dx of that function. So in order to get the dy over dx or the derivative of this function y, we have to evaluate the individual derivatives. this function so we have the first term that is the derivative of a constant 3 times a function x squared and also for the second term that is the derivative of a function times a constant a function x and a constant 2 and in this third term that is simply the derivative of a constant so if you wish to get the derivative of that so we have 3 times we're going to get d over dx of x squared plus 2 the derivative of the function x minus the derivative of the constant 1 so we're going to get that 3 and by power rule this is times 2 times x raised to 2 minus 1 and plus 2 that is times 1 because the exponent of this x is 1 so and times x raised to 1 oops 1 minus 1 minus the derivative of constant is again 0. So we have here 3 times 2 that is 6 x raised to 2 minus 1, that is simply x plus 2 times 1. This x raised to 1 minus 1 is x raised to 0, that is 1. So simply 2 times 1 times 1, that is 2. So we have now the derivative of this function. So another one, how about if we have this? So we have y is equals to... 2 thirds of x cubed minus 4x plus 86. So same formula, same approach. We have to get dy over dx. This is a constant and a function. This is a constant again and a function. So we have to get 2 thirds the derivative of x cubed minus 4 times the derivative of this function x plus d over dx of this a to the 6th constant. So 2 thirds, bring down 3 because that is the exponent, multiply it. Then x cubed minus 1, so power rule again. So minus 4, the derivative of this x is simply 1. Okay, we're going to evaluate it again. Okay, so plus the derivative of this constant, again, the derivative of constant is, yes, you are right, that is 0. So as you can see here, 2 thirds times 3, simply the 3 will cancel. And then we will be having 2 times x raised to 3 minus 1, that is x squared minus 4. And that shall be our answer. Okay? So the fifth rule is the derivative of a product. So it says here that our formula is dy over dx. You wish to find the derivative of product d over dx of u times v where u and v are both a function. So we have to apply this formula u is it times the dv over dx plus the v du over dx. Okay, so we have this u and v. So first we have to get or to copy first the function u. Then let's try to derivative x. That will be our first term plus. copy v then we're going to get the derivative of u with respect to x so one example is this okay is x plus 1 times x plus 2 okay we can use FOIL in order to get this to perform this product but using this derivative of a product we can simply use the formula let's say dy over dx so what are we going to do is this will be our u and this will be our v so it says in the formula we need to copy the first one x plus 1 then we're going to get the derivative get the derivative of the second one x plus 2 plus copy the v this is our v x plus 2 Then we're going to get the derivative of the u, which is actually x plus 1. So in other words, copy the first, this first, and then get the derivative of this, multiply it. That's your first term. Plus, copy the second term, and then what we're going to do is multiply it by the derivative of the first. So in this one, we have x plus 1 and the derivative of this is actually what? The derivative of x is 1 and the derivative of constant is actually 0. So plus x plus 2, the derivative of x is actually 1 and the derivative of 1 is actually, oops, I'm sorry, I need to erase this. Then the derivative of 1 or this 1 is actually 0. So that is a constant. So again, we get the derivative of x which is equal to 1. Get the derivative of 1 which is equal to 0. So simplifying, we have x plus 1 simply. This will be multiplied to 1 because 0 has nothing to do with that. So x plus 1 plus... Simplifying. Simply, x plus 2 is multiplied to 1, so we will be having x plus 2 here. So to simplify that, we have x plus 1 plus x plus 2. So we have the answer, combined like terms, x plus x, that is 2x, plus 1 plus 2, that is equal to 3. So another way to get this is what I'm telling you a while ago, is to get the... product of this using the pool method Okay, so if we're going to get the derivative of that, we simplify that first. So into a single function, we multiply this 2. So what are we going to have is the FOIL. Apply the FOIL method. First, x times x, that is equal to x squared. Outer, x times 2, that is plus 2x. And our inner is 1 times x, that is equal to x. and last one times two that is plus two then we're going to get the derivative of this so if you get the derivative of this dy over dx so we apply the power formula here so 2x plus 2 okay plus 1 why Power formula for here, x squared, 2x, then this is the derivative of a constant times a function x. So derivative of constant, we're going to get it out of the derivative, then derivative, the function x will result to 2. And the derivative of this x is 1, plus the derivative of this 2 is actually a constant that is equal to 0. So to simplify, we have now the answer. Simply, 2x plus 2 plus 1 is actually equal to 2x plus 3, which is the same as this. Okay, but sometimes there will be a given function in terms of y that is really hard to multiply or really, I would say, time-consuming. you would really want to use this product rule okay rather than getting first the product then apply this formula of power rule and the derivative of a constant as a function so let's try another formula we have y is equals to x cube plus 2x times 2x minus 1 So if we're going to get a derivative of this, y prime, that is equivalent also to dy over dx. Simply, this is our u, this is our v. So simply copy the u first, or the first term, x cubed plus 2x times the d over dx of 2x minus 1, that is the derivative of the second, plus, copy the second. which is 2x minus 1 then we're going to get the derivative of u or the first okay so we have x cube plus 2x so here we go so we have x cube plus 2x times the derivative of 2x is actually what it is equal to 2 okay and the derivative of this Negative 1 or minus 1, since that is a constant, that is minus 0. Plus 2x minus 1 times the derivative of this. So we can apply power formula in this one. So we'll be having 3x squared plus 2. So the derivative of this is actually 2 because this is a function x. multiplied by constant the derivative of this is 1 2 times 1 that is equivalent to 2 and this is a product power rule I mean okay so simplifying our answer we have x cubed plus 2x times 2 plus 2x minus 1 times 3x squared plus 2 okay so in order for us to find that so we're going to simplify 2x cubed plus 2 times 2x that is 4x plus we're going to find the foil of this okay so first 2x times 3x squared that is 6x cubed outer 2x times 2 that is equal to 4x inner negative 1 times 3x squared that is negative 3x squared then finally last term you negative 1 times 2 that is equivalent to minus so combining like terms So we have here 2x cubed plus 6x cubed. That is simply 8x cubed. So we have a second degree here. So we can write it here. So another one, 4x plus 4x. Again, we're combining like terms here. 4x plus 4x, that is 8x. And then finally, the constant term is minus 2. So our answer should be... 8x cubed minus 3x squared plus 8x minus 2. This shall be our answer. So another one for product rule, we have here the given x cubed minus 3x squared multiplied by x squared plus 4x plus 2. So same concept, we will apply the product rule here. So again, copy first the first, that shall be our u and this shall be our v. First term, minus 3x squared times we're going to get a derivative with respect to x of the v, which is x squared plus 4x plus 2 plus copy the v. or the second term plus 4x plus 2 then we're going to get the derivative of x cubed minus 3x squared that is the derivative of the first so this case we should copy only here then we're going to get the derivative of this this is a combination of a product or a power rule and this is a derivative of constant times a function and this is the derivative of a constant. So the derivative of x squared is simply by power rule 2x and the derivative of 4x is simply 4, 4. And the derivative of this constant 2 is actually equal to 0. So plus x squared plus 4x plus 2 times the derivative of this since this is a power rule so 3x squared yes you got it right so minus this is a combination of power rule and a constant so we have 3 times 2 because of power rule 6x raised to 2 minus 1 so we will be having 6x okay so simplify x cubed minus 3x squared we have 2x plus 4 and then for this we have x squared plus 4x plus 2 times 3x squared 3x squared minus 6x so in order to simplify we need to apply FOIL first x cubed times 2x that is equivalent to 2x raised to 4 outer x cubed times 4 that is plus 4x cubed inner negative 3x squared times 2x that is equivalent to negative 6x cubed and last negative 3x squared times 4 that is minus 12x squared so it gets the computation gets long okay so plus that is actually void or not void but simply the normal computation for product of these two functions so we have 3x squared times x squared that is 3x raised to 4 and 3x squared times 4x that is plus 12x cubed so 3x squared times 2 that is plus 6x squared okay then negative 6x times x squared that is equivalent to that is equivalent to negative x raised to 3 so negative 6x times 4x that is negative 24x squared and lastly negative 6x times 2 that is negative 12x so to simplify we're going to combine like terms okay so we have first the highest exponent we can get it here so this is the highest exponent so 3x raised to 4. so next is we have raised to 4 that is uh oh my mistake so that should be no we are wrong this should be we can simplify to x raised to 4 then plus 3x raised to 4 that is equivalent to 5x raised to 4 then we have here cube plus 4x cube minus 6x cube that is negative 2x cube plus 12x cube that is 10x cube minus 6x cube that is plus 4x cube okay And for our x squared, negative 12x squared plus 6x squared, that is equivalent to negative 6x squared, minus 24x squared, negative 6x squared minus 24x squared, that yields to negative 30x squared. And finally, for our x, so we should copy because we don't have anything to simplify with x. This should be our final answer. Okay? So let's go now to our rule. So our last rule for this video is actually the derivative of... a quotient. So this will be our formula for that dy over dx is equal to the derivative with respect to f of a function that is in quotient form is equals to v times du over dx minus u times dv over dx. all over by v squared okay so this will be our formula wherein we have a the numerator u and the denominator v so what are we going to do first is that we're going to get first the the the denominator then multiply it to the derivative of the numerator minus the the numerator multiplied by the derivative the derivative of the denominator dv over dx all over we're going to square the denominator or in other words that is we have a new money for that low derivative of high minus high derivative of low divided by low which is the denominator square So let's try some problems. So for example, we have y is equal to x over x plus 1. So it is obvious that our u here is the x and our b here is the denominator which is x plus 1. So in order to get a derivative of our function, so we're going to apply this formula or simply this. So what is that? Low, the denominator, times the derivative of the high, which is actually the numerator, minus copy the high, which is the numerator, times derivative of low, which is x plus 1, divided by low squared, which is x plus 1. squared so let's recap so we have here low which is the denominator low times the derivative of high minus high which is the numerator times the derivative of low so divided by the low squared or the b squared the denominator raised to okay so Low, derivative of high. High, derivative of low. divided by x plus 1 squared. So we shall have the answer x plus 1 times the derivative of x which is simply 1 minus x times the derivative of x is 1 plus the derivative of 1 is 0 divided by x plus 1 squared. So we have the answer x plus 1 times 1 which is simply x plus 1 minus x times 1 which is x divided by x plus 1 squared to simplify we x will cancel x minus x so we are left with 1 over x plus 1 squared so this will be our answer okay so another problem derivative of a quotient we have 2x x squared plus 1 so again we will be applying the low d high i suggest uh you memorize this low d high minus high d low divided by low squared okay so first we need to get the low which is the denominator x squared plus 1 you Multiply it by the derivative of the height, d over dx of 2x minus high, which is 2x, times the derivative of low, d over dx, x squared plus 1, divided by low squared, x squared plus 1, raised to. So, we have here x squared plus 1 times the derivative of. 2x is simply 2 minus 2x the derivative of this we have we need to evaluate first this x squared is 2x plus the derivative of the constant is 0 so divided by still x squared plus 1 squared so we have now x squared plus 1 times 2 that is 2x squared plus 2 minus 2x times 2x that is 4x squared divided by x squared plus 1 squared so we have now the answer 2x squared minus 4x squared that is negative 2x squared plus 2 divided by x squared plus 1 squared Okay, so we can simplify this further by factoring out 2. So we have 1 minus x squared divided by x squared plus 1 squared. So this shall be our answer. If you find this video helpful, don't forget to click like and subscribe to my channel. Thank you so much for listening and God bless you.

So here is the formula d over dx of any constant c is actually equal to zero. Okay, so if you're asked to find the derivative, for example, let's say again, y is equal to 1000. The derivative of y with respect to x is always equal to 0. Okay, so for our number 2 rule is what we call the power rule. So the power rule has this. formula dy oh okay d over dx of x raised to any number n then it is actually equal to n times x raised to n minus 1 okay that is we're going to get the power then multiply it to our original x but we're going to subtract 1 from its original exponent.

So let's say for example we have y is equals to x cubed. If we're going to get the derivative of y with respect to x we get the exponent 3 then multiply it by the function x cubed but we have to apply minus 1 because the formula says n times x raised to n minus 1 so that we have this 3x raised to 3 minus 1 is actually 3x squared and we now get the derivative of this function y is equals to x cubed so another example is for example we have a function that has x raised to the fifth okay so we have dy over dx is equal to 5 times x raised to 5 minus 1 or in other words simplifying this 5x times x raised to the fourth okay so very simple formula okay the power rule for our third is the derivative constant times a function times a function so it says here the formula dy over dx is actually equal to d over dx of the function constant times a certain f of X So we have here, since the derivative of constant is, we have to, since this is constant c, we can actually isolate that and get the derivative of that f of x. Okay? So for example, we have here an example.

So for example, find the derivative of y is equals to 3 times x raised to 6. So this is our constant 3. So all we have to do to find the derivative of that is to apply that formula. So what are we going to do is 3 times dy over dx of x raised to 6. So in order to get that, we have 3 apply power formula 6 times x raised to 6 minus 1. so we have applied the power formula So to simplify, we have 3 times 6, that is 18. x raised to 6 minus 1 is 18 times x raised to 5. So if we have a constant, simply we're going to get the derivative of this function and isolate this constant. Okay, so another example is this.

For example, we have here y is equal to 2x squared. Okay. In order to get the derivative of this, we have to leave the constant itself and get the derivative of the function x squared.

So we have here 2 times the derivative of x squared by power formula. We get the exponent. We multiply it. Then the original function, x raised to 2 minus 1, that will be equal to 2 times 2, that is 4. times x 2 minus 1 is x raised to 1 or in other words x so that we have the answer for x okay so how about the fourth rule the derivative of sum and difference So in this case, we shall be applying the formula dy over dx is equal to d over dx times u plus or minus v. And that is equal to the du over dx plus or minus dv over dx, wherein u and v are two terms. So for example, we have here this.

y is equal to 3x squared plus 2x minus 1 and find the dy over dx of that function. So in order to get the dy over dx or the derivative of this function y, we have to evaluate the individual derivatives. this function so we have the first term that is the derivative of a constant 3 times a function x squared and also for the second term that is the derivative of a function times a constant a function x and a constant 2 and in this third term that is simply the derivative of a constant so if you wish to get the derivative of that so we have 3 times we're going to get d over dx of x squared plus 2 the derivative of the function x minus the derivative of the constant 1 so we're going to get that 3 and by power rule this is times 2 times x raised to 2 minus 1 and plus 2 that is times 1 because the exponent of this x is 1 so and times x raised to 1 oops 1 minus 1 minus the derivative of constant is again 0. So we have here 3 times 2 that is 6 x raised to 2 minus 1, that is simply x plus 2 times 1. This x raised to 1 minus 1 is x raised to 0, that is 1. So simply 2 times 1 times 1, that is 2. So we have now the derivative of this function.

So another one, how about if we have this? So we have y is equals to... 2 thirds of x cubed minus 4x plus 86. So same formula, same approach. We have to get dy over dx. This is a constant and a function.

This is a constant again and a function. So we have to get 2 thirds the derivative of x cubed minus 4 times the derivative of this function x plus d over dx of this a to the 6th constant. So 2 thirds, bring down 3 because that is the exponent, multiply it.

Then x cubed minus 1, so power rule again. So minus 4, the derivative of this x is simply 1. Okay, we're going to evaluate it again. Okay, so plus the derivative of this constant, again, the derivative of constant is, yes, you are right, that is 0. So as you can see here, 2 thirds times 3, simply the 3 will cancel. And then we will be having 2 times x raised to 3 minus 1, that is x squared minus 4. And that shall be our answer.

Okay? So the fifth rule is the derivative of a product. So it says here that our formula is dy over dx. You wish to find the derivative of product d over dx of u times v where u and v are both a function. So we have to apply this formula u is it times the dv over dx plus the v du over dx.

Okay, so we have this u and v. So first we have to get or to copy first the function u. Then let's try to derivative x. That will be our first term plus. copy v then we're going to get the derivative of u with respect to x so one example is this okay is x plus 1 times x plus 2 okay we can use FOIL in order to get this to perform this product but using this derivative of a product we can simply use the formula let's say dy over dx so what are we going to do is this will be our u and this will be our v so it says in the formula we need to copy the first one x plus 1 then we're going to get the derivative get the derivative of the second one x plus 2 plus copy the v this is our v x plus 2 Then we're going to get the derivative of the u, which is actually x plus 1. So in other words, copy the first, this first, and then get the derivative of this, multiply it.

That's your first term. Plus, copy the second term, and then what we're going to do is multiply it by the derivative of the first. So in this one, we have x plus 1 and the derivative of this is actually what? The derivative of x is 1 and the derivative of constant is actually 0. So plus x plus 2, the derivative of x is actually 1 and the derivative of 1 is actually, oops, I'm sorry, I need to erase this. Then the derivative of 1 or this 1 is actually 0. So that is a constant.

So again, we get the derivative of x which is equal to 1. Get the derivative of 1 which is equal to 0. So simplifying, we have x plus 1 simply. This will be multiplied to 1 because 0 has nothing to do with that. So x plus 1 plus...

Simplifying. Simply, x plus 2 is multiplied to 1, so we will be having x plus 2 here. So to simplify that, we have x plus 1 plus x plus 2. So we have the answer, combined like terms, x plus x, that is 2x, plus 1 plus 2, that is equal to 3. So another way to get this is what I'm telling you a while ago, is to get the... product of this using the pool method Okay, so if we're going to get the derivative of that, we simplify that first.

So into a single function, we multiply this 2. So what are we going to have is the FOIL. Apply the FOIL method. First, x times x, that is equal to x squared.

Outer, x times 2, that is plus 2x. And our inner is 1 times x, that is equal to x. and last one times two that is plus two then we're going to get the derivative of this so if you get the derivative of this dy over dx so we apply the power formula here so 2x plus 2 okay plus 1 why Power formula for here, x squared, 2x, then this is the derivative of a constant times a function x. So derivative of constant, we're going to get it out of the derivative, then derivative, the function x will result to 2. And the derivative of this x is 1, plus the derivative of this 2 is actually a constant that is equal to 0. So to simplify, we have now the answer.

Simply, 2x plus 2 plus 1 is actually equal to 2x plus 3, which is the same as this. Okay, but sometimes there will be a given function in terms of y that is really hard to multiply or really, I would say, time-consuming. you would really want to use this product rule okay rather than getting first the product then apply this formula of power rule and the derivative of a constant as a function so let's try another formula we have y is equals to x cube plus 2x times 2x minus 1 So if we're going to get a derivative of this, y prime, that is equivalent also to dy over dx. Simply, this is our u, this is our v. So simply copy the u first, or the first term, x cubed plus 2x times the d over dx of 2x minus 1, that is the derivative of the second, plus, copy the second. which is 2x minus 1 then we're going to get the derivative of u or the first okay so we have x cube plus 2x so here we go so we have x cube plus 2x times the derivative of 2x is actually what it is equal to 2 okay and the derivative of this Negative 1 or minus 1, since that is a constant, that is minus 0. Plus 2x minus 1 times the derivative of this.

So we can apply power formula in this one. So we'll be having 3x squared plus 2. So the derivative of this is actually 2 because this is a function x. multiplied by constant the derivative of this is 1 2 times 1 that is equivalent to 2 and this is a product power rule I mean okay so simplifying our answer we have x cubed plus 2x times 2 plus 2x minus 1 times 3x squared plus 2 okay so in order for us to find that so we're going to simplify 2x cubed plus 2 times 2x that is 4x plus we're going to find the foil of this okay so first 2x times 3x squared that is 6x cubed outer 2x times 2 that is equal to 4x inner negative 1 times 3x squared that is negative 3x squared then finally last term you negative 1 times 2 that is equivalent to minus so combining like terms So we have here 2x cubed plus 6x cubed. That is simply 8x cubed. So we have a second degree here.

So we can write it here. So another one, 4x plus 4x. Again, we're combining like terms here.

4x plus 4x, that is 8x. And then finally, the constant term is minus 2. So our answer should be... 8x cubed minus 3x squared plus 8x minus 2. This shall be our answer.

So another one for product rule, we have here the given x cubed minus 3x squared multiplied by x squared plus 4x plus 2. So same concept, we will apply the product rule here. So again, copy first the first, that shall be our u and this shall be our v. First term, minus 3x squared times we're going to get a derivative with respect to x of the v, which is x squared plus 4x plus 2 plus copy the v. or the second term plus 4x plus 2 then we're going to get the derivative of x cubed minus 3x squared that is the derivative of the first so this case we should copy only here then we're going to get the derivative of this this is a combination of a product or a power rule and this is a derivative of constant times a function and this is the derivative of a constant. So the derivative of x squared is simply by power rule 2x and the derivative of 4x is simply 4, 4. And the derivative of this constant 2 is actually equal to 0. So plus x squared plus 4x plus 2 times the derivative of this since this is a power rule so 3x squared yes you got it right so minus this is a combination of power rule and a constant so we have 3 times 2 because of power rule 6x raised to 2 minus 1 so we will be having 6x okay so simplify x cubed minus 3x squared we have 2x plus 4 and then for this we have x squared plus 4x plus 2 times 3x squared 3x squared minus 6x so in order to simplify we need to apply FOIL first x cubed times 2x that is equivalent to 2x raised to 4 outer x cubed times 4 that is plus 4x cubed inner negative 3x squared times 2x that is equivalent to negative 6x cubed and last negative 3x squared times 4 that is minus 12x squared so it gets the computation gets long okay so plus that is actually void or not void but simply the normal computation for product of these two functions so we have 3x squared times x squared that is 3x raised to 4 and 3x squared times 4x that is plus 12x cubed so 3x squared times 2 that is plus 6x squared okay then negative 6x times x squared that is equivalent to that is equivalent to negative x raised to 3 so negative 6x times 4x that is negative 24x squared and lastly negative 6x times 2 that is negative 12x so to simplify we're going to combine like terms okay so we have first the highest exponent we can get it here so this is the highest exponent so 3x raised to 4. so next is we have raised to 4 that is uh oh my mistake so that should be no we are wrong this should be we can simplify to x raised to 4 then plus 3x raised to 4 that is equivalent to 5x raised to 4 then we have here cube plus 4x cube minus 6x cube that is negative 2x cube plus 12x cube that is 10x cube minus 6x cube that is plus 4x cube okay And for our x squared, negative 12x squared plus 6x squared, that is equivalent to negative 6x squared, minus 24x squared, negative 6x squared minus 24x squared, that yields to negative 30x squared.

And finally, for our x, so we should copy because we don't have anything to simplify with x. This should be our final answer. Okay?

So let's go now to our rule. So our last rule for this video is actually the derivative of... a quotient. So this will be our formula for that dy over dx is equal to the derivative with respect to f of a function that is in quotient form is equals to v times du over dx minus u times dv over dx. all over by v squared okay so this will be our formula wherein we have a the numerator u and the denominator v so what are we going to do first is that we're going to get first the the the denominator then multiply it to the derivative of the numerator minus the the numerator multiplied by the derivative the derivative of the denominator dv over dx all over we're going to square the denominator or in other words that is we have a new money for that low derivative of high minus high derivative of low divided by low which is the denominator square So let's try some problems.

So for example, we have y is equal to x over x plus 1. So it is obvious that our u here is the x and our b here is the denominator which is x plus 1. So in order to get a derivative of our function, so we're going to apply this formula or simply this. So what is that? Low, the denominator, times the derivative of the high, which is actually the numerator, minus copy the high, which is the numerator, times derivative of low, which is x plus 1, divided by low squared, which is x plus 1. squared so let's recap so we have here low which is the denominator low times the derivative of high minus high which is the numerator times the derivative of low so divided by the low squared or the b squared the denominator raised to okay so Low, derivative of high. High, derivative of low. divided by x plus 1 squared.

So we shall have the answer x plus 1 times the derivative of x which is simply 1 minus x times the derivative of x is 1 plus the derivative of 1 is 0 divided by x plus 1 squared. So we have the answer x plus 1 times 1 which is simply x plus 1 minus x times 1 which is x divided by x plus 1 squared to simplify we x will cancel x minus x so we are left with 1 over x plus 1 squared so this will be our answer okay so another problem derivative of a quotient we have 2x x squared plus 1 so again we will be applying the low d high i suggest uh you memorize this low d high minus high d low divided by low squared okay so first we need to get the low which is the denominator x squared plus 1 you Multiply it by the derivative of the height, d over dx of 2x minus high, which is 2x, times the derivative of low, d over dx, x squared plus 1, divided by low squared, x squared plus 1, raised to. So, we have here x squared plus 1 times the derivative of.

2x is simply 2 minus 2x the derivative of this we have we need to evaluate first this x squared is 2x plus the derivative of the constant is 0 so divided by still x squared plus 1 squared so we have now x squared plus 1 times 2 that is 2x squared plus 2 minus 2x times 2x that is 4x squared divided by x squared plus 1 squared so we have now the answer 2x squared minus 4x squared that is negative 2x squared plus 2 divided by x squared plus 1 squared Okay, so we can simplify this further by factoring out 2. So we have 1 minus x squared divided by x squared plus 1 squared. So this shall be our answer. If you find this video helpful, don't forget to click like and subscribe to my channel.

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Transcript for:Essential Rules for Calculating Derivatives

Transcript for:
Essential Rules for Calculating Derivatives