In this video, we're going to talk about how to find the derivative using the definition of the derivative formula. So basically, we need to find the derivative of a function using the limit process. And frime of x represents the derivative of f ofx. And it's equal to the limit as h approaches zero of f of x + h minus f ofx / h. So that's the formula that we need to use. So let's say if f ofx is a linear function 5x - 4, what is frime of x? What's the first derivative of this function? So go ahead and try that. Now what is f ofx plus h? How can we find that? Well, if f ofx is 5x - 4, then f ofx + h is going to be 5 * x + h - 4. So all you need to do is in f ofx, wherever you see an x, replace it with x plus h. So now let's plug in everything into this formula. So it's going to be the limit as h approaches zero and then it's going to be 5 x + h - 4 and then minus f ofx which fx is 5x - 4. And so we're going to divide all of this by h. Now let's simplify. So first we need to distribute five to x + h. And don't forget to rewrite the limit expression until you get your final answer. So it's going to be 5x + 5 hus 4 and then distribute the negative sign. So it's going to be - 5x + 4 / h. Now 5x and 5x they add up to zero. -4 + 4 is equal to zero. So we're left over with the limit as h approaches zero 5h / h. And the h terms cancel. hid h is one. And so the limit as h approaches zero for a constant like five since there's no h in this expression anymore. This is simply equal to 5. So that's the derivative of 5x - 4. It's five. Now let's try another example. So let's say if f ofx is equal to x^2, what is the first derivative of the function? Go ahead and use the definition of the derivative to find frime of x. So first let's use the formula frime of x is equal to the limit as h approaches zero of fx + h - fx / h. So if f ofx is equal to x^2 what is f ofx plus h? That's the first thing you need to decide. So all we have to do is replace x with x + h. So it's going to be x + h^2. So we have the limit as h approaches zero x + h^2us fx which is originally just x^2 / h. Now we need to foil x + h^2. So I'm going to expand it for now. So what we really have is x + h times another x + h. So let's foil x * x. That's going to be x^2. And since I'm running out of space, I'm going to put this on the top and then x * h that's xh and then we have to multiply h * x which is also xh and finally h * h that's going to be h 2 and then - x^2 all / h. Now let's combine like terms and let's cancel. x^2 and x^2 adds up to zero. And xh + xh well that's 2xh. And so we're going to have the limit as h approaches zero for 2xh + h ^2 / h. Now what do you think we need to do at this point? Once you get to this step, what we need to do is factor out an h. If we take out an h, the GCF 2xh / h is 2x and h ^2 / h, well that's h. Now, we could cancel the h variables on the outside. hid h is one. So, this gives us the limit as h approaches zero for 2x + h. So now we're going to apply this limit expression. We're going to replace h with zero. So it's going to be 2x + 0 which is 2x. So if f ofx is equal to x^2, the derivative frime of x is equal to 2x based on the definition formula of the derivative. And that's the answer. Now let's work on another example. So let's say if we have 1 /x, what is the first derivative of 1 /x? By the way, feel free to pause the video and try these examples yourself. So first, let's find f ofx + h. So if we replace x with x + h, we're going to get 1 / x plus h. Now let's start with the formula. frime of x is equal to the limit as h approaches zero fx + h - fx / h. Now f ofx plus h we said it's 1 / x + h and f ofx is 1 /x and this is all divided by h. So here we have a complex fraction. What should we do in a situation like this? If you have a complex fraction, multiply the top and the bottom by the common denominator of these two fractions. So the common denominator is going to be x * x + h. Whatever you do to the top, you must also do to the bottom of the complex fraction. Now we need to multiply. So if we take this fraction and multiply by x * x + h, the x + h terms will cancel, eliminating this fraction. And what we have left over is simply x. So right now we're going to have the limit as h approaches zero and on top we have an x right now. Now let's take the second fraction multiply by that uh term. So if we do that x will cancel and we're going to get x + h with a negative sign in front of it. So this is going to be x + h. And on the bottom, we simply just need to write these things together. So, it's going to be h * x * x + h. Now, let's distribute the negative sign. So, we're going to have the limit as h approaches zero x - x - h x + x adds up to zero. And so this is going to give us the limit as h approaches zeroh / h * x * x + h / h is1. So now we have the limit as h approaches zero -1 /x * x + h. As soon as you cancel that h, you can now use direct substitution. So we can apply this limit expression. Let's replace h with zero. Now x + 0 is x. So we have x * x which is x^2. So the answer is -1 / x^2 and that is the derivative of 1 /x. Now let's say that f ofx is equal to the roo of x. What is the first derivative frime of x? Go ahead and try that. So let's start with the formula. frime of x is equal to the limit as h approaches zero fx + h minus fx over h. Now in this problem, what do you think f ofx plus h represents? Now don't forget, all we need to do is replace x with x plus h. So it's going to equal the square root of x plus h. and f ofx is simply the root of x. Now what should we do if we have radicals in a fraction? To simplify this expression, you need to multiply the top and the bottom by the conjugate of the numerator. So the conjugate is going to have the same expression square x + h and x. The difference is instead of a negative sign, we are going to have a positive sign. So now let's foil what we have on top. So the square root of x + h times itself. The square roots will cancel and you're going to get x + h. And then we have these two terms which is going to be plus x x + h. And then if we multiply those two terms, x * x + h. And finally the of x * the of x is simply x. And don't forget about the negative sign. And on the bottom, don't distribute. Simply just I would recommend just rewriting it. Now the square root of x times the square root of x + h. Um, we have a positive term and a negative one. So they're going to add up and cancel to zero. X and negative X will also add up to zero. So now we're left with the limit as h approaches zero and it's h / h * the of x + h + x. And as you know h / h is 1. So we're left with the limit as h approaches zero and we have this expression. Now what we need to do is direct substitution. We need to replace h with zero at this point. So x + 0 will simply be x. Now the square roo of x plus the of x, there's a coefficient of 1. 1 + 1 is 2. So you should get 1 / 2 x. By the way, this limit expression should no longer be here. When you replace h with zero, this expression disappears. So it shouldn't be in this step anymore. This is the final answer. So that's the first derivative of the square root of x. Now what if the square root is in the bottom of a fraction? Like if we have 8 over the roo of x, what's the first derivative of this function? Well, first let's find f ofx plus h. So it's going to equal that. Now let's use the limit process to find the derivative. So f of x + h is going to be 8 over the of x + h. and I'm forgetting the limit expression. This is why it's always good to rewrite the formula just to avoid making mistakes. So we have the limit as h approaches zero and then f x plus h is 8 over the of x + h and then minus fx and then / h. So now we have a complex fraction with radicals. So basically we have two problems combined into one. We have the 1 /x problem with the square root of x problem. So what I recommend is eliminating the fractions within the larger fractions. So I'm going to multiply by the common denominator just to begin. So it's going to be the roo of x * the of x + h. So these terms will cancel leaving behind 8 * the square of x and then when I multiply this fraction by that term the square root of x will cancel leaving behind these two. So it's going to be -8 x + h. And on the bottom simply rewrite these things together. So it's going to be h * the of x * the of x + h. Now what should we do at this point? What's our next move here? So since we have radicals in the numerator, we need to multiply the top and the bottom by the conjugate of the numerator. And so that's going to be 8 x + 8 x + h / the same thing. So let's foil the stuff on top. So 8x * itself. What is that going to give us? Well, we know 8 * 8 is 64. And the square of x * the x is x. Next, we have 8 * 8, which is 64. And then x x + h. And then we have the two middle terms. This is going to be -64. x x + h and then finally 8 * 8 and then square x + h * itself which is just going to be x + h. And on the bottom just rewrite everything. Now this is a very long problem but if you have a problem like this on a test this is what you have to do. So first we could cancel these two terms and then let's distribute -64 to x plus h. So what we have now is 64x and then - 64x and - 64h divided by everything on the bottom. Now let's cancel those two terms. And so we're left with limit as h approaches zero -64h / everything that is in the denominator which is a lot of stuff. Now in the next step we could cancel h because as soon as we can do that we can use direct substitution. We can get rid of the other h variables in the bottom. But right now we have the limit as h approaches zero -64 over x x + h and so forth. So now I'm going to replace h with zero. So I'm no longer going to write the limit expression. So it's -64 and then we have x. Now this is going to be x + 0 which is also square x and then 8 * the of x and then we're going to replace h with zero there. So that's going to be 8 x as well. Now the square roo of x * the of x is x and then 8 + 8 is 16. And then we could divide -64 by 16. So we're going to get -4 over x x. And so you can leave the answer like that if you want to. x x is also x^ 3. You could write it that way too, but that's the answer. Now the last example I'm going to go over is a polomial function. So let's say we have f(x)= x^2 - 5x + 9. What's the first derivative of this function? So first what's f x + h? Let's decide that. So instead of writing x^2 we're going to write x + h 2. Instead of writing 5x is going to be 5 * x + h. So then frime of x is going to be the limit as h approaches zero and then f ofx + h that's x + h 2 - 5 x + h + 9 and then minus f ofx which is x^2 - 5 x + 9. Don't forget to distribute the negative sign and this is all divided by h. So first we need to expand x + h^2 already. Now we've done that earlier in this video. So x + h * x + h. We found that it was x^2 + 2 xh + h 2. Now we need to distribute the 5. So it's going to be -5x - 5 h + 9 and then distribute this negative sign. So it's going to be x^2 + 5x - 9 all / h. So x^2 and x^2 can be cancelled. -5x and 5x would disappear and 9 and9 we can get rid of. So now we have left over the limit as h approaches zero. So we have h^2 plus 2xh - 5 h / h. Now let's factor out the greatest common factor in the numerator which is h. So h ^2 / h is h. 2xh / h is 2x5h / h is5. So now we could cancel those two. And so we have the limit as h approaches zero h + 2x - 5. Now let's use direct substitution. Let's replace h with zero. So 0 + 2x - 5 is simply 2x - 5. So that's the first derivative of x^2 - 5 x + 9. It's 2x - 5.