 In this video, we'll provide the solution to question number 13 for practice exam number 2 for math 12-10, in which case we're asked to find the derivative of the function f of x equals 1 over the square root of x using the definition of the derivative. It's very important here you recognize the instructions say the definition of the derivative. Later on in a calculus course, it's typical to use to learn techniques such as the power rule or the quotient rule, which might make this a little bit easier to compute absolutely, but to get full credit, to get any credit, honestly, we have to follow the instructions given here. We have to provide the calculation by the definition. So what that means is to compute f prime of x right here, we take the limit of the difference quotient, take the limit as h approaches 0 of f of x plus h minus f of x over h. If we're not taking the limit of a difference quotient, we won't get any credit on this one because we didn't follow the instructions. Now, if we apply f of x plus h to this specific question, taking the limit as h approaches 0 here, we're going to get 1 over the square root of x plus h minus then we get f of x, which is 1 over the square root of x, and then this will all sit above an h right here. So this function has a couple of issues going on with it. It has both rational functions and square roots. We're going to have to treat them separately. So let's first deal with the fact we have fractions inside of fractions. We have this nested fraction. So identifying the denominators of the baby's fractions, we're going to times the mother fraction by their least common multiple. So we're going to multiply the square root of x times the square root of h, but we have to do it to the top and bottom of the big fraction so that it remains proportional. Don't bother distributing that is multiplying the square root of x with the square root of x plus h. Leave things factored. It's perfectly fine to leave things factored. In the numerator, you would distribute these things through, in which case we then get the limit as h approaches 0. You're going to get the square root of x times the square root of x plus h all over the square root of x plus h. You're going to get minus the square root of x times the square root of x plus h over the square root of x. And then the denominator we now get h times the square root of x times the square root of x plus h. Do not multiply out the denominator. It will not be to your benefit to do so. It's not going to help you, so don't do it. Now, the reason we multiply by the blue fraction above is that you'll now notice there's some simplification of fractions. The square root of x plus h cancels here, and then on the second one the square root of x will cancel out right here. So if we write this in simplified form to see where we are, we take the limit as h goes to zero, we're going to get the square root of x minus the square root of x plus h all over h times the square root of x times the square root of x plus h. Leave the denominator factored, it's better to do so. Also don't be tempted by, oh, there's a square root of x and a square root of x right here. We can't cancel out the square root of x because there's not a common factor across the top and bottom. While the denominator is divisible at the square root of x, the numerator is not because not everything in the numerator is divisible at the square root of x. Same thing can be said about the square root of x plus h right there. So in order to proceed from here, since we have a square root of x and a square root of x plus h to the numerator, my advice is to rationalize the numerator. So we're going to multiply again by a strategic number one. This time we're going to multiply by the square root of x plus the square root of x plus h. Make sure we do it to the denominator as well so that the fraction stays proportional to its original form. All right, when we do that we're going to foil out the numerator in which case you're going to get the square root of x and the square root of x is the first one. Then you're going to get plus the square root of x times the square root of x plus h. In the next one you're going to have minus the square root of x plus h times the square root of x. And then lastly you're going to get a minus the square root of x plus h squared. And this all sits above. Now we have h square root of x square root of x plus h. And then you have the square root of x plus the square root of x X plus H, despite any temptations to do so, do not multiply out the dominer, simplify the numerator. That's what our goal is right now. So some things to note, we have the square root of X times square of X plus H. We have the square root of X plus H times square of X. The order of multiplication doesn't matter. One is positive, one is negative. So these guys are gonna cancel each other out. So they're gone. That's why we multiplied by the conjugate in the first place. If you take the square root of X square, that'll just become an X. If you take the square root of X plus H square, that'll just become an X plus H. Do pay attention to the signs though. You're going to get X minus X plus H all over this massive denominator. There's a lot going on here. Taking the limit as H goes to zero right here. So see what happens is this negative sign would distribute, right? And then you have an X minus X, so they cancel out. In which case, then we're at the moment, we're gonna get the limit as H goes to zero. I'm gonna write the denominator first because it's so huge. I wanna make sure the fraction bar's big enough for it. H times the square root of X plus H times the square root of X plus the square root of X plus H. In the denominator, we now just have a negative H. Took a lot of effort there. Now we can do it. Now we have a multiple of H on top and a multiple of H on the bottom. We can cancel those things out. Our limit then became the limit as H approaches zero. Again, I'm gonna write out the denominator, not multiply it out. We get the square root of X, we get the square root of X plus H. We're going to get the square root of X plus the square root of X plus H. And this is all under negative one. Now you'll notice that although there are H's in the denominator, if we were to plug in H equals zero, there's not gonna be any division by zero. And therefore, we're now ready to evaluate the limit at zero there. So in doing so, we're gonna get in the denominator square root of X, square root of X plus zero. We're gonna get the square root of X plus the square root of X plus zero, negative one there. So getting rid of the zeros, we're gonna get the square root of X times the square root of X times the square root of X plus the square root of X, negative one right there. For which case if we get a, if we get square root of X times square root of X, that's equal to a X right there. So negative one over X. If you take the square root of X plus the square root of X, you're gonna get two times the square root of X. And this turned out to be our derivative. We're gonna end up with negative one over two times X square root of X. And that is the derivative of one over the square root of X calculated by using the definition of the derivative. By all means, if you know things like the power rule or the quotient rule or something, you might have also written the answer as negative one over two times X to the three halves. That's appropriate or you could have done something like negative one half times X to the negative three halves. Those are all equivalent in form. And so if you were to check your answer using the power rule, assuming you know the power rule, you'd verify this is the correct derivative. But for any credit on this question, the derivative must be calculated using the definition of the derivative, which means that we need to start off at this process right here. We need to know this. In particular, we're not just simplifying a difference quotient, it's a limit of a difference quotient. If there's no reference that you're taking the limit of a different quotient, if you never use the words limit anywhere, then you won't get full credit on this. So use proper limit notation as you're computing this one.