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Let $f(x)=1 / \sqrt{x} .$ Compute $f^{\prime}(5)$ by showing that\begin{equation}\frac{f(5+h)-f(5)}{h}=-\frac{1}{\sqrt{5} \sqrt{5+h}(\sqrt{5+h}+\sqrt{5})}\end{equation}

$-\frac{1}{10 \sqrt{5}}$

Calculus 1 / AB

Chapter 3

DIFFERENTIATION

Section 1

Definition of the Derivative

Derivatives

Differentiation

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Okay, so this question gives us f of X equals one over squared of X, and it's asking us to find f crime of five. So to do this, let's find the derivative. So limit is H approaches zero of f of X plus each minus f of X all over stage. So in this case, F of X is one over square root of X. So let's plug in X plus h into that minus one over the function. And just for simplicity's sake, let's rewrite this over H in the front here as one over age. So now we have limits. As H approaches zero, it is combining fractions here. Square root of ax minus square root of X plus h all divided by square root of ax times square root of X plus h oh, times one over h. So now from here, we'll do a little trick with square roots that you might remember from Algebra two and we'll multiply by the conjugal. All right, so here's where the conjugated comes in and we could only do this if we multiply by it on top and bottom. So we maintain our factor one and now we see that this is just the difference of squares formula. So so the square roots on the top vanish. So we're just left with X minus X plus h over square root of ax square root of X plus h times that conjugated factor and we still have the one over h out here. So now you can see that these axes cancel, which leaves a negative H on top, which cancels with this age in front. So now we're just left with limit as h approach zero of one negative one over square root of ax square root of ax plus H times X in the square root plus square root of X plus h. And now we can finally evaluate this limit by plugging in zero. So remember, if we do that, this h goes away in this H goes away because they go to zero, so that turns out to be negative. One over square root of ax times square root of ax times square root of X plus square root of X, which is just negative. One over X times, two square root of acts or negative one over two acts to the three halves so we have f prime of axe equals negative one over two X to the three halves. So f prime of five is negative one over two times 5 to 3 halves, which turns out to be negative one over 10 times the square root of five after simplification. So this was a much more complicated function to find the derivative of than what we're used to, mainly because of this sneaky trick with the contra git that you really have to be careful of. But after that, we get the h is to cancel, and we get a nice closed form of our derivative.

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