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Find the derivative. Simplify where possible.

$ y = \cosh^{-1} \sqrt x $

$\frac{1}{2 \sqrt{x(x-1)}}$

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Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

All right, let's go ahead and solve this problem. The question is find the ark Coach of Squared of Exes derivative with respect to X. Okay, so there's a couple of ways you can approach this, of course. And it also depends on what you remember. So if you remember the derivative off our coach, which is equal to one over the square, root off X squared minus one when X is a number greater than or equal to one. Oh, strictly greater than one, because we don't want the denominator to be zero. Then we can use this fact. So I'm going to approach this the first way. Okay, so if I use that fact, then we're simply going to use changeable. It is going to look like one over square root off the input squared. So the squared off X quantity squared minus one times the derivative off the square root of X, which we know that it is equal to one over to root X. Okay, Now, one thing that you need to keep in mind here is that square root of X quantity squared is equal to the absolute value of X. Okay, so if we want to make sure that things work out nicely. For example, here we know that the denominator may contain a zero, or we know that inside the square root we want to avoid imaginary number negative numbers so that we don't get imaginary numbers. So if we use those kind of assumptions, we can try to say things such as Okay, what if X is greater than one? Well, under this condition, we know that absolute value of X will end up being X. So if we want to make the calculations look simpler, we want to add these assumptions to make this look like the absolute value off to root X times the square root off X minus one, assuming that X is greater than one. So that's one way to solve this thing. The other way to approach this problem is if you remember that Coach inverse of X is equal to the natural log off X plus square root of X squared minus one where X is greater than or equal toe one. Then we're going to take the derivative off a natural log basically, and then we're going to approach it with the channel multiple times so I'm not gonna go into detail. But simply put, we're gonna have it in the form one over X plus the square root off X squared minus one instead of ex you would be using the square root of X. And the same argument is going to appear with this expression. And you will be multiplying the derivative off the input, which is going to be the derivative off this expression with, of course, the square works the square root substitute and okay, and then we should get the exact same result. And that's how you find the derivative off our coach.

University of California, Berkeley