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Find the derivative of the function. Simplify where possible.

$ y = x \sin^{-1} x + \sqrt{1 -x^2} $

$$\sin ^{-1} x$$

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Baylor University

University of Michigan - Ann Arbor

University of Nottingham

Boston College

in this problem. We are practicing our differentiation skills specifically with using inverse triggered a metric functions. And also, this is a great example of in very informal proof. And this is a great precursor to future math. Classes that you might take were given the function y equals x times, the inverse sign of X plus the square root of one minus X squared. So we have to differentiate this to prove that we get the inverse sign of X. We're told that's what we have to wind up with. So why prime are derivative would be equal to the derivative with respect to X of X sine inverse X plus the derivative of the square root of one minus X squared. So how am I going to do this? Well, this first term looks like a product, so we're going to use the product rule and the second term we're going to use chain rule. So when we apply, those rules will get that Why prime equals the inverse sign of X times one plus x times one over the square root of one minus X squared, plus 1/2 times the square root of one minus X squared times negative two x So hopefully that made sense how we went from product ruled chain rule to get this derivative here. And then we could simplify it. We could get why prime equals the inverse sign of X plus X over the square root of one minus X squared minus X over the square root of one minus X squared. While these two terms are the exact staying with opposite signs so they can cancel. So we're going to get that. Why Prime equals the inverse sign of X, just as we were supposed to. The problem tells us that that's what we have to find. So I hope that this problem helped you understand a little bit more about differentiation, how we can choose the rules to differentiate. And then I hope this helped you understand a little bit more about about an informal proof, or maybe what to do when a problem tells you to prove something

University of Denver