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Find the derivatives of the given functions.$$y=x^{2} \cos ^{-1} x+\sqrt{1-x^{2}}$$

$\frac{d y}{d x}=\frac{x^{2}}{\sqrt{1-x^{2}}}+2 x \cos ^{-1} x-\frac{x}{\sqrt{1-x^{2}}}$

Calculus 1 / AB

Chapter 27

Differentiation of Transcendental Functions

Section 3

Derivatives of the Inverse Trigonometric Functions

Derivatives

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Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

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All right. We want to find the derivative dy dx for the function Why is equal to X square our cousin X plus the square root of one minus x square. To solve this problem, we're going to rely on derivative shortcuts. We've learned throughout single variable calculus in particular the inverse trig derivatives that we've recently learned. So I've noted here the inverse trig derivatives. The most important one we need for this problem is DDX Arcos X equals negative 1/1 minus x squared square rooted. We also are going to have to use the chain rule in the product rule. So why can be written as X squared? Arcos X plus one minus X squared like half. Thus, we can take dy dx using the product in the first term chain rule, A second term as dy dx equals two. X. Archos sex taking the root of X squared minus x squared times one over root, one minus x square. This is derivative. Our coast plus one half, one minus X squared and one half times 92 X. The second two terms have the same denominator of square, one minus X squared plus. We can combine them to obtain final solution at the bottom. Dy dx is equal to X. Archos x minus x squared plus x over the square root of one minus x squared.

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