Show solutionHide solution+ Question 10 Let f(x)=tan−1(1−x2)f(x) = \tan^{-1}(\sqrt{1 - x^2}) for −1<x<1-1 < x < 1. Find f′(x)f'(x). Simplify your answer as much as possible. Question 10 - Solution We are given: f(x)=tan−1(1−x2)f(x) = \tan^{-1}(\sqrt{1 - x^2}) Let u=1−x2u = \sqrt{1 - x^2}, then by the chain rule: f′(x)=11+u2⋅u′f'(x) = \frac{1}{1 + u^2} \cdot u' Step 1: Compute u′u': u=(1−x2)1/2⇒u′=12(1−x2)−1/2⋅(−2x)=−x1−x2u = (1 - x^2)^{1/2} \Rightarrow u' = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} Step 2: Compute 1+u21 + u^2: 1+u2=1+(1−x2)=2−x21 + u^2 = 1 + (1 - x^2) = 2 - x^2 Step 3: Combine everything: f′(x)=11+u2⋅u′=12−x2⋅−x1−x2=−x(2−x2)1−x2f'(x) = \frac{1}{1 + u^2} \cdot u' = \frac{1}{2 - x^2} \cdot \frac{-x}{\sqrt{1 - x^2}} = \boxed{ \frac{-x}{(2 - x^2)\sqrt{1 - x^2}} }