Question 8 Let f(x)=tan−1(1−x2)f(x) = \tan^{-1}\left( \sqrt{1 - x^2} \right), where −1<x<1-1 < x < 1. (a) Compute f′(x)f'(x). (b) Simplify your result as much as possible. Show solutionHide solution+Question 8 - Solution We are given: f(x)=tan−1(1−x2)f(x) = \tan^{-1}\left( \sqrt{1 - x^2} \right) (a) Differentiate using the chain rule: Let u=1−x2u = \sqrt{1 - x^2}, so: f(x)=tan−1(u)⇒f′(x)=11+u2⋅dudxf(x) = \tan^{-1}(u) \Rightarrow f'(x) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} Now compute each part: u=(1−x2)1/2,dudx=12(1−x2)−1/2⋅(−2x)=−x1−x2u = (1 - x^2)^{1/2}, \quad \frac{du}{dx} = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} Now plug into the derivative: f′(x)=11+(1−x2)⋅−x1−x2=12−x2⋅−x1−x2f'(x) = \frac{1}{1 + (1 - x^2)} \cdot \frac{-x}{\sqrt{1 - x^2}} = \frac{1}{2 - x^2} \cdot \frac{-x}{\sqrt{1 - x^2}} f′(x)=−x(2−x2)1−x2f'(x) = \boxed{ \frac{-x}{(2 - x^2)\sqrt{1 - x^2}} } (b) Final simplified form: f′(x)=−x(2−x2)1−x2\boxed{f'(x) = \frac{-x}{(2 - x^2)\sqrt{1 - x^2}}}
