Question

If $y={\cos }^{-1}\left(\frac{9-x^2}{9+x^2}\right)$ then $y^{\prime }(-1)$ is equal to

• A. $\frac{-3}{5}$
• B. $\frac{3}{5}$
• C. $\frac{2}{7}$
• D. $\frac{3}{8}$

Answer 1

The correct option is A $\frac{-3}{5}$

$y={\cos }^{-1}\left(\frac{9-x^2}{9+x^2}\right)={\cos }^{-1}\left(\frac{1-{\left(x/3\right)}^2}{1+{\left(x/3\right)}^2}\right)$

$=\{\begin{array}{c}2{\tan }^{-1}\frac{x}{3},0\le x<\infty \\ -2{\tan }^{-1}\frac{x}{3},-\infty <x<0\end{array}$

$\Rightarrow y^{\prime }=\{\begin{array}{c}\frac{2}{3}.\frac{1}{1+{\left(x/3\right)}^2},0<x<\infty \\ -\frac{2}{3}.\frac{1}{1+{\left(x/3\right)}^2},-\infty <x<0\end{array}$

$\Rightarrow y^{\prime }(-1)=-\frac{3}{5}$