Question

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

• A. $\frac{2}{5}$
• B. $\frac{-2}{5}$
• C. $-\frac{1}{5}$
• D. None of these

Answer 1

Answer: (B)

$\frac{-2}{5}$

$f\left(x\right)={\cos }^{-1}\left(\frac{x-x^{-1}}{x+x^{-1}}\right)$
$=\left({\cos }^{-1}\frac{x^2-1}{x^2+1}\right)$
$={\cos }^{-1}\left(-\left(\frac{1-x^2}{1+x^2}\right)\right)$
$=(\pi -{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right))$ ....... $(\because {\cos }^{-1}x+{\cos }^{-1}(-x)=\pi )$
$f\left(x\right)=(\pi -2{\tan }^{-1}x)$ ....... $(\because 2{\tan }^{-1}x={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right))$
Diff. both sides w.r.t $x$
$f^{\prime }\left(x\right)=0-\frac{2}{1+x^2}$
$f^{\prime }(-2)=\frac{-2}{1+{(-2)}^2}$
$=\frac{-2}{5}$