Question

The derivative of $tan^{- 1}\left[\frac{\left(3 x^{2} - 1\right)}{\left(3 x - x^{3}\right)}\right]$ with respect to $sin^{- 1}\left[\frac{\left(x^{2} - 1\right)}{\left(x^{2} + 1\right)}\right]$ is

• A. $\frac{2}{3}$
• B. $\frac{- 2}{3}$
• C. $\frac{3}{2}$
• D. $-\frac{3}{23}$

Answer 1

Answer: (C)

$y=\left(tan\right)^{- 1}\left(\frac{3 x^{2} - 1}{3 x - x^{3}}\right)$
$=\frac{\pi }{2}-tan1\left(\frac{3 x - x^{3}}{3 x^{2} - 1}\right)=\frac{\pi }{2}+3\left(tan\right)^{1}x$
$\frac{d y}{d x}=\frac{3}{1 + x^{2}}$
$u=\left(sin\right)^{- 1}\left(\frac{x^{2} - 1}{x^{2} + 1}\right)=\frac{\pi }{2}-\left(cos\right)^{- 1}\left(\frac{x^{2} - 1}{x^{2} + 1}\right)$
$=-\frac{\pi }{2}+2tan^{- 1}x$
$\frac{d u}{d x}=\frac{2}{1 + x^{2}}$
$\frac{d y}{d x}=\frac{\frac{d y}{d u}}{\frac{d u}{d x}}=\frac{3}{2}$