Question

If $y={\tan }^{-1}\left(\frac{a-x}{1+ax}\right)$ then $\frac{dy}{dx}=$

• A. $\frac{1}{(1+x^2)}$
• B. $\frac{a}{(1+a{x}^2)}$
• C. $-\frac{1}{(1+x^2)}$
• D. $\frac{x}{(1+x^2)}$

Answer 1

Answer: (C)

Given , $y={\tan }^{-1}\left(\frac{a-x}{1+ax}\right)$
$\Rightarrow y={\tan }^{-1}a-{\tan }^{-1}x$
Taking derivative w.r.t. ${}^{\prime }{x}^{\prime }$ on both sides, we get
$\frac{dy}{dx}=-\frac{1}{1+x^2}$