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. $'x'$ on both sides, we get
$ \frac{dy}{dx} = - \frac{1}{1 + x^2}$