Q88 / 180 · COMEDK 2014 · Mathematics
If $y=\tan ^{-1}\left(\frac{a-x}{1+a x}\right)$, then $\frac{d y}{d x}=$
- A$\frac{1}{\left(1+x^{2}\right)}$
- B$\frac{a}{\left(1+a x^{2}\right)}$
- Correct answerC$-\frac{1}{\left(1+x^{2}\right)}$
- D$\frac{x}{\left(1+x^{2}\right)}$
Solution
Given, $y=\tan ^{-1}\left(\frac{a-x}{1+a x}\right)$
$$
\Rightarrow \quad y=\tan ^{-1} a-\tan ^{-1} x
$$
Taking derivative w.r.t. $x$ on both sides, we get
$$
\frac{d y}{d x}=-\frac{1}{1+x^{2}}
$$
Hence, the correct answer is option: 'C'.