Question

If $\tan^{-1}(x^2 + y^2) = \alpha $ then $\frac{dy}{dx} $ is equal to

• A. $\frac{-x}{y}$
• B. $xy$
• C. $\frac{x}{y}$
• D. $-xy$

Answer 1

Answer: (A)

We have, $\tan ^{-1}\left(x^{2}+y^{2}\right)=\alpha $
$\Rightarrow x^{2}+y^{2} \equiv \tan\, \alpha$
On differentiating both sides w.r.t. $x$, we get
$2 x+2 y \frac{d y}{d x} =0$
$\therefore \frac{d y}{d x} =\frac{-x}{y}$