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
$2x+2y\frac{dy}{dx}=0$
$\therefore \frac{dy}{dx}=\frac{-x}{y}$