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

The correct option is A $\frac{-x}{y}$

If ${\tan }^{-1}(x^2+y^2)=\alpha$

then $x^2+y^2=\tan \alpha$

Now differentiate both sides w.r.t. $x$

$2x+2y\frac{dy}{dx}=0$, note that $\alpha$ is some constant

$\Rightarrow x+y\frac{dy}{dx}=0$

$\Rightarrow y\frac{dy}{dx}=-x$

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