Question

If $x=y \sqrt{1-y^2}$, then $\frac{d y}{d x}$ is equal to

(A) 0

(B) $x$

(C) $\frac{\sqrt{1-y^2}}{1-2 y^2}$

(D) $\frac{\sqrt{1-y^2}}{1+2 y^2}$

Answer 1

On differentiating w.r.t. $x$, we get

$\begin{array}{l}1=\frac{d y}{d x} \sqrt{1-y^2}+y \cdot \frac{1}{2 \sqrt{1-y^2}}(-2 y) \frac{d y}{d x} \\\Rightarrow \quad 1=\frac{d y}{d x}\left[\frac{1-2 y^2}{\sqrt{1-y^2}}\right] \Rightarrow \frac{d y}{d x}=\frac{\sqrt{1-y^2}}{1-2 y^2}\end{array}$

So, The correct option will be (C).