If $x^2 y^2=\sin ^{-1} x+\cos ^{-1} x$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=1$ and $y=2$ is

Question

Aicharya · Accepted Answer

$\begin{aligned} & x^2 y^2 \ & = \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\end{aligned}$ Differentiating w.r.t. x, we get $\begin{aligned} & 2 x y^2+2 y x^2 \frac{d y}{d x}=0 \ 	herefore & \left.\frac{\mathrm{~d} y}{d x}ight|_{(1,2)}=-2 \end{aligned}$

If $x^2 y^2=\sin ^{-1} x+\cos ^{-1} x$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=1$ and $y=2$ is

Solution