**Question:**

If $\sin ^{2} y+\cos x y=k$, find $\frac{d y}{d x}$ at $x=1, y=\frac{\pi}{4}$

**Solution:**

We are given with an equation $\sin ^{2} y+\cos (x y)=k$, we have to find $\frac{d y}{d x}$ at $x=1, y=\frac{\pi}{4}$ by using the given

equation, so by differentiating the equation on both sides with respect to $x$, we get,

$2 \sin y \cos y \frac{d y}{d x}-\sin (x y)\left[(1) y+x \frac{d y}{d x}\right]=0$

$\frac{d y}{d x}[2 \sin y \cos y-x \sin (x y)]=y \sin (x y)$

$\frac{d y}{d x}=\frac{y \sin (x y)}{2 \sin y \cos y-x \sin (x y)}$

By putting the value of point in the derivative, which is $x=1, y=\frac{\pi}{4}$,

$\frac{\mathrm{dy}}{\mathrm{dx}}(\mathrm{x}=1, \mathrm{y}=\mathrm{r} / 4)=\frac{\frac{\pi}{4} \sin \left(\frac{\pi}{4}\right)}{2 \sin \frac{\pi}{4} \cos \frac{\pi}{4}-(1) \sin \frac{\pi}{4}}$

$\frac{d y}{d x}(x=1, y=\pi / 4)=\frac{\frac{\pi}{4 \sqrt{2}}}{1-\frac{1}{\sqrt{2}}}=\frac{\frac{\pi}{4 \sqrt{2}}}{\frac{\sqrt{2}-1}{\sqrt{2}}}=\frac{\pi}{4(\sqrt{2}-1)}$