Question

Find $\frac{dy}{dx}$ at $x=1,y=\frac{\pi }{4}$ if ${\sin }^{2}y+\cos xy=k$.

Answer 1

${\sin }^{2}y+\cos xy=k$

$2\sin y\cos y\frac{dy}{dx}+(-\sin xy)(y+x\frac{dy}{dx})=0$

Put $y=\frac{\pi }{4},x=1$

$2\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\frac{dy}{dx}-\frac{1}{\sqrt{2}}(\frac{\pi }{4}+\frac{dy}{dx})=0$

$\frac{dy}{dx}-\frac{1}{\sqrt{2}}\frac{dy}{dx}=\frac{\pi }{4\sqrt{2}}$

$\frac{dy}{dx}=\frac{\pi }{4(\sqrt{2}-1)}$