Question

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

Answer 1

We have,

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

By differentiating both sides with respect to x, we get

$$\begin{gathered} \frac{d}{dx}\left({\sin }^{2}y\right)+\frac{d}{dx}\left(\cos xy\right)=\frac{dk}{dx} \\ \Rightarrow 2\sin y\frac{d}{dx}\left(\sin y\right)+\left(-\sin xy\right)\frac{d}{dx}\left(xy\right)=0 \\ \Rightarrow 2\sin y\cos y-\sin xy\left(x\frac{dy}{dx}+y\frac{dx}{dx}\right)=0 \\ \Rightarrow 2\sin y\cos y-x\sin xy\frac{dy}{dx}-y\sin xy=0 \\ \Rightarrow x\sin xy\frac{dy}{dx}=2\sin y\cos y-y\sin xy \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{2\sin y\cos y-y\sin xy}{x\sin xy} \\ \Rightarrow {\left[\frac{dy}{dx}\right]}_{\left(x=1,y=\frac{\mathrm{\pi }}{4}\right)}=\frac{2\sin y\cos y-y\sin xy}{x\sin xy} \\ =\frac{2\sin {\displaystyle \frac{\mathrm{\pi }}{4}}\cos {\displaystyle \frac{\mathrm{\pi }}{4}}-{\displaystyle \frac{\mathrm{\pi }}{4}}\sin \left(1\times {\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{1\times \sin \left(1\times {\displaystyle \frac{\mathrm{\pi }}{4}}\right)} \\ =\frac{2\times {\displaystyle \frac{1}{\sqrt{2}}}\times {\displaystyle \frac{1}{\sqrt{2}}}-{\displaystyle \frac{\mathrm{\pi }}{4}}\times {\displaystyle \frac{1}{\sqrt{2}}}}{1\times {\displaystyle \frac{1}{\sqrt{2}}}} \\ =\frac{2\times {\displaystyle \frac{1}{2}}-{\displaystyle \frac{\mathrm{\pi }}{4\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}} \\ =\frac{{\displaystyle \frac{1}{1}}-{\displaystyle \frac{\mathrm{\pi }}{4\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}} \\ =\frac{{\displaystyle \frac{4\sqrt{2}-\mathrm{\pi }}{4\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}} \\ =\frac{4\sqrt{2}-\mathrm{\pi }}{4} \\ =\sqrt{2}-\frac{\mathrm{\pi }}{4} \end{gathered}$$