Question

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

Answer 1

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

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

Differentiate with respect to $x$,

$\Rightarrow 2\sin y\cos y\left(\frac{dy}{dx}\right)–\sin xy(y+x\frac{dy}{dx})=0$ $\left[\because \frac{d}{dx}\left(f\left(u\right)\right)=\frac{d}{du}f\left(u\right)\times \frac{du}{dx}\right]$

$\Rightarrow \left(\frac{dy}{dx}\right)[2\sin y\cos y–x\sin xy]=y\sin xy$

$\Rightarrow \frac{dy}{dx}=\frac{y\sin xy}{\sin 2y–x\sin xy}$ $\left[\because \sin 2\theta =2\sin \theta \cos \theta \right]$

Hence, the differentiation of ${\sin }^{2}y+\cos xy=k$ with respect to $x$ is $\frac{\mathbf{y}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{y}}{\mathbf{sin}\mathbf{}\mathbf{2}\mathbf{y}\mathbf{}\mathbf{–}\mathbf{}\mathbf{x}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{y}}$.