Question

If  $${(\sin x)}^y=x+y$$ , prove that $$\frac{dy}{dx}=\frac{1-(x+y)y\cot x}{(x+y)\log \sin x-1}$$ ?

 

Answer 1

$$\text{ We have },{(\sin x)}^y=x+y$$

 Taking log on both the sides,

$$\log {(\sin x)}^y=\log (x+y)$$

$$\Rightarrow y\log (\sin x)=\log (x+y)$$

Differentiating with respect to x using chain rule, 

$$\frac{d}{dx}\{y\log (\sin x)\}=\frac{d}{dx}\{\log (x+y)\}$$
$$\Rightarrow y\frac{d}{dx}\log \sin x+\log \sin x\frac{dy}{dx}=\frac{1}{x+y}\frac{d}{dx}(x+y)$$
$$\Rightarrow \frac{y}{\sin x}\frac{d}{dx}(\sin x)+\log \sin x\frac{dy}{dx}=\frac{1}{(x+y)}[1+\frac{dy}{dx}]$$
$$\Rightarrow \frac{y(\cos x)}{(\sin x)}+\log \sin x\frac{dy}{dx}=\frac{1}{(x+y)}+\frac{1}{(x+y)}\frac{dy}{dx}$$
$$\Rightarrow \frac{dy}{dx}(\log \sin x-\frac{1}{x+y})=\frac{1}{(x+y)}-y\cot x$$
$$\Rightarrow \frac{dy}{dx}\left\{\frac{(x+y)\log \sin x-1}{(x+y)}\right\}=\frac{1-y(x+y)\cot x}{x+y}$$
$$\Rightarrow \frac{dy}{dx}=\left\{\frac{1-y(x+y)\cot x}{(x+y)\log \sin x-1}\right\}$$