Question

If $y={\left(\sin x\right)}^{{\left(\sin x\right)}^{{\left(\sin x\right)}^{...\infty }}}$, prove that $\frac{y^2\cot x}{\left(1-y\log \sin x\right)}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y={\left(\sin x\right)}^{{\left(\sin x\right)}^{{\left(\sin x\right)}^{....\infty }}} \\ \Rightarrow y={\left(\sin x\right)}^y \end{gathered}$$
Taking log on both sides,
$$\begin{gathered} \log y=\log {\left(\sin x\right)}^y \\ \Rightarrow \log y=y\log \left(\sin x\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{1}{y}\frac{dy}{dx}=y\frac{d}{dx}\left\{\log \left(\sin x\right)\right\}+\log \sin x\frac{dy}{dx} \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=y\left(\frac{1}{\sin x}\right)\frac{d}{dx}\left(\sin x\right)+\log \sin x\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}\left(\frac{1}{y}-\log \sin x\right)=\frac{y}{\sin x}\left(\cos x\right) \\ \Rightarrow \frac{dy}{dx}\left(\frac{1-y\log \sin x}{y}\right)=y\cot x \\ \Rightarrow \frac{dy}{dx}=\frac{y^2\cot x}{\left(1-y\log \sin x\right)} \end{gathered}$$