Question

If $y=(\sin x{)}^{(\sin x{)}^{\left(\sin x\right)...\infty }}$ then show that $\frac{dy}{dx}=\frac{y^2\cot x}{1-y\log \sin x}$

Answer 1

$y=(\sin x{)}^{(\sin x{)}^{\left(\sin x\right)...\infty }}$

$\Rightarrow y=(\sin x{)}^y$
$\Rightarrow \log y=y\log \sin x$
Differentiating both side w.r.t $x$
$\frac{1}{y}\frac{dy}{dx}=\log \sin x\frac{dy}{dx}+y\frac{\cos x}{\sin x}$

$(\frac{1}{y}-\log \sin x)\frac{dy}{dx}=y\frac{\cos x}{\sin x}$

$\therefore \frac{dy}{dx}=\frac{y^2\cot x}{1-y\log \sin x}$