Question

Find $\frac{dy}{dx}$, if $y=x^{\sin x}+(\sin x{)}^{\cos x}$

Answer 1

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

let $x^{\sin x}=h$

applying $\ln$ on both sides

$\sin x\ln x=\ln h$

differiating both sides w.r.t $x$

$\cos x\ln x+\frac{\sin x}{x}=\frac{1}{h}\frac{dh}{dx}$

$\therefore \frac{dh}{dx}=x^{\sin x}(\cos x\ln x+\frac{\sin x}{x})$

let$(\sin x{)}^{\cos x}=t$

apply $\ln$ on both sides

$\cos x\ln \left(sinx\right)=\ln t$

differentiatite on both sides w.r.t $x$

$-\sin x\ln x\left(\sin x\right)+\frac{\cos x}{\sin x}=\frac{1}{t}\frac{dt}{dx}$

$\therefore \frac{dt}{dx}=(\sin x{)}^{\cos x}(\frac{\cos x}{\sin x}-\sin x\ln x(\sin x\left)\right)$

$y=h+t⟹\frac{dy}{dx}=\frac{dh}{dx}+\frac{dt}{dx}$

$\therefore \frac{dy}{dx}=x^{\sin x}(\cos x\ln x+\frac{\sin x}{x})+(\sin x{)}^{\cos x}(\frac{\cos x}{\sin x}-\sin x\ln x(\sin x\left)\right)$