Question

Differentiate $x^{sinx}$ w.r.t. $(sinx{)}^x.$

Answer 1

Differentiate $x^{\sin x}$ w.r.t $(\sin x{)}^x$

i.e, $\frac{d\left(x^{{\sin }^x}\right)}{d(\sin x{)}^x}=\frac{d\left(x^{\sin x}\right)}{dx}.\frac{dx}{d(\sin x{)}^x}=$

$=\frac{d\left(x^{\sin x}\right)}{dx}.\frac{1}{\frac{d(\sin x{)}^x}{dx}}----\left(1\right)$

Let $y_1=x^{\sin x}$

Applying $\log$ both sides

$\log y_1=\log x^{\sin x}=\sin x\log x$

Integrating w.r.t

$\frac{1}{y_1}\frac{d{y}_1}{dx}=\frac{d}{dx}\left(\sin x\log x\right)=\sin x.\frac{1}{x}+\cos x\log x$

$\Rightarrow \frac{d{y}_1}{dx}=y_1(\frac{\sin x}{x}+\cos x\log x)=x^{\sin x}(\frac{\sin x}{x}+\cos x\log x)----\left(ii\right)$

Again Let $y_2=(\sin x{)}^x$

$\log y_2=\log (\sin x{)}^x=x\log \sin x$

differentiating both sides w.r.t $x$

$\frac{1}{y_2}\frac{d{y}_2}{dx}=\frac{d}{dx}\left(x\log \sin x\right)=x\frac{d}{dx}\log \sin x+\log \sin x.\frac{dx}{dx}$

$=x.\frac{1}{\sin x}\cos x+\log \sin x$

$=x.\cot x=\log \sin x$

$\frac{d{y}_2}{dx}=y_2(x\cot x+\log \sin x)=\left(\sin x{)}^x\right(x\cot x+\log \sin x)----(iii)$

Putting these $\left(ii\right)$ and $\left(iii\right)$ in $\left(1\right)$ we have

$\frac{d(x{)}^{\sin x}}{d(\sin x{)}^x}=x^{\sin x}(\frac{\sin x}{x}+(\cos x\log x\left)\right)\times \frac{1}{\left(\sin x{)}^x\right(xdx+\log \sin x)}$