Question

If $y=(\sin x{)}^{\log x}+x^{\sin x}$ then find $\frac{dy}{dx}$.

Answer 1

We are given that $y=(sinx{)}^{logx}+x^{sinx}$

We have to find $\frac{dy}{dx}$,

let $y_1=\left(sinx\right)logx$ & $y_2=x^{sinx}$

$\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}=\frac{dy}{dx}$

now, we have $y_1=\left(sinx\right)logx$

$log{y}_1=\left(logx\right)\left(log\right(sinx\left)\right)$

$\frac{1}{y_1}\frac{d{y}_1}{dx}=\frac{1}{x}\left(log\right(sinx\left)\right)+\frac{logx}{\left(sinx\right)}\left(cosx\right)$

$\frac{d{y}_1}{dx}=y_1[\frac{log\left(sinx\right)}{x}+(logx\left)cotx\right]$

now, $y_2=x^{sinx}\Rightarrow log{y}_2=\left(sinx\right)\left(logx\right)$

$\frac{1}{y_2}\frac{d{y}_2}{dx}=\frac{sinx}{x}+\left(cosx\right)\left(logx\right)$

$\frac{d{y}_2}{dx}=y_2[\frac{sinx}{x}+(logx\left)\right(cosx\left)\right]$

$\frac{dy}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

$\frac{dy}{dx}=(sinx{)}^{logx}[\frac{log\left(sinx\right)}{x}+(logx\left)\right(cotx\left)\right]+x^{sinx}[\frac{sinx}{x}+(logx\left)\right(cosx\left)\right]$