Question

If $y=x^{\tan x}+(sinx{)}^{\cos x}$, find $\frac{dy}{dx}$.

Answer 1

We are given, $y=x^{tanx}+(sinx{)}^{cosx}$

let $y_1=x^{tanx}$ & $y_2=(sinx{)}^{cosx}$

we have to find $\frac{dy}{dx}\Rightarrow$

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

So, $y_1=x^{tanx}$

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

$\frac{1}{y_1}\frac{d{y}_1}{dx}=\frac{tanx}{x}+\left(se{c}^{2}x\right)\left(logx\right)$

So, $\frac{d{y}_1}{dx}=y_1(\frac{tanx}{x}+(se{c}^{2}x\left)\right(logx\left)\right)$

So, similarly, $y_2=(sinx{)}^{cosx}$

$log{y}_2=cosx\left(log\right(sinx\left)\right)$

$\frac{1}{y_2}\frac{d{y}_2}{dx}=\frac{cosx}{sinx}.cosx+(-sinx)\left(log\right(sinx\left)\right)$

$\frac{d{y}_2}{dx}=y_2(cotx.cosx-(sinx\left)\right(log\left(sinx\right)\left)\right)$

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

$\frac{dy}{dx}=\left(x^{tanx}\right)(\frac{tanx}{x}+(se{c}^{2}x\left)\right(logx\left)\right)+\left(\right(sinx{)}^{cosx}\left)\right(cotx.cosx-\left(sinx\right)\left(log\right(sinx\left)\right))$