Question

Differentiate the given functions w.r.t. x.

$x^x-2^{sinx}$.

Answer 1

Ley y = $x^x-2^{sinx}$

Let $u=x^{x}andv=2^{sinx}$

$\therefore$ y = u - v

Differentiating both sides w.r.t. x, we get

$\frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}$

Now, $u=x^x$

Taking log on both sides,

$\Rightarrow logu=log{x}^x\Rightarrow logu=xlogx\Rightarrow \frac{1}{u}\frac{du}{dx}=x\frac{d}{dx}logx+logx\frac{d}{dx}x(Differentiatew.r.t,x)\Rightarrow \frac{1}{u}\frac{du}{dx}=x\times \frac{1}{x}+logx\Rightarrow \frac{1}{u}\frac{du}{dx}=1+logx,\frac{du}{dx}=x^x(1+logx)v=2^{sinx}Takinglogonbothsides,\Rightarrow logv=log\left(2^{sinx}\right)\Rightarrow logv=\left(sinx\right)log2\Rightarrow \frac{1}{v}\frac{dv}{dx}=cosx\left(log2\right)(Differentiatew.r.t.x)\Rightarrow \frac{dv}{dx}=v\left[cosxlog2\right]=2^{sinx}\left[cosxlog2\right]Now,puttingthevaluesof\frac{du}{dx}and\frac{dv}{dx}inEq.\left(i\right),\frac{dy}{dx}=x^x[1+logx]-2^{\left[sinx\right]}\left[cosxlog2\right]$