Question

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

${(x+\frac{1}{x})}^x+x^{(1+\frac{1}{x})}$

Answer 1

Let y = ${(x+\frac{1}{x})}^x+x^{(1+\frac{1}{x})}$

Again let $u={(x+\frac{1}{x})}^x,v=x^{(1+\frac{1}{x})}$

$\therefore$ y = u + v

Differentiating w.r.t. x

$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$............(i)

Now, $u={(x+\frac{1}{x})}^x$

Taking log on both sides

$\Rightarrow logu=log{(x+\frac{1}{x})}^x\Rightarrow logu=xlog(x+\frac{1}{x})[\therefore log{m}^n=nlogm]$

Differentiating w.r.t. x

$\Rightarrow \frac{1}{u}\frac{du}{dx}=x\frac{d}{dx}log(1+\frac{1}{x})+log(x+\frac{1}{x})\frac{d}{dx}\left(x\right)\left[Usingchainrule\right]\Rightarrow \frac{1}{u}\frac{du}{dx}=\frac{x\times 1}{(x+\frac{1}{x})}[1-\frac{1}{x^2}]+log(x+\frac{1}{x})\Rightarrow \frac{1}{u}\frac{du}{dx}=\frac{x}{x^2+1}\left[\frac{x^2-1}{x^2}\right]+log(x+\frac{1}{x})\Rightarrow \frac{1}{u}\frac{du}{dx}=[\frac{(x^2-1)}{(x^2+1)}+log(x+\frac{1}{x})]\Rightarrow \frac{du}{dx}=u[\frac{(x^2-1)}{(x^2+1)}+log(x+\frac{1}{x})]\Rightarrow \frac{du}{dx}={(x+\frac{1}{x})}^x[\frac{(x^2-1)}{(x^2+1)}+log(x+\frac{1}{x})]$

Now, $v=x^{(1+\frac{1}{x})}$

Taking log on both sides,

$\Rightarrow logv=log{x}^{(1+\frac{1}{x})}\Rightarrow logv=(1+\frac{1}{x})logx$

Differentiating w.r.t., x,

$\Rightarrow \frac{1}{v}\frac{dv}{dx}=(1+\frac{1}{x})\frac{d}{dx}\left\{logx\right\}+logx\frac{d}{dx}(1+\frac{1}{x})\left(Usingchainrule\right)\Rightarrow \frac{1}{v}\frac{dv}{dx}=(1+\frac{1}{x})\times \frac{1}{x}+logx[0-\frac{1}{x^2}]\Rightarrow \frac{1}{v}\frac{dv}{dx}=\left[\frac{x+1}{x^2}\right]-\frac{1}{x^2}logx\Rightarrow \frac{1}{v}\frac{dv}{dx}=\left[\frac{x+1-logx}{x^2}\right]\Rightarrow \frac{dv}{dx}=v\left[\frac{x+1-logx}{x^2}\right]\Rightarrow \frac{dv}{dx}=x^{(1+\frac{1}{x})}\left[\frac{x+1-logx}{x^2}\right]Puttingthevaluesof\frac{du}{dx}and\frac{dv}{dx}inEq.\left(i\right),\therefore \frac{dy}{dx}={(1+\frac{1}{x})}^x[\frac{(x^2-1)}{(x^2+1)}+log(x+\frac{1}{x})]+x^{(1+\frac{1}{x})}\left[\frac{x+1-logx}{x^2}\right]$