Question

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

$(logx{)}^x+x^{logx}$

Answer 1

Let y = $(logx{)}^x+x^{logx}$

Let $u=(logx{)}^x,v=x^{logx}$

$\therefore$ y = u + v

Differentiating w.r.t. x, $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$ ......(i)

Now, $u=(logx{)}^x$

Taking log on both sides,

$\Rightarrow logu=log(logx{)}^x\Rightarrow logu=xlog(logx)$

Differentiating w.r.t., x we get

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

Again, $v=x^{logx}$

Taking log on both sides,

$\Rightarrow logv=log{x}^{logx}\Rightarrow logv=\left(logx\right)\left(logx\right)\Rightarrow logv=(logx{)}^2$

Differentiaing w.r.t. x, we get

$\Rightarrow \frac{1}{v}\frac{dv}{dx}=2logx\frac{d}{dx}\left(logx\right)=2logx\times \frac{1}{x}(\therefore \frac{d}{dx}[f\left(x\right){]}^2=2f\left(x\right)\frac{d}{dx}f\left(x\right))\Rightarrow \frac{dv}{dx}=v\left[\frac{2logx}{x}\right]\Rightarrow \frac{dv}{dx}=x^{logx}\left[\frac{2logx}{x}\right]Now,puttingthevaluesof\frac{du}{dx}and\frac{dv}{dx}inEa.\left(i\right),weget\frac{dy}{dx}=(logx{)}^x[\frac{1}{logx}+log(logx\left)\right]+x^{logx}\left[\frac{2logx}{x}\right]\frac{dy}{dx}=(logx{)}^{x-1}[1+logxlog(logx\left)\right]+2x^{logx-1}.logx$