Question

Differentiate given problems w.r.t.x.
$x^x+x^a+a^x+a^a$, for some fixed a>0 and x>0.

Answer 1

$\frac{d}{dx}{x}^x+x^a+a^x+a^a$

$\Rightarrow$ $\frac{d}{dx}\left(x^x\right)+\frac{d}{dx}\left(x^a\right)+\frac{d}{dx}\left(a^x\right)+\frac{d}{dx}\left(a^a\right)$

=$\left(x^x\right)\frac{d}{dx}\left(xlogx\right)+a{x}^{a-1}+a^x$ log a

$(\because \frac{d}{dx}(u^v)=(u^v\left)\frac{d}{dx}\right(vlogu\left)\right)$

=$x^x(x\frac{1}{x}+logx.1)+a{x}^{a-1}+a^x$log a

=$x^x(1+logx)+a{x}^{a-1}+a^x$ log a

Alternative method

Let $y=x^x+x^a+a^x+a^a$,

Let $u=x^x\Rightarrow y=u+x^a+a^x+a^a$,

$\Rightarrow$ $\frac{dy}{dx}=\frac{du}{dx}+\frac{d}{dx}+x^a\frac{d}{dx}{a}^x+\frac{d}{dx}{a}^a$

$\Rightarrow$ $\frac{dy}{dx}=\frac{du}{dx}+a{x}^{a-1}+a^x$ log a

Now, u=$x^x$

Taking log on both sides,

$\Rightarrow$ log u = log $x^x$ = x log x $[\because log{m}^n=nlogm]$

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

$\Rightarrow \frac{1}{u}=\frac{du}{dx}logx+x\left(\frac{1}{x}\right)=1+logx\Rightarrow \frac{du}{dx}=u(1+logx)=x^x(1+logx)$

Putting $\frac{du}{dx}$ in Eq. (i), we get $\frac{dy}{dx}=x^x$(1+log x) + $a{x}^{a-1}$+log a