Question

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

Answer 1

Let $y=x^x+x^a+a^x+a^a$
Also let, $x^x=u,x^a=v,a^x=w,$ and $a^a=s$
$\therefore y=u+v+w+s$
$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}+\frac{ds}{dx}$ .....(1)
$u=x^x$
$\Rightarrow \log u=\log x^x$
$\Rightarrow \log u=x\log x$
Differentiating both sides with respect to x, we obtain
$\frac{1}{u}\frac{du}{dx}=\log x.\frac{d}{dx}\left(x\right)+x.\frac{d}{dx}\left(\log x\right)$
$\Rightarrow \frac{du}{dx}=u[\log x.1+x\frac{1}{x}]$
$\Rightarrow \frac{du}{dx}=x^x[\log x+1]=x^x(1+logx)$ .....(2)
$v=x^a$
$\therefore \frac{dv}{dx}=\frac{d}{dx}\left(x^a\right)$
$\Rightarrow \frac{dv}{dx}=a{x}^{a-1}$ .....(3)
$w=a^x$
$\Rightarrow \log w=\log a^x$
$\Rightarrow \log w=x\log a$
Differentiating both sides with respect to x, we obtain
$\frac{1}{w}.\frac{dw}{dx}=\log a.\frac{d}{dx}\left(x\right)$
$\Rightarrow \frac{dw}{dx}=w\log a$
$\Rightarrow \frac{dw}{dx}=a^x\log a$ .....(4)
$s=a^a$
Since $a$ is constant, $a^a$ is also a constant.
$\therefore \frac{ds}{dx}=0$ .....(5)
From (1), (2), (3), (4) and (5) we obtain
$\frac{dy}{dx}=x^x(1+\log x)+a{x}^{a-1}+a^x\log a+0$
$=x^x(1+\log x)+a{x}^{a-1}+a^x\log a$