Question

Differentiate the function w.r.t.x. : $x^{x}+a^{a}$

Answer 1

we have to differentiate $y=x^x+a^a$

here a is constant so, differentiate $a^a$ is also constant and hence differentiation of $a^a$ = 0


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

$y=e^{xlogx}+a^a$

differentiate both sides,

$\frac{dy}{dx}=\frac{d\{e^{xlogx}\}}{dx}+\frac{da^a}{dx}$

$\frac{dy}{dx}=e^{xlogx}\left[x\frac{d(logx)}{dx}+logx\frac{dx}{dx}\right]+0$

$\frac{dy}{dx}=x^x\left[x\times\frac{1}{x}+logx\right]$

$\frac{dy}{dx}=x^x[1+logx]$