Question

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

Answer 1

The correct answer is $=x^x(1+logx)+ax^{a-1}+a^xloga$
Let $y=x^x+x^a+a^x+a^a$
Also,let $x^x=u,x^a=v,a^x=w$ and $a^a=s$
$∴y=u+v+w+s$
$⇒\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}+\frac{ds}{dx} ......(1)$
$u=x^x$
$⇒logu=logx^x$
$⇒logu=xlogx$
Differentiating both sides with respect to $x$,we obtain
$\frac{1}{u}\frac{du}{dx}=logx.\frac{d}{dx}(x)+x.\frac{d}{dx}(logx)$
$⇒\frac{du}{dx}=u[logx.1+x.\frac{1}{x}]$
$⇒\frac{du}{dx}=x^x[logx+1]=x^x(1+logx) .....(2)$
$v=x^a$
$∴\frac{dv}{dx}=\frac{d}{dx}(x^n)$
$⇒\frac{dv}{dx}=ax^{a-1} ......(3)$
$w=a^x$
$⇒logw=loga^x$
$⇒logw=xloga$
Differentiating both sides with respect to $x$,we obtain
$\frac{1}{w}.\frac{dw}{dx}=loga.\frac{d}{dx}(x)$
$⇒\frac{dw}{dx}=w\,loga$
$⇒\frac{dw}{dx}=a^xloga ......(4)$
$s=a^a$
Since $a$ is constant,$a^a$ is also a constant.
$∴\frac{ds}{dx}=0 ......(5)$
From (1),(2),(3),(4),and (5),we obtain
$\frac{dy}{dx}=x^x(1+logx)+ax^{a-1}+a^xloga+0$
$=x^x(1+logx)+ax^{a-1}+a^xloga$