Question

Differentiate the given function w.r.t. $x$:
$x^{x^2-3}+(x-3{)}^{x^2}$, for $x>3$

Answer 1

Let $y=x^{x^2-3}+(x-3{)}^{x^2}$
Also, let $u=x^{x^2-3}$ and $v=(x-3{)}^{x^2}$
$\therefore y=u+v$
Differentiating both sides with respect to x, we obtain
$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$ .....(1)
$u=x^{x^2-3}$
$\therefore \log u=\log (x^{x^2-3}$)
$\Rightarrow \log u=(x^2-3)\log x$
Differentiating with respect to x, we obtain
$\frac{1}{u}\frac{du}{dx}=\log x.\frac{du}{dx}=\log x.\frac{d}{dx}(x^2-3)+(x^2-3).\frac{d}{dx}\left(\log x\right)$
$\Rightarrow \frac{1}{u}\frac{du}{dx}=\log x.2x+(x^2-3).\frac{1}{x}$
$\frac{du}{dx}=x^{x^2-3}[\frac{x^2-3}{x}+2x\log x]$
Also,
$v=(x-3{)}^{x^2}$
$\therefore \log v=\log (x-3{)}^{x^2}$
$\Rightarrow \log v=x^2\log (x-3)$
Differentiating both sides with respect to x, we obtain
$\frac{1}{v}\frac{dv}{dx}=\log (x-3).\frac{d}{dx}\left(x^2\right)+x^2.\frac{d}{dx}\left[\log \right(x-3\left)\right]$
$\Rightarrow \frac{1}{v}\frac{dv}{dx}=\log (x-3).2x+x^2.\frac{1}{x-3}.\frac{d}{dx}(x-3)$
$\Rightarrow \frac{dv}{dx}=v\left[2x\log \right(x-3)+\frac{x^2}{x-3}.1]$
$\Rightarrow \frac{dv}{dx}=(x-3{)}^{x^2}[\frac{x^2}{x-3}+2x\log (x-3\left)\right]$
Substituting the expressions of $\frac{du}{dx}$ and $\frac{dv}{dx}$ in equation (1), we obtain
$\frac{dy}{dx}=x^{x^2-3}[\frac{x^2-3}{x}+2x\log x]+(x-3{)}^{x^2}[\frac{x^2}{x-3}+2x\log (x-3\left)\right]$