Question

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

Answer 1

The correct answer is: $x^{x^2-3}.[\frac{x^2-3}{x}+2xlogx]+(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]$
Let $y=x^{x^2-3}+(x-3)^{x^2}$
Also,let $u=x^{x^2-3}$ and $v=(x-3)^{x^2}$
$∴y=u+v$
Differentiating both sides with respect to $x$,we obtain
$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx} .....(1)$
$u=x^{x^2-3}$
$∴logu=log(x^{x^2-3})$
$logu=(x^2-3)logx$
Differentiating with respect to $x$,we obtain
$\frac{1}{u}.\frac{du}{dx}=logx.\frac{d}{dx}(x^2-3)+(x^2-3).\frac{d}{dx}(logx)$
$⇒\frac{1}{u}\frac{du}{dx}=logx.2x+(x^2-3).\frac{1}{x}$
$⇒\frac{du}{dx}=x^{x^2-3}.[\frac{x^2-3}{x}+2x\,logx]$
Also,
$v=(x-3)^{x^2}$
$∴logv=log(x-3)^{x^2}$
$⇒logv=x^2log(x-3)$
Differentiating both sides with respect to $x$,we obtain
$\frac{1}{v}.\frac{dv}{dx}=log(x-3).\frac{d}{dx}(x^2)+x^2.\frac{d}{dx}[log(x-3)]$
$⇒\frac{1}{v}\frac{dv}{dx}=log(x-3).2x+x^2.\frac{1}{x}-3.\frac{d}{dx}(x-3)$
$⇒\frac{dv}{dx}=v[2xlog(x-3)+\frac{x^2}{x-3}.1]$
$⇒\frac{dv}{dx}=(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]$
Substituting the expressions of $\frac{du}{dx}$ and $\frac{dv}{dx}$ in equation (1), we obtain
$\frac{dy}{dx}=x^{x^2-3}.[\frac{x^2-3}{x}+2xlogx]+(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]$