Question

Differentiate given problems w.r.t.x.
$x^{x^2-3}+(x-3{)}^{x^2},forx>3$

Answer 1

Let y=$x^{x^2-3}+(x-3{)}^{x^2},forx>3$

$\Rightarrow$ y=u+v

Differentiating both sides w.r.t. x, we get

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

Now, u=$x^{x^2-3}$

Taking log on both sides,

log u=log $x^{x^2-3}$

Differentiating both sides w.r.t.x, we get

$\frac{1}{u}\frac{du}{dx}=(x^2-3).\frac{1}{x}+2xlogx$

$\frac{du}{dx}=u[\frac{x^2-3}{x}+2xlogx]$

$\frac{du}{dx}=x^{x^2-3}[\frac{x^2-3}{x}+2xlogx]$

Now, v=$(x-3{)}^{x^2}$

Takign log on both sides,

log v=log $(x-3{)}^{x^2}=x^{2}log(x-3\left)\right(\because log{m}^n=nlogm)$

Differentiating both sides w.r.t.x, we get

$\frac{1}{v}\frac{dv}{dx}=\left(x^2\right)\frac{d}{dx}log(x-3)+log(x-3)\frac{d}{dx}{x}^2=x^2\frac{1}{x-3}+2xlog(x-3)$

$\frac{dv}{dx}=v[\frac{x^2}{x-3}+2xlog(x-3\left)\right]$

$\frac{dv}{dx}=x^{x^2-3}[\frac{x^2}{x-3}+2xlog(x-3\left)\right]$

Putting the values of $\frac{du}{dx}$ and $\frac{dv}{dx}$ in Eq. (i), we get

$\frac{dv}{dx}=x^{x^2-3}[\frac{x^2-3}{x}+2xlogx]+(x-x{)}^{x^2}[\frac{x^2}{x-3}+2xlog(x-3\left)\right]$