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

## 1 Answer

Toolbox:
• $[(uv)'=u'v+uv']$
Step 1:
It is of the form $u+v$
$\Rightarrow u=x^{\large x^2-3}$
$v=(x-3)^{\large x^2}$
Now $u=x^{\large x^2-3}$
Taking $\log$ on both sides
$\log u=\log x^{\large x^2-3}=(x^2-3)\log x$
Step 2:
Differentiating with respect to $x$
$\large\frac{1}{u}\frac{du}{dx}$$=2x\log x+(x^2-3)\large\frac{1}{x} \large\frac{du}{dx}$$=u\big(2x\log x+\large\frac{x^2-3}{x}\big)$
$\quad\;=x^{\large x^2-3}[2x\log x+\large\frac{x^2-3}{x}]$
Step 3:
Now $v=(x-3)^{\Large x^{\Large 2}}$
Taking $\log$ on both sides
Now $\log v=\log(x-3)^{\Large x^{\Large 2}}$
Differentiating both sides
$\large\frac{1}{v}\frac{dv}{dx}=$$[2x\log(x-3)+\large\frac{x^2}{x-3}] \large\frac{dv}{dx}$$=v.[2x\log(x-3)+\large\frac{x^2}{x-3}]$
$\large\frac{dv}{dx}$$=(x-3)^{\Large x^{\Large 2}}.[2x\log(x-3)+\large\frac{x^2}{x-3}] Step 4: \large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx} \quad\;=x^{\Large x^{\Large 2}- 3}[$$2x\log x+\large\frac{x^2-3}{x}]$$+(x-3)^{\Large x^{\Large 2}}.[2x\log(x-3)+\large\frac{x^2}{x-3}]$
answered May 14, 2013

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