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Question:

$x^{x^{2}-3}+(x-3)^{x^{2}}$, for $x>3$

Solution:

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

$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$   ...(1)

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

$\therefore \log u=\log \left(x^{x^{2}-3}\right)$

$\log u=\left(x^{2}-3\right) \log x$

Differentiating with respect to x, we obtain

$\frac{1}{u} \cdot \frac{d u}{d x}=\log x \cdot \frac{d}{d x}\left(x^{2}-3\right)+\left(x^{2}-3\right) \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{1}{u} \frac{d u}{d x}=\log x \cdot 2 x+\left(x^{2}-3\right) \cdot \frac{1}{x}$

$\Rightarrow \frac{d u}{d x}=x^{x^{2}-3} \cdot\left[\frac{x^{2}-3}{x}+2 x \log x\right]$

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

$\frac{1}{v} \cdot \frac{d v}{d x}=\log (x-3) \cdot \frac{d}{d x}\left(x^{2}\right)+x^{2} \cdot \frac{d}{d x}[\log (x-3)]$

$\Rightarrow \frac{1}{v} \frac{d v}{d x}=\log (x-3) \cdot 2 x+x^{2} \cdot \frac{1}{x-3} \cdot \frac{d}{d x}(x-3)$

$\Rightarrow \frac{d v}{d x}=v\left[2 x \log (x-3)+\frac{x^{2}}{x-3} \cdot 1\right]$

$\Rightarrow \frac{d v}{d x}=(x-3)^{x^{2}}\left[\frac{x^{2}}{x-3}+2 x \log (x-3)\right]$

Substituting the expressions of $\frac{d u}{d x}$ and $\frac{d v}{d x}$ in equation (1), we obtain

$\frac{d y}{d x}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log x\right]+(x-3)^{x^{2}}\left[\frac{x^{2}}{x-3}+2 x \log (x-3)\right]$