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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability5 Marks
Question
Differentiate the function $x^{x^{2}-3}+(x-3)^{x^{2}}, \text { for } x>3$ w.r.t to x.
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Answer

Let $y = x^{x^{2}-3}+(x-3)^{x^{2}}$
And let $u = x^{x^{2}-3}$ and $v = (x-3)^{x^{2}}$
$\therefore y = u + v$
Differentiating both sides w.r.t. x we get
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}.....(i)$
Now,
$u = x^{x^{2}-3}$
Taking logarithm both sides
$\log u = \log   x^{x^{2}-3}$
$\Rightarrow \log u = (x^2 - 3) \log x$
Differentiating w.r.t. x, we get
$\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\log _{\mathrm{X}} \times \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}-3\right)+\left(\mathrm{x}^{2}-3\right) \times \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})$
$\Rightarrow \frac{d u}{d x}=u\left[\log x \times 2 x+\left(x^{2}-3\right) \times \frac{1}{x}\right]$
$\Rightarrow \frac{d u}{d x}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log x\right]$.....(ii)
Also,
$v = (x-3)^{x^{2}}$
Taking logarithm both sides
$\log v = \log (x-3)^{x^{2}}$
$\Rightarrow \log v = x^2 \log(x-3)$
Differentiating both sides w.r.t. x
$\frac{1}{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{dx}}=\log (\mathrm{x}-3) \times \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)+\mathrm{x}^{2} \times \frac{\mathrm{d}}{\mathrm{dx}}[\log (\mathrm{x}-3)]$
$\Rightarrow \frac{d v}{d x}=v\left[\log (x-3) \times 2 x+x^{2} \times \frac{1}{(x-3)} \times \frac{d}{d x}(x-3)\right]$
$\Rightarrow \frac{d v}{d x}=(x-3)^{x^{2}}\left[2 x \log (x-3)+\frac{x^{2}}{(x-3)} \times 1\right]$
$\Rightarrow \frac{d v}{d x}=(x-3)^{x^{2}}\left[\frac{x^{2}}{(x-3)}+2 x \log (x-3)\right]$....(iii)
Substituting (ii) and (iii) in (i)
$\therefore \frac{d y}{d x}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log x\right]$ + $(x-3)^{x^{2}}\left[\frac{x^{2}}{(x-3)}+2 x \log (x-3)\right]$

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