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Continuity and Differentiability — MATHS STD 12 Science — Question

Rajasthan 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^2log(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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