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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following functions with respect to x:
$\text{x}^{\text{x}^2-3}+(\text{x}-3)^{\text{x}^2}$
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Answer

Let $\text{y}=\text{x}^{\text{x}^2-3}+(\text{x}-3)^{\text{x}^2}$
Also, let $\text{u}=\text{x}^{\text{x}^2-3}\text{ and v}=(\text{x}-3)^{\text{x}^2}$
$\therefore \text{y}=\text{u}+\text{v}$
Differentiating both sides with respect to x, we obtain
$\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ .....(\text{i})$
$\text{u}=\text{x}^{\text{x}^2-3}$
$\log\text{u}=(\text{x}^2-3)\log\text{x}$
Differentiating with respect to x, we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log\text{x}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^2-3\big)+\big(\text{x}^2-3\big)\times\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log\text{x}\times2\text{x}+(\text{x}^2-3)\times\frac{1}{\text{x}}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}^2-3}\times\Big[\frac{\text{x}^2-3}{\text{x}}+2\text{x}\log\text{x}\Big]$
Also,
$\text{v}=(\text{x}-3)^{\text{x}^2}$
$\therefore\log\text{v}=\log(\text{x}-3)^{\text{x}^2}$
$\Rightarrow\log\text{v}=\text{x}^2\log(\text{x}-3)$
Differentaiting both sides with respect to x, we obtain
$\frac{1}{\text{v}}\times\frac{\text{dv}}{\text{dx}}=\log(\text{x}-3)\times\frac{\text{d}}{\text{dx}}(\text{x}^2)+\text{x}^2\times\frac{\text{d}}{\text{dx}}[\log(\text{x}-3)]$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\log(\text{x}-3)\times2\text{x}+\text{x}^2\times\frac{1}{\text{x}-3}\times\frac{\text{d}}{\text{dx}}(\text{x}-3)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[2\text{x}\log(\text{x}-3)+\frac{\text{x}^2}{\text{x}-3}\times1\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=(\text{x}-3)^{\text{x}^2}\Big[\frac{\text{x}^2}{\text{x}-3}+2\text{x}\log(\text{x}-3)\Big]$
Substituting the expressions of $\frac{\text{du}}{\text{dx}}$ and $\frac{\text{dv}}{\text{dx}}$ in equation (1), we obtain
$\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}^2-3}\Big[\frac{\text{x}^2-3}{\text{x}}+2\text{x}\log\text{x}\Big] \\ +(\text{x}-3)^{\text{x}^2}\Big[\frac{\text{x}^2}{\text{x}-3}+2\text{x}\log(\text{x}-3)\Big]$

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