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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate w.r.t. x the function in Exercise:
$\text{x}^{\text{x}^2-3}+(\text{x}-3)^{\text{x}^2},$  for x > 3
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Answer

Let 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
$\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ \ \dots(1)$
$\text{u}=\text{x}^{\text{x}^2-3}$
$\therefore\ \log\text{u}=\log(\text{x}^{\text{x}^2-3})$
$\log\text{u}=(\text{x}^2-3)\log\text{x}$
Differentiating with respect to x, we obtain
$\frac{1}{\text{u}}\cdot\frac{\text{du}}{\text{dx}}=\log\text{x}.\frac{\text{d}}{\text{dx}}(\text{x}^2-3)+(\text{x}^2+3)\cdot\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log\text{x}\cdot2\text{x}+(\text{x}^2-3)\cdot\frac{1}{\text{x}}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}^2-3}\cdot\Bigg[\frac{\text{x}^2-3}{\text{x}}+2\text{x}\log\text{x}\Bigg]$
Also,
$\text{v}=(\text{x}-3)^{\text{x}^2}$
$\therefore\ \log\text{v}=\log(\text{x}-3)^{\text{x}^2}$
$\Rightarrow\ \log\text{v}=\text{x}^2\log(\text{x}-3)$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\cdot\frac{\text{dv}}{\text{dx}}=\log(\text{x}-3)\cdot\frac{\text{d}}{\text{dx}}(\text{x}^2)+\text{x}^2\cdot\frac{\text{d}}{\text{dx}}\big[\log(\text{x}-3)\big]$
$\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\log(\text{x}-3)\cdot2\text{x}+\text{x}^2\cdot\frac{1}{\text{x}-3}\cdot\frac{\text{d}}{\text{dx}}(\text{x}-3)$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{v}\Bigg[2\text{x}\log(\text{x}-3)+\frac{\text{x}^2}{\text{x}-3}\cdot1\Bigg]$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=(\text{x}-3)^{\text{x}^2}\Bigg[\frac{\text{x}^2}{\text{x}-3}+2\text{x}\log(\text{x}-3)\Bigg]$
Substituting the expressions of $\frac{\text{du}}{\text{dx}}\text{ and }\frac{\text{dv}}{\text{dx}}$ in equation (1), we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}^2-3}\Bigg[\frac{\text{x}^2-3}{\text{x}}+2\text{x}\text{logx}\Bigg]+(\text{x}-3)^\text{x} \Bigg[\frac{\text{x}^{\text{x}}}{\text{x}-3}+2\text{x}\log(\text{x}-3)\Bigg]$

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