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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Differentiate the following w.r.t. x:
$\tan^{-1}\Big(\frac{3\text{a}^2\text{x}-\text{x}^3}{\text{a}^3-3\text{ax}^2}\Big),\frac{-1}{\sqrt{3}}<\frac{\text{x}}{\text{a}}<\frac{1}{\sqrt{3}}$
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Answer

Let $\text{y}=\tan^{-1}\Big(\frac{3\text{a}^2\text{x}-\text{x}^3}{\text{a}^3-3\text{ax}^2}\Big)$
Put $\text{x}=\text{a}\tan\theta\Rightarrow\ \theta=\tan^{-1}\frac{\text{x}}{\text{a}}$
$\therefore\ \text{y}=\tan^{-1}\bigg[\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\bigg]$ $\bigg[\because\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\bigg]$
$=\tan^{-1}(\tan3\theta)=3\theta$
$=3\tan^{-1}\frac{\text{x}}{\text{a}}\Big[\because\theta=\tan^{-1}\frac{\text{x}}{\text{a}}\Big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=3\cdot\frac{\text{d}}{\text{dx}}\tan^{-1}\frac{\text{x}}{\text{a}}$ $=3\cdot\Bigg[\frac{1}{1+\frac{\text{x}^2}{\text{a}^2}}\Bigg]\cdot\frac{\text{d}}{\text{dx}}\cdot\Big(\frac{\text{x}}{\text{a}}\Big)$
$=3\cdot\frac{\text{a}^2}{\text{a}^2+\text{x}^2}.=\frac{1}{\text{a}}=\frac{3\text{a}}{\text{a}^2+\text{x}^2}$

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