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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\sqrt{1+\text{a}^2\text{x}^2-1}}{\text{ax}}\Big),\text{x}\neq0$
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Answer

Let $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{a}^2\text{x}^2}-1}{\text{ax}}\Big)$
Put $\text{ax}=\tan\theta$
$\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{a}^2\text{x}^2}-1}{\text{ax}}\Big)$
$=\tan^{-1}\Big(\frac{\sec\theta-1}{\tan\theta}\Big)$
$=\tan^{-1}\Big(\frac{1-\cos\theta}{\sin\theta}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{2\sin^2\theta}{2}}{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}\bigg)$
$\text{y}=\tan^{-1}\Big(\frac{\tan\theta}{2}\Big)$
$=\frac{\theta}{2}$
$\text{y}=\frac{1}{2}\tan^{-1}(\text{ax})$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\Big(\frac{1}{1+(\text{ax})^2}\Big)\frac{\text{d}}{\text{dx}}(\text{ax})$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2(1+\text{a}^2\text{x}^2)}(\text{a})$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}}{2(1+\text{a}^2\text{x}^2)}$

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