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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:
$\tan^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big\},-\text{a}<\text{x}<\text{a}$
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Answer

Let $\text{y}=\tan^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big\}$ Put $\text{x}=\text{a}\sin\theta$ $\text{y}=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\sqrt{\text{a}^2-\text{a}^2\sin^2\theta}}\Big\}$ $\text{y}=\tan^{-1}\bigg\{\frac{\text{a}\sin\theta}{\sqrt{\text{a}^2(1-\sin^2\theta)}}\bigg\}$ $\text{y}=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}\cos\theta}\Big\}$ $\text{y}=\tan^{-1}(\tan\theta)\ .....(\text{i})$ Here, $-\text{a}<\text{x}<\text{a}$ $\Rightarrow-1<\frac{\text{x}}{\text{a}}<1$ $\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$ From equation (i), $\text{y}=\theta$ $\Big[\text{Since},\tan^{-1}(\tan\theta)=\theta,\text{ if }\theta \in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$ $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)\big[\text{Since x}=\text{a }\sin \theta\big]$ Differentiating it with respect to x, $\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\big(\frac{\text{x}}{\text{a}}\big)^2}}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{a}}\Big)$ $=\frac{\text{a}}{\sqrt{\text{a}^2-\text{x}^2}}\times\big(\frac{1}{\text{a}}\big)$$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}$

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