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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$\tan^{-1}\Big\{\frac{\text{x}}{\text{a}+\sqrt{\text{a}^2-\text{x}^2}}\Big\},-\text{a}<\text{x}<\text{a}$
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Answer

Let $\tan^{-1}\Big\{\frac{\text{x}}{\text{a}+\sqrt{\text{a}^2-\text{x}^2}}\Big\}$
Put $\text{x}=\text{a}\sin\theta,\text{ So}$
$\text{y}=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}+\sqrt{\text{a}^2+\text{a}^2\sin^2\theta}}\Big\}$
$\tan^{-1}\bigg\{\frac{\text{a}\sin\theta}{\text{a}+\sqrt{\text{a}^2(1-\sin^2\theta)}}\bigg\}$
$=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}+\text{a}\cos\theta}\Big\}$
$=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}(1+\cos\theta)}\Big\}$
$=\tan^{-1}\Big(\frac{\sin\theta}{1+\cos\theta}\Big)$
$=\tan^{-1}\bigg(\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\times\frac{\theta}{2}}\bigg)$
$\text{y}=\tan^{-1}\Big(\tan\frac{\theta}{2}\Big)\ .....\text{(i)}$
Here, $-\text{a}<\text{x}<\text{a}$
$\Rightarrow -1<\frac{\text{x}}{\text{a}}<1$
$\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$
$\Rightarrow-\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4}$
So, from equation (i),
$\text{y}=\frac{\theta}{2}\ \Big[\text{Since}, \tan^{-1}(\tan\theta)=\theta,\text{ if }\theta\in \Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{1}{2}+\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big) \big[\text{Since, x}=\text{a}\sin\theta\big]$
Differentiating it with respect to x using chian rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\frac{1}{\sqrt{1-\big(\frac{\text{x}}{\text{a}}\big)^2}}\frac{\text{d}}{\text{dx}}\big(\frac{\text{x}}{\text{a}}\big)$
$=\frac{\text{a}}{2\sqrt{\text{a}^2-\text{x}^2}}\Big(\frac{1}{\text{a}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}$

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