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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:
$\sin^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big\}$
✓

Answer

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

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