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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big), 0<\text{x}<\infty$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
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Answer

Let $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{\sqrt{1+\text{x}^2}}\Big)+\cos^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big)$
Put $\text{x}=\tan\theta$
$\therefore\text{y}=\sin^{-1}\Big(\frac{\tan\theta}{\sqrt{1+\tan^2\theta}}\Big)+\cos^{-1}\Big(\frac{1}{\sqrt{1+\tan^2\theta}}\Big)$
$\Rightarrow \text{y}=\sin^{-1}\bigg(\frac{\frac{\sin\theta}{\cos\theta}}{\sec\theta}\bigg)+\cos^{-1}\Big(\frac{1}{\sec\theta}\Big)$
$\Rightarrow\text{y}=\sin^{-1}\bigg(\frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}}\bigg)+\cos^{-1}(\cos\theta)$
$\Rightarrow\text{y}=\sin^{-1}(\sin\theta)+\cos^{-1}(\cos\theta)\ .....(\text{i})$
Here, $0<\text{x}<\infty$
$\Rightarrow 0<\tan\theta<\infty$
$\Rightarrow 0 <\theta<\frac{\pi}{2}$
So, from equation (i),
$\text{y}=\theta+\theta$
$\begin{bmatrix} \text{Since}, \sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in \Big[-\frac{\pi}{2},\frac{\pi}{2}\Big] \\ \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi] \end{bmatrix}$
$\Rightarrow\text{y}=2\theta$
$\Rightarrow\text{y}=2\tan^{-1}\text{x}\ \big[\text{Since, x}=\tan\theta\big]$
Differentiate it with respect to x,
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2}{1+\text{x}^2}$

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