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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following functions with respect to x:
$\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
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Answer

Let $\text{y}=\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dy}}\Big(\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Big)$
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
[Using chain rule]
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Bigg[\frac{(1-\text{x}^2)\frac{\text{d}}{\text{dx}}(1+\text{x}^2)-(1+\text{x}^2)\frac{\text{d}}{\text{dx}}(1-\text{x}^2)}{(1-\text{x})^2}\Bigg]$
[Using chain rule]
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\bigg[\frac{(1-\text{x}^2)(2\text{x})-(1+\text{x}^2)(-2\text{x})}{(1-\text{x}^2)^2}\bigg]$
$=\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Big[\frac{2\text{x}-2\text{x}^3+2\text{x}+2\text{x}^3}{(1-\text{x}^2)^2}\Big]$
$=\frac{4\text{x}}{\big(1-\text{x}^2\big)^2}\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
So,
$\frac{\text{d}}{\text{dx}}\Big(\sin\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)\Big)=\frac{4\text{x}}{\big(1-\text{x}^2\big)^2}\cos\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$

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