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

Answer

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

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