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

Answer

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

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