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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}^2}}\Big)$
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Answer

Let $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}^2}}\Big)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}^2}}\Big)\Big\}$
$=\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)$
[Using chain rule and quotient rule]
$=\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)^2}}\times\Bigg[\frac{(\text{x}^2+\text{a}^2)^\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x})-\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{a}^2)^\frac{1}{2}}{\Big[(\text{x}^2+\text{a}^2)^\frac{1}{2}\Big]^2}\Bigg]$
$=\frac{\sqrt{\text{x}^2+\text{a}^2}}{\sqrt{\text{x}^2+\text{a}^2-\text{x}^2}}\times\Bigg[\frac{\sqrt{\text{x}^2+\text{a}^2}-\frac{\text{x}}{2\sqrt{\text{x}^2+\text{a}^2}}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{a}^2)}{(\text{x}^2+\text{a}^2)}\Bigg]$
$=\frac{\sqrt{\text{x}^2+\text{a}^2}}{\text{a}(\text{x}^2+\text{a}^2)}\times\Big[\sqrt{\text{x}^2+\text{a}^2}-\frac{\text{x}}{2\sqrt{\text{x}^2+\text{a}^2}}\times2\text{x}\Big]$
$=\frac{\sqrt{\text{x}^2+\text{a}^2}}{\text{a}(\text{x}^2+\text{a}^2)}\times\Big[\frac{\text{x}^2+\text{a}^2-\text{x}^2}{\sqrt{\text{x}^2+\text{a}^2}}\Big]$
$=\frac{\text{a}^2}{\text{a}(\text{x}^2+\text{a}^2)}$
$=\frac{\text{a}}{(\text{x}^2+\text{a}^2)}$
So,
$\frac{\text{d}}{\text{dx}}\Big\{\sin^{-1}\Big(\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}\Big)\Big\}=\frac{\text{a}^2}{\text{a}(\text{x}^2+\text{a}^2)}$

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