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

Answer

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

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