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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:
$\tan^{-1}\Big(\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big)$
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Answer

Let $\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\text{x}-\text{a}}{\text{x}}}{\frac{\text{x}+\text{a}}{\text{x}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{x}}{\text{x}}-\frac{\text{x}}{\text{x}}}{\frac{\text{x}}{\text{x}}+\frac{\text{a}}{\text{x}}}\bigg)$
$=\tan^{-1}\bigg(\frac{1-\frac{\text{x}}{\text{x}}}{1+1\times\frac{\text{a}}{\text{x}}}\bigg)$
$\text{y}=\tan^{-1}(1)-\tan^{-1}\big(\frac{\text{a}}{\text{x}}\big)$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=0-\frac{1}{1+\big(\frac{\text{a}}{\text{x}}\big)^2}\frac{\text{d}}{\text{dx}}\big(\frac{\text{a}}{\text{x}}\big)$
$=-\frac{\text{x}^2}{\text{x}^2+\text{a}^2}\Big(\frac{-\text{a}}{\text{x}^2}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}}{\text{a}^2+\text{x}^2}$

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