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

Let $\text{y}=\tan^{-1}\Big(\frac{\text{a}+\text{bx}}{\text{b}-\text{ax}}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}+\text{bx}}{\text{b}}}{\frac{\text{b}-\text{ax}}{\text{b}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}}{\text{b}}+\frac{\text{bx}}{\text{b}}}{\frac{\text{b}}{\text{a}}-\frac{\text{ax}}{\text{b}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{a}}{\text{b}}+\text{x}}{1-\big(\frac{\text{a}}{\text{b}}\big)\text{x}}\bigg)$
$\text{y}=\tan^{-1}\big(\frac{\text{a}}{\text{b}}\big)+\tan^{-1}\text{x}$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+\frac{1}{1+\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$

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