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Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{a}+\text{b}\tan\text{x}}{\text{b}-\text{a}\tan\text{x}}\Big)$
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Answer

Let, $\text{y}=\tan^{-1}\Big[\frac{\text{a}+\text{b}\tan\text{x}}{\text{b}-\text{a}\tan\text{x}}\Big]$
$\Rightarrow\text{y}=\tan^{-1}\bigg[\frac{\frac{\text{a}+\text{b}\tan\text{x}}{\text{b}}}{\frac{\text{b}-\text{a}\tan\text{x}}{\text{b}}}\bigg]$
$\Rightarrow\text{y}=\tan^{-1}\bigg[\frac{\frac{\text{a}}{\text{b}}+\tan\text{x}}{1-\frac{\text{a}}{\text{b}}\tan\text{x}}\bigg]$
$\Rightarrow\text{y}=\tan^{-1}\Bigg[\frac{\tan\Big(\tan^{-1}\frac{\text{a}}{\text{b}}\Big)+\tan\text{x}}{1-\tan\Big(\tan^{-1}\frac{\text{a}}{\text{b}}\Big)\times\tan\text{x}}\Bigg]$
$\Rightarrow\text{y}=\tan^{-1}\Big[\tan\Big(\tan^{-1}\frac{\text{a}}{\text{b}}+\text{x}\Big)\Big]$
$\Rightarrow\text{y}=\tan^{-1}\big(\frac{\text{a}}{\text{b}}\big)+\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+1$
$\therefore\ \frac{\text{dy}}{\text{dx}}=1$

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