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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:
$\tan^{-1}\Big(\frac{2^{\text{x}+1}}{1-4^{\text{x}}}\Big),-\infty<\text{x}<0$
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Answer

Let $\text{y}=\tan^{-1}\Big\{\frac{2^{\text{x}+1}}{1-4^{\text{x}}}\Big\}$
Put $2\text{x}=\tan\theta,\text{ so}$
$\text{y}=\tan^{-1}\Big\{\frac{2^{\text{x}}\times2}{1-(2^\text{x})^2}\Big\}$
$=\tan^{-1}\Big\{\frac{2\tan\theta}{1-\tan^2\theta}\Big\}$
$\text{y}=\tan^{-1}\big\{\tan(2\theta)\big\}\ .....(\text{i})$
Here, $-\infty<\text{x}<0$
$\Rightarrow 2^{-\infty}<2^{\text{x}}<2^{0}$
$\Rightarrow 0<2^{\text{x}}<1$
$\Rightarrow 0< \theta< \frac{\pi}{4}$
$\Rightarrow 0 < (2\theta) <\frac{\pi}{2}$
From equation (i),
$\text{y}=2\theta \Big[\text{Since}, \tan^{-1}(\tan\theta)=\theta, \text{ if }\theta\in \Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\text{y}=2\tan^{-1}(2^\text{x})$
Differentiate it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{2}{1+(2^\text{x})^2}\frac{\text{d}}{\text{dx}}(2^\text{x})$
$=\frac{2\times2^\text{x}\log2}{1+4^{\text{x}}}$
$\frac{\text{dy}}{\text{dx}}=\frac{2^{\text{x}+1}\log2}{1+4^{\text{x}}}$

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