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

Let $\text{y}=\tan^{-1}\Big(\frac{2\text{a}^{\text{x}}}{1-\text{a}^{2\text{x}}}\Big)$
Put $\text{a}^{\text{x}}=\tan\theta$
$\Rightarrow\text{y}=\tan^{-1}\Big\{\frac{2\times\text{a}^\text{x}}{1-(\text{a}^{\text{x}})^2}\Big\}$
$\Rightarrow\text{y}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)$
$\Rightarrow\text{y}=\tan^{-1}(\tan2\theta)\ .....(\text{i})$
Here, $-\infty<\text{x}<0$
$\Rightarrow\text{a}^{-\infty}<\text{a}^{\text{x}}<2^{0}$
$\Rightarrow 0<\tan\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{4}$
$\Rightarrow 0<2\theta<\frac{\pi}{2}$
So, from equation (i),
$\text{y}=2\theta\Big[\text{Since},\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\big(-\frac{\pi}{2},\frac{\pi}{2}\big)\Big]$
$\Rightarrow\text{y}=2\tan^{-1}(\text{a}^{\text{x}})$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{2}{1+(\text{a}^{\text{x}})^2}\frac{\text{d}}{\text{dx}}(\text{a}^{\text{x}})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\times\text{a}^{\text{x}}\log_\text{e}\text{a}}{1+\text{a}^{2\text{x}}}$
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{2\text{a}^{\text{x}}\log_\text{e}\text{a}}{1+\text{a}^{2\text{x}}}$

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