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

Let $\text{y}=\cos^{-1}\Big(\frac{1-\text{x}^{2\text{n}}}{1+\text{x}^{2\text{n}}}\Big)$
Put $\text{x}=\tan\theta,\text{ so}$
$\text{y}=\cos^{-1}\bigg(\frac{1-(\text{x}^{\text{n}})^2}{1+(\text{x}^{\text{n}})^2}\bigg)$
$=\cos^{-1}\Big(\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big)$
$\text{y}=\cos^{-1}(\cos2\theta)\ .....(\text{i})$
Here, $0<\text{x}<\infty$
$\Rightarrow 0<\text{x}^{\text{n}}<\infty$
$\Rightarrow 0<\theta<\frac{\pi}{2}$
$\Rightarrow 0<(2\theta)<\pi$
So, from equation (i),
$\text{y}=2\theta\big[\text{Since}, \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi]\big]$
$\text{y}=2\tan^{-1}\big(\text{x}^{\text{n}}\big)$
Differantiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=2\Big(\frac{1}{1+(\text{x}^{\text{n}})^2}\Big)\frac{\text{d}}{\text{dx}}(\text{x}^{\text{n}})$
$=\frac{2}{1+\text{x}^{2\text{n}}}\times(\text{nx}^{\text{n}-1})$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{nx}^{\text{n}-1}}{1+\text{x}^{2\text{n}}}$

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