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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
Differentiate $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ with respect to $\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big),$ if $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
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Answer

Let, $\text{u}=\sin^{-1}(2\text{x}\sqrt{1-\text{x}^2})$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\sin\theta\sqrt{1-\sin^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\sin\theta\cos\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let $\text{v}=\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}}}\Big)$
$\Rightarrow\text{v}=\tan^{-1}\Big(\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}\Big)$
$\Rightarrow\text{v}=\tan^{-1}\Big(\frac{\sin\theta}{\cos\theta}\Big)$
$\Rightarrow\text{v}=\tan^{-1}(\tan\theta)\ .....(\text{ii})$
Here, $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}= 2\theta\bigg[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]\bigg]$
$\Rightarrow\text{u}=2\sin^{-1}\text{x}[\text{Since, x}=\sin\theta]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta\bigg[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\bigg]$
$\Rightarrow\text{v}=\sin^{-1}\text{x}[\text{since, x}=\sin\theta]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....\text{(iv)}$
Dividing equation (iii) by (iv)
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{1}$
$\therefore\frac{\text{du}}{\text{dv}}=2$

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