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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cot^2\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big),$ if 0 < x < 1.
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Answer

Let $\text{u}=\sin^{-1}\Big(\sqrt{1-\text{x}^2}\Big)$
Put $\text{x}=\cos\theta\Rightarrow\theta=\cos^{-1}\text{x},\text{ so}$
$\text{u}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
And,
Let $\text{v}=\cot^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
$=\cot^{-1}\Big(\frac{\cot\theta}{\sqrt{1-\cos^2\theta}}\Big)$
$=\cot^{-1}\Big(\frac{\cos\theta}{\sin\theta}\Big)$
$\text{v}=\cot^{-1}(\cot\theta)\ .....(\text{ii})$
Here, $0<\text{x}<1$
$\Rightarrow0<\cos\theta<1$
$\Rightarrow0<\theta<\frac{\pi}{2}$
So, from equation (i),
$\text{u}=\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{u}=\cos^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta\big[\text{Since,}\cot^{-1}(\cot\theta)=\theta,\text{if }\theta\in(0,\pi) \big]$
$\text{v}=\cos^{-1}\text{x}$
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),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-1}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{-1}$
$\frac{\text{du}}{\text{dv}}=1$

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