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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cos^{-1}\text{x},$ if
$\text{x}\in(0, 1)$
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Answer

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

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