Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferentiation5 Marks
Question
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cos^{-1}\text{x},$ if
$\text{x}\in(-1,0)$
✓
Answer
Let $\text{u}=\sin^{-1}\sqrt{1-\text{x}^2}$
Put $\text{x}=\cos\theta, \text{So},$
$\Rightarrow\text{u}=\sin^{-1}\sqrt{1-\cos^2\theta}$
$\Rightarrow\text{u}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
And, $\text{v}=\cos^{-1}\text{x}\ .....(\text{ii})$
Here,
$\text{x}\in(-1,0)$
$\Rightarrow\cos\theta\in(-1, 0)$
$\Rightarrow\theta\in\Big(\frac{\pi}{2},\pi\Big)$
So, from equation (i),
$\text{u}=\pi-\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\pi-\theta,\theta\in\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big)\Big]$
$\text{u}=\pi-\cos^{-1}\text{x }\big[\text{Since}\text{x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=0-\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{du}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
And, 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)
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{1}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{-1}$
$\frac{\text{du}}{\text{dv}}=-1$
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