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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate $\tan^{-1}\bigg(\frac{\sqrt{1 - \text{x}^{2}}}{\text{x}}\bigg)$with respect to $\cos^{-1}(2 \text{x}\sqrt{1- \text{x}^{2}}),$ when $\text{x}\neq0.$
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Answer

let $\text{u} = \tan^{-1}\bigg(\frac{\sqrt{1-\text{x}^{2}}}{\text{x}}\bigg),\text{v} = \cos^{-1}\bigg(2 \text{x}\sqrt{1- \text{x}^{2}}),\text{x} = \cos\theta\therefore\theta =\cos^{-1}\text{x}$
$\therefore\text{u} = \tan^{-1}\bigg(\frac{\sqrt{1-\cos^{2}\theta}}{\cos\theta}\bigg) = \tan^{-1}(\tan\theta) = \theta = \cos^{-1}\text{x}$
and $\text{v} = \cos^{-1}( 2\cos\theta\sqrt{1 -\cos^{2}\theta}) = \cos^{-1}(\sin2\theta) = \cos^{-1}\bigg(\cos\bigg(\frac{\pi}{2} - 2 \theta\bigg)\bigg)$
$ = \frac{\pi}{2} - 2\theta =\frac{\pi}{2} - 2\cos^{-1}\text{x}$
$\frac{\text{du}}{\text{dx}} = \frac{-1}{\sqrt{1-\text{x}^{2}}},\frac{\text{dv}}{\text{dx}} = \frac{2}{\sqrt{1-\text{x}^{2}}}$
$\therefore\frac{\text{du}}{\text{dv}} =\frac{-1}{\sqrt{1- \text{x}^{2}}}\times\frac{\sqrt{1-\text{x}^{2}}}{2} =\frac{-1}{2}$
( In case, If x = sin θ then answer is $\frac{1}{2}).$

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