    Question

# Differential coefficient of ${\mathrm{sin}}^{-1}\left(x\right)$ w.r.t.${\mathrm{cos}}^{-1}\left(\sqrt{\left(1-{x}^{2}\right)}\right)$ is

A

$1$

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B

$2$

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C

$\frac{1}{\left(1+{x}^{2}\right)}$

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D

none of these

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Solution

## The correct option is A $1$Explanation for correct option:Step-1: Differentiate the ${\mathrm{sin}}^{-1}\left(x\right)$ with respect to $x$.Let the given, $u={\mathrm{sin}}^{-1}\left(x\right)$$⇒\frac{du}{dx}=\frac{1}{\sqrt{1-{x}^{2}}}..........\left(i\right)$Step-2: Differentiate the ${\mathrm{cos}}^{-1}\left(\sqrt{\left(1-{x}^{2}\right)}\right)$ with respect to $x$.Let the given, $v={\mathrm{cos}}^{-1}\left(\sqrt{\left(1-{x}^{2}\right)}\right)$put, $x=\mathrm{sin}\left(\theta \right)\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{sin}}^{-1}\left(x\right)$$⇒v={\mathrm{cos}}^{-1}\left(\sqrt{\left(1-{\left(\mathrm{sin}\left(\theta \right)\right)}^{2}\right)}\right)\phantom{\rule{0ex}{0ex}}⇒v={\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(\theta \right)\right)\phantom{\rule{0ex}{0ex}}⇒v=\theta \phantom{\rule{0ex}{0ex}}⇒v={\mathrm{sin}}^{-1}\left(x\right)$Differentiate with respect to $x$.$⇒\frac{dv}{dx}=\frac{1}{\sqrt{1-{\left(x\right)}^{2}}}...........\left(ii\right)$Step-3: Divide equation $\left(i\right)$ by equation $\left(ii\right)$.$⇒\frac{\left(\frac{du}{dx}\right)}{\left(\frac{dv}{dx}\right)}=\frac{\frac{1}{\sqrt{\left(1-{x}^{2}\right)}}}{\frac{1}{\sqrt{\left(1-{x}^{2}\right)}}}\phantom{\rule{0ex}{0ex}}⇒\frac{du}{dv}=1$Hence, correct answer is option $\left(A\right)$.  Suggest Corrections  4      Similar questions
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