The differential coefficient of {cos ^{ - 1}} | vedclass

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Question

(C) Let $y = {\cos ^{ - 1}}(\sqrt x )$ and $z = \sqrt {1 - x} $. First,differentiate $y$ with respect to $x$: $\frac{dy}{dx} = -\frac{1}{\sqrt{1 - (\s

Vedclass · Accepted Answer

$\frac{1}{\sqrt x }$

Vedclass · Answer

(C) Let $y = {\cos ^{ - 1}}(\sqrt x )$ and $z = \sqrt {1 - x} $. First,differentiate $y$ with respect to $x$: $\frac{dy}{dx} = -\frac{1}{\sqrt{1 - (\sqrt{x})^2}} 	imes \frac{d}{dx}(\sqrt{x}) = -\frac{1}{\sqrt{1 - x}} 	imes \frac{1}{2\sqrt{x}}$. Next,differentiate $z$ with respect to $x$: $\frac{dz}{dx} = \frac{1}{2\sqrt{1 - x}} 	imes \frac{d}{dx}(1 - x) = \frac{1}{2\sqrt{1 - x}} 	imes (-1) = -\frac{1}{2\sqrt{1 - x}}$. Now,find the differential coefficient of $y$ with respect to $z$: $\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{-\frac{1}{\sqrt{1 - x}} 	imes \frac{1}{2\sqrt{x}}}{-\frac{1}{2\sqrt{1 - x}}} = \frac{1}{\sqrt{x}}$. Thus,the correct option is $C$.

The differential coefficient of ${\cos ^{ - 1}}(\sqrt x )$ with respect to $\sqrt {1 - x} $ is

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