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

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Question

(A) Let $y = \cos^{-1} \left( \sqrt{\frac{1+x}{2}} ight)$.Substitute $x = \cos(2	heta)$,which implies $	heta = \frac{1}{2} \cos^{-1}(x)$.Then,$\sq

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$-\frac{1}{2\sqrt{1-x^2}}$

Vedclass · Answer

(A) Let $y = \cos^{-1} \left( \sqrt{\frac{1+x}{2}} ight)$.Substitute $x = \cos(2	heta)$,which implies $	heta = \frac{1}{2} \cos^{-1}(x)$.Then,$\sqrt{\frac{1+x}{2}} = \sqrt{\frac{1+\cos(2	heta)}{2}} = \sqrt{\frac{2\cos^2(	heta)}{2}} = \cos(	heta)$.Thus,$y = \cos^{-1}(\cos(	heta)) = 	heta$.Substituting back for $	heta$,we get $y = \frac{1}{2} \cos^{-1}(x)$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{2} \left( -\frac{1}{\sqrt{1-x^2}} ight) = -\frac{1}{2\sqrt{1-x^2}}$.

The differential coefficient of $\cos^{-1} \left( \sqrt{\frac{1+x}{2}} \right)$ with respect to $x$ is

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