frac{d}{d x}left(cos ^{-1}left(frac{4 x^3}{27 | vedclass

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Question

(C) Let $y = \cos ^{-1}\left(\frac{4 x^3}{27}-x\right)$.We can rewrite the expression as $y = \cos ^{-1}\left(4\left(\frac{x}{3}\right)^3 - 3\left(\fr

Vedclass · Accepted Answer

$\frac{-3}{\sqrt{9-x^2}}$

Vedclass · Answer

(C) Let $y = \cos ^{-1}\left(\frac{4 x^3}{27}-xight)$.We can rewrite the expression as $y = \cos ^{-1}\left(4\left(\frac{x}{3}ight)^3 - 3\left(\frac{x}{3}ight)ight)$.Let $\frac{x}{3} = \cos A$,then $A = \cos ^{-1}\left(\frac{x}{3}ight)$.Substituting this into the expression,we get $y = \cos ^{-1}(4 \cos ^3 A - 3 \cos A)$.Using the trigonometric identity $\cos(3A) = 4 \cos ^3 A - 3 \cos A$,we have $y = \cos ^{-1}(\cos 3A) = 3A$.Thus,$y = 3 \cos ^{-1}\left(\frac{x}{3}ight)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 3 	imes \left(-\frac{1}{\sqrt{1 - (x/3)^2}}ight) 	imes \frac{d}{dx}\left(\frac{x}{3}ight)$.$\frac{dy}{dx} = 3 	imes \left(-\frac{1}{\sqrt{(9-x^2)/9}}ight) 	imes \frac{1}{3} = 3 	imes \left(-\frac{3}{\sqrt{9-x^2}}ight) 	imes \frac{1}{3} = -\frac{3}{\sqrt{9-x^2}}$.

$\frac{d}{d x}\left(\cos ^{-1}\left(\frac{4 x^3}{27}-x\right)\right)=$

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