If y = frac{5x}{sqrt[3]{(1 - x)^2}} + cos^2(2 | vedclass

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(A) Given $y = 5x(1 - x)^{-2/3} + \cos^2(2x + 1)$.Applying the product rule and chain rule:$\frac{dy}{dx} = 5 \left[ (1 - x)^{-2/3} + x \cdot \left( -

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$\frac{5(3 - x)}{3(1 - x)^{5/3}} - 2\sin(4x + 2)$

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(A) Given $y = 5x(1 - x)^{-2/3} + \cos^2(2x + 1)$.Applying the product rule and chain rule:$\frac{dy}{dx} = 5 \left[ (1 - x)^{-2/3} + x \cdot \left( -\frac{2}{3} ight)(1 - x)^{-5/3}(-1) ight] + 2\cos(2x + 1)(-\sin(2x + 1))(2)$.$\frac{dy}{dx} = 5(1 - x)^{-2/3} + \frac{10x}{3(1 - x)^{5/3}} - 4\cos(2x + 1)\sin(2x + 1)$.Using $2\sin	heta\cos	heta = \sin(2	heta)$,we have $4\cos(2x + 1)\sin(2x + 1) = 2\sin(4x + 2)$.$\frac{dy}{dx} = \frac{5(1 - x) + \frac{10x}{3}}{(1 - x)^{5/3}} - 2\sin(4x + 2) = \frac{5 - 5x + \frac{10x}{3}}{(1 - x)^{5/3}} - 2\sin(4x + 2)$.$\frac{dy}{dx} = \frac{5(3 - 3x + 2x)}{3(1 - x)^{5/3}} - 2\sin(4x + 2) = \frac{5(3 - x)}{3(1 - x)^{5/3}} - 2\sin(4x + 2)$.

If $y = \frac{5x}{\sqrt[3]{(1 - x)^2}} + \cos^2(2x + 1)$,then $\frac{dy}{dx} = $

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