If y = (x cot^3 x)^{3/2},then frac{dy}{dx} = | vedclass

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(A) Given $y = (x \cot^3 x)^{3/2}$.Applying the chain rule,$\frac{dy}{dx} = \frac{3}{2}(x \cot^3 x)^{1/2} \cdot \frac{d}{dx}(x \cot^3 x)$.Using the pr

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$\frac{3}{2}(x \cot^3 x)^{1/2} [\cot^3 x - 3x \cot^2 x \csc^2 x]$

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(A) Given $y = (x \cot^3 x)^{3/2}$.Applying the chain rule,$\frac{dy}{dx} = \frac{3}{2}(x \cot^3 x)^{1/2} \cdot \frac{d}{dx}(x \cot^3 x)$.Using the product rule for the derivative of $(x \cot^3 x)$:$\frac{d}{dx}(x \cot^3 x) = \frac{d}{dx}(x) \cdot \cot^3 x + x \cdot \frac{d}{dx}(\cot^3 x)$.$= 1 \cdot \cot^3 x + x \cdot (3 \cot^2 x \cdot \frac{d}{dx}(\cot x))$.$= \cot^3 x + x \cdot (3 \cot^2 x \cdot (-\csc^2 x))$.$= \cot^3 x - 3x \cot^2 x \csc^2 x$.Therefore,$\frac{dy}{dx} = \frac{3}{2}(x \cot^3 x)^{1/2} [\cot^3 x - 3x \cot^2 x \csc^2 x]$.

If $y = (x \cot^3 x)^{3/2}$,then $\frac{dy}{dx} = $

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