If y = frac{1}{4}u^4 and u = frac{2}{3}x^3 + | vedclass

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(C) Given $y = \frac{1}{4}u^4$ and $u = \frac{2}{3}x^3 + 5$. Using the chain rule,$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. First,different

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$\frac{2}{27}x^2(2x^3 + 15)^3$

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(C) Given $y = \frac{1}{4}u^4$ and $u = \frac{2}{3}x^3 + 5$. Using the chain rule,$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. First,differentiate $y$ with respect to $u$: $\frac{dy}{du} = \frac{1}{4} \cdot 4u^3 = u^3$. Next,differentiate $u$ with respect to $x$: $\frac{du}{dx} = \frac{2}{3} \cdot 3x^2 = 2x^2$. Now,substitute these into the chain rule formula: $\frac{dy}{dx} = u^3 \cdot 2x^2$. Substitute $u = \frac{2x^3 + 15}{3}$ into the expression: $\frac{dy}{dx} = \left( \frac{2x^3 + 15}{3} \right)^3 \cdot 2x^2$. $\frac{dy}{dx} = \frac{(2x^3 + 15)^3}{27} \cdot 2x^2 = \frac{2}{27}x^2(2x^3 + 15)^3$.

If $y = \frac{1}{4}u^4$ and $u = \frac{2}{3}x^3 + 5$,then $\frac{dy}{dx} = $

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