If a spherical balloon has a variable diamete | vedclass

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(C) The diameter of the sphere is $d = 3x + \frac{9}{2} = \frac{3}{2}(2x + 3)$.The radius $r$ is $\frac{d}{2} = \frac{3}{4}(2x + 3)$.The volume $V$ of

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$\frac{27\pi}{8}(2x + 3)^2$

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(C) The diameter of the sphere is $d = 3x + \frac{9}{2} = \frac{3}{2}(2x + 3)$.The radius $r$ is $\frac{d}{2} = \frac{3}{4}(2x + 3)$.The volume $V$ of a sphere is $V = \frac{4}{3}\pi r^3$.Substituting $r$, we get $V = \frac{4}{3}\pi \left[ \frac{3}{4}(2x + 3) \right]^3 = \frac{4}{3}\pi \cdot \frac{27}{64}(2x + 3)^3 = \frac{9\pi}{16}(2x + 3)^3$.Differentiating $V$ with respect to $x$ using the chain rule:$\frac{dV}{dx} = \frac{9\pi}{16} \cdot 3(2x + 3)^2 \cdot \frac{d}{dx}(2x + 3) = \frac{27\pi}{16}(2x + 3)^2 \cdot 2 = \frac{27\pi}{8}(2x + 3)^2$.

If a spherical balloon has a variable diameter $d = 3x + \frac{9}{2}$ units, then the rate of change of its volume $V$ with respect to $x$ is:

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