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(B) Given the diameter $ d = \frac{3}{2}(2x + 3) $.The radius $ r = \frac{d}{2} = \frac{3}{4}(2x + 3) $.Differentiating $ r $ with respect to $ x $,we

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

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

$A$ balloon,which always remains spherical,has a variable diameter of $ \frac{3}{2}(2x + 3) $. Find the rate of change of its volume with respect to $ x $.

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