QPaperGen

A balloon, which always remains spherical, has a variable diameter given by
\[d = \dfrac{3}{2}(2x + 1).\]
We are to find the rate of change of its volume with respect to \(x\).

The volume of a sphere is given by:
\[V = \dfrac{4}{3}\pi r^3\]

Since the diameter is \(d = \dfrac{3}{2}(2x + 1)\), the radius is:

\[r = \dfrac{d}{2} = \dfrac{3}{4}(2x + 1)\]

Substitute into the volume formula:

\[V = \dfrac{4}{3}\pi \left( \dfrac{3}{4}(2x + 1) \right)^3 = \dfrac{4}{3}\pi \cdot \dfrac{27}{64} (2x + 1)^3 = \dfrac{9}{16}\pi (2x + 1)^3\]

Now differentiate with respect to \(x\):

\[\dfrac{dV}{dx} = \dfrac{9}{16}\pi \cdot \dfrac{d}{dx}(2x + 1)^3 = \dfrac{9}{16}\pi \cdot 3(2x + 1)^2 \cdot 2 = \dfrac{27}{8}\pi (2x + 1)^2\]

Final Answer:
\[\dfrac{dV}{dx} = \dfrac{27}{8}\pi (2x + 1)^2\]