A balloon which always remains spherical, has a variable diameter $\frac{3}{2}(5x+7)$. Then the rate of change of its volume with respect to x is

The balloon has a diameter of $\frac{3}{2}(5x+7)$.

The radius is half the diameter:

$r = \frac{1}{2} \times \frac{3}{2}(5x+7)$

$r = \frac{3}{4}(5x+7)$

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The volume of a sphere is:

$V = \frac{4}{3}\pi r^3$

Substituting $r = \frac{3}{4}(5x+7)$:

$V = \frac{4}{3}\pi \left[\frac{3}{4}(5x+7)\right]^3$

$V = \frac{4}{3}\pi \times \frac{27}{64}(5x+7)^3$

$V = \frac{108}{192}\pi(5x+7)^3$

$V = \frac{9}{16}\pi(5x+7)^3$

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Differentiating with respect to $x$ using the chain rule:

$\frac{dV}{dx} = \frac{9\pi}{16} \times 3(5x+7)^2 \times 5$

$\frac{dV}{dx} = \frac{135\pi}{16}(5x+7)^2$

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The rate of change of volume with respect to $x$ is:

$\frac{dV}{dx} = \frac{135\pi}{16}(5x+7)^2$