# How do you differentiate #V=4/3pi^3+8pir^2#?

Assuming

Therefore,

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To differentiate ( V = \frac{4}{3}\pi r^3 + 8\pi r^2 ) with respect to ( r ), use the power rule for differentiation:

[ \frac{dV}{dr} = \frac{d}{dr} \left( \frac{4}{3}\pi r^3 + 8\pi r^2 \right) = \frac{d}{dr} \left( \frac{4}{3}\pi r^3 \right) + \frac{d}{dr} \left( 8\pi r^2 \right) ]

Now, differentiate each term separately:

[ \frac{d}{dr} \left( \frac{4}{3}\pi r^3 \right) = \frac{4}{3}\pi \frac{d}{dr} (r^3) = \frac{4}{3}\pi \times 3r^2 = 4\pi r^2 ]

[ \frac{d}{dr} \left( 8\pi r^2 \right) = 8\pi \frac{d}{dr} (r^2) = 8\pi \times 2r = 16\pi r ]

So, the derivative of ( V ) with respect to ( r ) is:

[ \frac{dV}{dr} = 4\pi r^2 + 16\pi r ]

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