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

Jan 21, 2017

Assuming $r$ is the variable, it is 16πr.

#### Explanation:

The power rule states that $\left({x}^{n}\right) ' = n {x}^{n - 1}$. We also know that if $y$ is a function of $x$, then for any real nonzero constant $c$, $\left(c y\right) ' = c \left(y\right) '$. Also recall that the derivative of a constant by itself is zero, and that the derivative of a sum is the sum of derivatives of the terms.

Since 4/3π^3 is a constant, its derivative is $0$.

(8πr^2)' = 8π(r^2)' = 8π(2r) = 16πr.

Therefore,

(dV)/(dr) = 0 + 16πr = 16πr.