Question

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

Answer 1

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

Explanation:

The power rule states that `(x^n)' = nx^(n-1)`. We also know that if `y` is a function of `x`, then for any real nonzero constant `c`, `(cy)' = c(y)'`. 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`.