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

1 Answer
Jan 21, 2017

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.