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#.