Question

How do you find the derivative of `72000-320(pi)r^3/9(pi)r^2`?

Answer 1

`-1600/9pi^2r^4`

Explanation:

`f(r)=72000-320(pi)r^3/9(pi)r^2`

Let's rationalise this first
(we can ignore the leading constant as its derivative will be zero)

`f(r) = c - 320/9pi^2r^5`

Applying the power rule:
`f'(r) = 0-320/9pi^2 * 5r^4` (Since `320/9pi^2` is just a constant)

`=-1600/9pi^2r^4`