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

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Answer 1 · 2016-09-24T01:59:18

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

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

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)

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

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