Question #b5a72

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Answer 1 · 2018-01-18T20:09:34

Simplify the expression first

$((216πr^2-2πr^5)/(3r^2))+2πr^2$

Fisrt get rid of the r^2 in the denominator:

$(216π−2πr^3)/(3)+2πr^2$

It seems easier now:

$(216π)/3−(2πr^3)/(3)+2πr^2$

So the derivate will be:

$0- (2pi*3r^2)/3+2pi*2*r$

$0- 2pir^2+4pir$