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the derivative of 1/2bh?

yes i know the product rule

so not respect to b?

the change in Area with respect to time is also known as the rate
you want dA/dt , and so on..

I'm confused with finding the derivative.

alli know is this formula: a=pi^2 r but im guessing it's not relevant for this problem.

and also a=1/2*b*h

right

to take the derivative of A with respect to time you write a little d in front, and divide by dt

dA/dt

yes that is the left side
you have
dA/dt = 0.5 * d/dt ( B * H)
use the product rule on B*H

oh okay so we ignore the 0.5 for now?

so here: dA/dt=1/2(dh/dt)+(db/dt)

except that is not the product rule
http://www.1728.org/chainrul.htm

the product rule of d/dt (x * y) = x * dy/dt + y * dx/dt

?

i understand what you're saying
up there.

is this right? dA/dt= 1/2 (b*dh/dt)+(h*db/dt)