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i found the derivative. 4(t-1)^3 3(t+5)^2

is it just \[(t-1)^4(t+5)^2\] ?

with the chain rule those are not the derivatives...

yes but instead of 2 as an exponent its 3. i know im suppose to = 0 then im not sure what to do.

wait that is right, t is single power, sorry.

okay so i did it wrong can we review step by step

yes

kk

so i get.... 4(t-1)^3 3(t+5)^2

(t-1)^3 (t+5)^2 (17+7 t)

how?

I'm trying to figure that out, I used Wolframalpha to check my work and it got something different.

\[s(t)=y\]
To find the values of y, sub in the values of t into the original equation.

how did you get to this step: =(t−1)3(t+5)2[4t+20+3t−3]