Question

g(t)=t^2e^-t+(ln(t))^2
differentiate the above function

Answer 1

\[g(t)=t^2e^{-t}+\ln^2(t)\]?

Answer 2

yes

Answer 3

looks like you need the product rule for the first one, and the chain rule for the second one

Answer 4

\[\frac{d}{dx}[\ln^2(x)]=2\ln(x)\times \frac{1}{x}=\frac{2\ln(x)}{x}\] via the chain rule
first one is a straight forward product rule

Answer 5

use \((fg)'=f'g+g'f\) with \(f(t)=t^2,f'(t)=2t,g(t)=e^{-t},g'(t)=-e^{-t}\)

Answer 6

why is g'(t)=-e^-t