1. ## Let f(t)...

Let f(t)=300(lnt-e^-4t)^3.Calculate f'(1) accurate to three decimal places.

2. One way to do it is first to calculate the derivative, and then expand the derivative in a power series about some point close to 1. Then use Cauchy's formula for the error term to decide how many terms of the series you need to sum to get that accuracy.

3. Originally Posted by mathcalculushelp
Let f(t)=300(lnt-e^-4t)^3.Calculate f'(1) accurate to three decimal places.
$\displaystyle f(t)= 300(ln(t)- e^{-4t})^3$? Are you allowed to use a calculator? Bruno J. is suggesting you expand in a power series so you can calculate by hand.

If you are allowed to use a calculator, just find f'(t) and evaluate f'(1)!

4. ## derivative using chain rule?

Am getting stuck at the f'(x)

300(ln t-e^-4t)^3

let ln t -e^-4t=u
1/t+4e^-4t dt = du
dt=t du/1+4e^-4t

hence
300(u)^3 . t du/1+4e^-4t

900u^2 . t du/1+4e^-4t

900(ln t -e^-4t)^2 . t du/1+4e^-4t

but this doesn't seem quite right.Any suggestions?

5. ## derivative using substitution

it isnt chain rule but substitution in above post.sorry for the typo.