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

Results 1 to 5 of 5

- December 11th 2009, 05:07 PM #1

- Joined
- Oct 2009
- Posts
- 92

- December 11th 2009, 05:26 PM #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.

- December 12th 2009, 04:38 AM #3

- Joined
- Apr 2005
- Posts
- 14,992
- Thanks
- 1129

- December 14th 2009, 12:09 PM #4

- Joined
- Oct 2009
- Posts
- 92

## 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?

- December 14th 2009, 12:11 PM #5

- Joined
- Oct 2009
- Posts
- 92