Hi everyone as per title, Im doinf a new math course and struggle with some ideas and concepts. Im very grateful for a site like this, it seems very knowledgable!

Printable View

- June 18th 2014, 06:05 PMNED1New member , eager to learn but i might need some help!
Hi everyone as per title, Im doinf a new math course and struggle with some ideas and concepts. Im very grateful for a site like this, it seems very knowledgable!

- June 18th 2014, 06:58 PMJeffMRe: New member , eager to learn but i might need some help!
Hi. There are many willing and able to help here.

State the problem exactly and completely.

If you have done some work, show it so we can see where you are stuck.

If you have an answer, but are unsure, show your answer PLUS your work so, IF you made a mistake, we can see where.

Otherwise, ask for a hint to get started. - June 18th 2014, 08:11 PMNED1Re: New member , eager to learn but i might need some help!
DY/DX e^5x sec (x^3).

I have to differentiate this. I belive I have to use the product rule then the chain rule.

f'x = e^5x d/dx ((sec (x^3)) + (sec (x^3)) d/dx ((e^5x).

g(u) = sec (u) where u = f(x) = (x^3)

now g'(u) = sec x tan x and f'x = 3x^2

Composite rule = k'(x) = g'(f(x)) f' (x)

= g' (u) f'(x)

= sec x tan x . 3x^2

f'x = e^5x (sec x tan x 3x^2) + (sec (x^3)) 5e^5x

Thats it. I find this quite difficult. - June 18th 2014, 08:54 PMpeterwilliamRe: New member , eager to learn but i might need some help!
Hi Me too new here and just joined the things.

- June 18th 2014, 09:02 PMromsekRe: New member , eager to learn but i might need some help!
- June 18th 2014, 09:27 PMNED1Re: New member , eager to learn but i might need some help!
- June 18th 2014, 09:39 PMromsekRe: New member , eager to learn but i might need some help!
D[ e^(5x) sec (x^3),x] - Wolfram|Alpha

I believe so - June 18th 2014, 09:39 PMNED1Re: New member , eager to learn but i might need some help!
Thanks very much.