# New member , eager to learn but i might need some help!

• Jun 18th 2014, 05:05 PM
NED1
New 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!
• Jun 18th 2014, 05:58 PM
JeffM
Re: 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.

Otherwise, ask for a hint to get started.
• Jun 18th 2014, 07:11 PM
NED1
Re: 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.
• Jun 18th 2014, 07:54 PM
peterwilliam
Re: New member , eager to learn but i might need some help!
Hi Me too new here and just joined the things.
• Jun 18th 2014, 08:02 PM
romsek
Re: New member , eager to learn but i might need some help!
Quote:

Originally Posted by NED1
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.

close but not quite.

\$\dfrac d {dx} \sec(x^3) = 3x^2 \sec(x^3)\tan(x^3)\$ not \$3x^2 \sec(x)\tan(x)\$ as you've written.
• Jun 18th 2014, 08:27 PM
NED1
Re: New member , eager to learn but i might need some help!
Quote:

Originally Posted by romsek
close but not quite.

\$\dfrac d {dx} \sec(x^3) = 3x^2 \sec(x^3)\tan(x^3)\$ not \$3x^2 \sec(x)\tan(x)\$ as you've written.

thank you! otherwise it is correct?
• Jun 18th 2014, 08:39 PM
romsek
Re: New member , eager to learn but i might need some help!
Quote:

Originally Posted by NED1
thank you! otherwise it is correct?

D[ e^(5x) sec (x^3),x] - Wolfram|Alpha

I believe so
• Jun 18th 2014, 08:39 PM
NED1
Re: New member , eager to learn but i might need some help!
Thanks very much.