# help with composite rule question

• Jun 3rd 2009, 04:08 AM
emmalou264
help with composite rule question
Hey can anyone help me with the following question....

use the composite rule to differentiate f(x) = e^cosx+sinx

thank you
• Jun 3rd 2009, 04:25 AM
Rapid_W
composite rule is another name for chain rule isn't it?

differentiate the seperate terms

so $e^{cos(x)}$ diff's to $-sin(x)(e^{cos(x)})$ using the chain rule (or composite rule (i.e. you diff the first term, then multiply by the diff of term inside brackets)) and $sin(x)$ diff's to $cos(x)$ using standard rules. then just add them together at the end

$f'(x)=-sin(e^{cos(x)})+cos(x)$ and to be more artistic you would swap them so you don't start with a negative sign.
• Jun 3rd 2009, 05:01 AM
Amer
if you mean

$f(x)=e^{cosx+sinx}$

$f'(x)=\left(\frac{d}{dx}(cosx+sinx)\right)e^{cosx+ sinx}$

$f'(x)=(-sinx+cosx)e^{cosx+sinx}$
• Jun 3rd 2009, 05:12 AM
Rapid_W
• Jun 3rd 2009, 05:49 AM
emmalou264
well using the product rule of

k'(x)= f'(x)g(x) + f(x) g'(x)

and the previous equation, can i show the derivitave of the function

g(x) = (cosx + sinx - 1) e^cosx+sinx

is g'(x) = (cos^2x - sin^2x)e^cosx +sinx
• Jun 3rd 2009, 06:03 AM
HallsofIvy
Quote:

Originally Posted by Rapid_W
composite rule is another name for chain rule isn't it?

differentiate the seperate terms

so $e^{cos(x)}$ diff's to $-sin(e^{cos(x)})$

The derivative of $e^{cos(x)}$ is -sin(x) times $e^{cos(x)}$, not -sin of $e^{cos(x)}$ as you write.

Quote:

using the chain rule (or composite rule (i.e. you diff the first term, then multiply by the diff of term inside brackets)) and $sin(x)$ diff's to $cos(x)$ using standard rules. then just add them together at the end

$f'(x)=-sin(e^{cos(x)})+cos(x)$ and to be more artistic you would swap them so you don't start with a negative sign.
• Jun 3rd 2009, 07:09 AM
Rapid_W
oops, missed the x out, simple error, have fixed my original post, thanks for pointing that out.