# Thread: help with composite rule question

1. ## 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

2. composite rule is another name for chain rule isn't it?

differentiate the seperate terms

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

$\displaystyle f'(x)=-sin(e^{cos(x)})+cos(x)$ and to be more artistic you would swap them so you don't start with a negative sign.

3. if you mean

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

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

$\displaystyle f'(x)=(-sinx+cosx)e^{cosx+sinx}$

4. Yup, take your pick depending on what your question is

5. 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

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

differentiate the seperate terms

so $\displaystyle e^{cos(x)}$ diff's to $\displaystyle -sin(e^{cos(x)})$
The derivative of $\displaystyle e^{cos(x)}$ is -sin(x) times $\displaystyle e^{cos(x)}$, not -sin of $\displaystyle e^{cos(x)}$ as you write.

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

$\displaystyle f'(x)=-sin(e^{cos(x)})+cos(x)$ and to be more artistic you would swap them so you don't start with a negative sign.

7. oops, missed the x out, simple error, have fixed my original post, thanks for pointing that out.