# composite / chain rule....differentiate

• May 28th 2009, 02:47 AM
redieeee_babieeee
composite / chain rule....differentiate
hi can someone lease help with the following question as i have done anythng this hard with the composite/chain rule

Q) use the composite rule to differentiate the function

f(x)= e^cos x+ sin x

thanx
• May 28th 2009, 02:59 AM
pickslides
is your equation $\displaystyle f(x) = e^{cos(x)}+sin(x)$ or $\displaystyle f(x) = e^{cos(x)+sin(x)}$ ?

either way you should employ the rule

$\displaystyle \frac{d}{dx}\left(e^{f(x)}\right) = f'(x)e^{f(x)}$
• May 28th 2009, 03:04 AM
redieeee_babieeee
• May 28th 2009, 03:20 AM
craig
Quote:

Originally Posted by redieeee_babieeee

As pickslides said, what you need to do is:

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

Can you differentiate $\displaystyle cos(x)+sin(x)$?
• May 29th 2009, 02:47 AM
redieeee_babieeee
so can i just check that the anwser will be

http://www.mathhelpforum.com/math-he...065983c7-1.gif-sin(x)+cos(x)

thanx
• May 29th 2009, 02:53 AM
craig
Quote:

Originally Posted by redieeee_babieeee
so can i just check that the anwser will be

http://www.mathhelpforum.com/math-he...065983c7-1.gif-sin(x)+cos(x)

thanx

Yep that's right, I'd write it with the $\displaystyle (\cos{x} - \sin{x})$ before the $\displaystyle e^{cos(x)+sin(x)}$ for clarities sake, but that's just me being pedantic ;)

$\displaystyle f'(x) = (\cos{x} - \sin{x})e^{cos(x)+sin(x)}$