If f(x)= 1/3( e^(3x) + e^ (-3x)), find f'(1)

G

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I can't get this:

If f(x) = 1/3( e^(3x) + e^(-3x) ), find f'(1).

So I did this:

. . .f(1) = 1/3(e^3 + e^-3)

. . .f'(1) = 1/3( e^3 - e^-3)

The book's answer is p'(1) = e^3 - e^-3

Where did the 1/3 go?

Thank you!
 
Where did the "p" come from? If the original function was f(x), the derivative should be f'(x), not p'(x). Are you sure you read the answer to this exercise?

Thank you.

Eliz.
 
You're forgetting the chain rule.

Don't forget the derivative of 3x; that's 3.

\(\displaystyle \L\\(3)(\frac{e^{3x}}{3})\)

Therefore,

\(\displaystyle \L\\\frac{d}{dx}[\frac{e^{3x}}{3}]=e^{3x}\)

\(\displaystyle \L\\\frac{d}{dx}[\frac{e^{-3x}}{3}]=-e^{-3x}\)

\(\displaystyle \L\\e^{3x}-e^{-3x}\)

Subbing in 1:

\(\displaystyle \L\\e^{3}-e^{-3}\)
 
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