Find f '(x) and f ''(x).
f(x)= x/9-e^x

- anonymous

Find f '(x) and f ''(x).
f(x)= x/9-e^x

- chestercat

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- anonymous

really having an issue with differentiables containing square roots and e^x

- anonymous

the derivative of e^x is always e^x, forever, it's its own derivative

- anonymous

what if a number is attached like the 9-e^x?

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## More answers

- anonymous

derivativave of 9 is 0,

- anonymous

ahh so you do seperate it... the derivative of 9 then e^x

- shadowfiend

So, when you take a derivative, you always take the separate derivative of parts that are added or subtracted together. For example:
f(x) = g(x) + h(x)
f'(x) = g'(x) + h'(x)
f''(x) = g''(x) + h''(x)
And:
f(x) = g(x) - h(x)
f'(x) = g'(x) - h'(x)
f''(x) = g''(x) - h''(x)

- anonymous

right

- shadowfiend

So let's take your problem in particular:
\[f(x) = \frac{x}{9} - e^x\]
You have to first take the derivative of \(\frac{x}{9}\), then that of \(e^x\). Then, you can subtract the latter from the former.

- anonymous

\[(x(e ^{x}))-(1(9+e^{x}))/(9+e^{x})^{2}\]

- shadowfiend

Is that a different problem?

- anonymous

its not subtract its divide

- anonymous

x divided by 9+e^{x}

- shadowfiend

Ah, got it.

- shadowfiend

So:
\[f(x) = \frac{x}{9 + e^x}\]?

- anonymous

yes and the other was the answer i got

- shadowfiend

I've always found it easiest to use product rule when I have division like that.
In this case, you're looking at:
\[f(x) = x(x + e^x)^{-1}\]

- shadowfiend

So in this case, we have product rule. Product rule says:
f(x) = h(x)g(x)
f'(x) = h'(x)g(x) + h(x)g'(x)

- anonymous

and thats the answer to fprime?

- anonymous

sorry to be the pain! lol im a non traditional student that never took calc in high school sad thing is im trying to major in it lmao

- anonymous

what is wrong with the answer i found?

- shadowfiend

Sorry, browser retardedness.

- anonymous

i feel ya

- shadowfiend

Also, what I meant was that you're looking at :
\[f(x) = x(9 + e^x)^{-1}\]

- shadowfiend

Looks like the only issue you had was a sign issue.

- shadowfiend

I ultimately got:
\[\frac{9 + e^x - xe^x}{(9 + e^x)^2}\]

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