## anonymous 5 years ago Find f '(x) and f ''(x). f(x)= x/9-e^x

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

derivativave of 9 is 0,

5. anonymous

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

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)

7. anonymous

right

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.

9. anonymous

$(x(e ^{x}))-(1(9+e^{x}))/(9+e^{x})^{2}$

Is that a different problem?

11. anonymous

its not subtract its divide

12. anonymous

x divided by 9+e^{x}

Ah, got it.

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

15. anonymous

yes and the other was the answer i got

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

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)

18. anonymous

and thats the answer to fprime?

19. 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

20. anonymous

what is wrong with the answer i found?

Sorry, browser retardedness.

22. anonymous

i feel ya

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

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

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