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

Mathematics
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really having an issue with differentiables containing square roots and e^x
the derivative of e^x is always e^x, forever, it's its own derivative
what if a number is attached like the 9-e^x?

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Other answers:

derivativave of 9 is 0,
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)
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.
\[(x(e ^{x}))-(1(9+e^{x}))/(9+e^{x})^{2}\]
Is that a different problem?
its not subtract its divide
x divided by 9+e^{x}
Ah, got it.
So: \[f(x) = \frac{x}{9 + e^x}\]?
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)
and thats the answer to fprime?
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
what is wrong with the answer i found?
Sorry, browser retardedness.
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: \[\frac{9 + e^x - xe^x}{(9 + e^x)^2}\]

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