6/x-9e^x find the antiderivative of f(x)

Question

tkhunny · Answer

$\dfrac{6}{x} - 9e^{x}\;\;?$

Which part troubles you, the logarithm or the exponential?

anonymous · Answer

the 9e^x. I'm quite confident for the first part the answer would come out as 6x^-1? right?

tkhunny · Answer

??  Isn't it already $6x^{-1}$?

Hint 1

$\dfrac{d}{dx}e^{x} = e^{x}$

Hint 2

$\dfrac{d}{dx}\ln(x) = 1/x$ <== Careful with this one, it doesn't quite work the other direction.

anonymous · Answer

so it'd be 9(1/x)? I'm looking through my notebook right now and I see a similar problem to it but I don't have a problem that has an e^x only just a e

abb0t · Answer

$\int\limits e^{-x}dx = -e^{-x}+C$, you can check this by taking the derivative of $e^{-x}$

abb0t · Answer

$-e^{-x}$

anonymous · Answer

but there's the 9 in front? So I don't think I can do anything like that to the e

abb0t · Answer

9 is a constant and you can simply factor it out.

abb0t · Answer

You can also check this by taking the derivaitve of $9[-e^{-x}]$ also.

anonymous · Answer

isnt the derivative of e^x just 1 though?

tkhunny · Answer

So, figure out that $\int e^x\;dx = e^{x} + C$.  Let the chain rule and the negative go!

abb0t · Answer

No. Use chain rule. derivative is simply $e^x$

anonymous · Answer

so the derivative of 9 is 9x and e^x is just e^x (1) but the 1 doesn't really matter so i dont put that so its 9xe^x?

abb0t · Answer

No! $\frac{ d }{ dx }(-9e^{-x})$ = $9[-1 \times e^{-x} \times \frac{ d }{ dx }(-x)] = 9[(-1 \times -1) \times e^{-x}] = 9e^{-x}$, which as you can see is just the anti-derivative.

anonymous · Answer

so 9e^-x is the anti-derivative of the 2nd part o.O?

abb0t · Answer

No, $-9e^{-x}$ is the anti-derivative of $9e^{-x}$

anonymous · Answer

ok