# Thread: chain rule & inverse functions...derivatives

1. ## chain rule & inverse functions...derivatives

I did this problem, and came up with the wrong answer. Just needed help on what i did wrong...thanks

f(x) = eln(e^2x2 + 3)

i got 4x(e^x) and it was way wrong...some help

2. Originally Posted by mathaction
I did this problem, and came up with the wrong answer. Just needed help on what i did wrong...thanks

f(x) = eln(e^2x2 + 3)

i got 4x(e^x) and it was way wrong...some help
do you mean $f(x) = e^{\ln \left( e^2 x^2 + 3\right)}$ or is it actually $f(x) = e \ln \left( e^2 x^2 + 3\right)$ ?

3. Originally Posted by Jhevon
do you mean $f(x) = e^{\ln \left( e^2 x^2 + 3\right)}$ or is it actually $f(x) = e \ln \left( e^2 x^2 + 3\right)$ ?
sorry, its

e^(lne^(2x^2)+3)

4. Originally Posted by mathaction
sorry, its

e^(lne^(2x^2)+3)
ok, so the +3 in the power of the first e is not a part of the power of the second e, got it. here's what we do. we would simplify first:

$y = e^{\ln e^{2x^2} + 3}$

$\Rightarrow y = e^{2x^2 + 3}$

$\Rightarrow y' = 4xe^{2x^2 + 3}$

5. ok...i just have to remember that whenever i have an e^x that expression will be the same in the derivative as the original function