# Thread: Derivative question involving both e and ln

1. ## Derivative question involving both e and ln

y = e^(-x) * ln x

I'm stumped on how to proceed here. Up until now I've only derived problems concerning ln alone. A step by step solution would be very much appreciated.

2. Originally Posted by Archduke01
y = e^(-x) * ln x

I'm stumped on how to proceed here. Up until now I've only derived problems concerning ln alone. A step by step solution would be very much appreciated.
Are you familiar with the product rule?

$\displaystyle \left(fg\right)^{\prime}=f^{\prime}g+g^{\prime}f$.

In your case, let $\displaystyle f(x)=e^{-x}$ and $\displaystyle g(x)=\ln x$.

Can you try to take it from here?

3. Originally Posted by Archduke01
y = e^(-x) * ln x

I'm stumped on how to proceed here. Up until now I've only derived problems concerning ln alone. A step by step solution would be very much appreciated.
$\displaystyle \frac{d}{dx}[e^{-x}\ln{x}]=e^{-x}\left(\frac{1}{x}\right)+(-e^{-x})\ln{x}$

This is the solution. The product rule is what was needed.