Different answers using product and quotient rule for differentiation

Question

I am trying to differentiate this:

$$y = \frac{x}{e^x}$$

via Quotient Rule:

$$y' = \frac{(e^x*1) - (x*e^x)}{e^{2x}}$$

$$ = \frac{e^x(1-x)}{e^{2x}}$$

$$ = \frac{(1-x)}{e^x}$$

via Product rule:

$$y = x*e^{-x}$$ so

$$ y' = x*e^{-x} + e^{-x}$$

$$ = e^{-x} * (x+1) $$

$$ = \frac{(x+1)}{e^x}$$ Obviously I'm making a mistake somewhere. Where is it?

Answer 1

When you differentiate $e^{-x}$ in the product rule version, you didn't put a minus sign in front. It should differentiate to $-e^{-x}$.

---

Edit:

$$\begin{align}y &= x\cdot e^{-x}\\\frac{dy}{dx}&=\frac{d}{dx}(x)\cdot e^{-x}+x\cdot \frac{d}{dx}(e^{-x})\\&=1\cdot e^{-x}-xe^{-x}\\&=\frac{1-x}{e^{-x}}\end{align}$$

Answer 2

$$\frac{d}{dx}e^{-x} = -e^{-x}$$

Answer 3

Heren you made a mistake :

$$y' = xe^{-x} + e^{-x}$$

It should be :

$$y' = -xe^{-x} + e^{-x}$$

$$y' = e^{-x}(1-x)=\frac {1-x} {e^x}$$