Derivative of an exponential function I am trying to solve
$$\frac{1}{e^{x}}$$
I first tried using the quotient rule, and ended up with:
$$\frac{e^{x}}{(e^{x})^2}$$
That was not the right answer, so I took a look at wolfram, and wolfram rewrites 
$$\frac{1}{e^{x}}$$ as 
$$e^{-x}$$
And then apply the chain rule, but that is irrelevant to me right now. I am looking for someone to explain how 
$$\frac{1}{e^{x}} = e^{-x}$$
Any help would be great, I want to solve this but I need to understand what just happened.
 A: Two things:
1) your method with the quotient rule works just fine, you just made a mistake. The quotient rule says
$$
\frac{d}{dx} \frac{f(x)}{g(x)} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}.
$$
It looks like you missed the negative sign in the numerator. Your answer is equivalent to the right answer if you put it.
2) Wolfram's method is also correct. The reason $e^{-x} = \frac{1}{e^x}$ (a fact that you definitely need to know to succeed in calculus!) is that
$$
e^x \cdot e^{-x} = e^{x - x} = e^0 = 1.
$$
A: $$y = \dfrac 1{e^x} = \dfrac {e^0}{e^x} = e^{0-x} = e^{-x}$$ $$y'= (-1)e^{-x} = -\dfrac {1}{e^x}$$
This is by using the chain rule for the exponential function: $$\left(e^{f(x)}\right)' = f'(x)e^{f(x)}$$
In this case, for $y = e^{-x}$, we have $e^{f(x)}$, where $f(x) = -x$, and $f'(x) = -1$.
A: There's more than one reason why $e^{-x} = \dfrac 1 {e^x}$, but here's one:
When you have a fraction of exponents, you subtract the power of the bottom from the power of the top:
$\dfrac {e^x}{e^y} = e^{x - y}$
Here, you have $\dfrac 1 {e^x} = \dfrac {e^0}{e^x}$.
edit: I like Hunter's explanation better, but think about why the explanations are equivalent.
