Derivative of $e^{-x}$ I know that derivative of $e^{x}$ is $e^{x}$ itself.
Example:
$$f(x)=4e^{x}+3e^{-x}$$
I calculated the derivative to be the same as the example itself, but the real answer (at the end of my book) is:
$$4e^{x}-3e^{-x}$$
Can somone please tell me why it has $-3$?
 A: Note that $e^{-x}$ is a composition: we are composing the function $g(x) = -x$ with the function $f(u)=e^u$. By the Chain Rule, we have:
$$
\frac{d}{dx}e^{-x} = \frac{d}{dx}f\Bigl(g(x)\Bigr)
= f'(g(x))g'(x).$$
Now, $g(x)=-x$, so $g'(x) =-1$. And $f(u) = e^u$, so $f'(u) = e^u$. Hence
$$\frac{d}{dx}e^{-x} = f'(g(x))g'(x) = e^{g(x)}(-1) = e^{-x}(-1) = -e^{-x}.$$
Therefore,
$$\frac{d}{dx}(4e^x + 3e^{-x}) = 4\frac{d}{dx}e^x +3\frac{d}{dx}e^{-x} = 4e^x + 3(-e^{-x}) = 4e^x-3e^{-x}.$$
A: Firstly, we can use chain rule, which is:
$$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$$
where $f(x)=e^x$ and $g(x)=-x$ (so that $f(g(x))=e^{-x}$).  We know that $\frac{d}{dx}e^x=e^x$ and $\frac{d}{dx}-x=-1$ so we get the derivative:
$\frac{d}{dx}e^{-x}=e^{-x}*-1=-e^{-x}$
To know why this occurs, we use first principles:
$$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$
so thus
$$\frac{d}{dx}e^{-x}=\lim_{h\rightarrow 0}\frac{e^{-(x+h)}-e^{-x}}{h} = \lim_{h\rightarrow 0}e^{-x}(\frac{e^{-h}-1}{h})$$
now, from the (a?) definition of $e$:
$$\lim_{h\rightarrow 0}\frac{e^{-h}-1}{h}=-1$$
and thus
$$\frac{d}{dx}e^{-x}=-e^{-x}$$
A: An answer has been accepted, but I want to add that the reason that $\frac{d}{dx}e^{-x} = -e^{-x}$ can also be explained with the product rule.
We know $e^x e^{-x} = 1$. So if we differentiate each side, we preserve equality. Using the product rule on the left, and knowing the derivative of a constant is zero gives
$ e^x e^{-x} + e^x \frac{d}{dx}(e^{-x}) = 0 $
We can rewrite this as
$ e^x \frac{d}{dx}(e^{-x}) = -1 $
Or, finally,
$ \frac{d}{dx}(e^{-x}) = -e^{-x} $
A: Though there are so many answers, I couldn't resist.
$$ \frac{d}{dx}\left(e^{f(x)}\right) =  \left(e^{f(x)}\right) \frac{d}{dx}\left(f(x)\right) $$
Whether $a>0$ or $a<0$
$$ \frac{d}{dx}\left(e^{ax}\right) =  a \left(e^{ax}\right)$$
Therefore, particularly when $a=-1$
$$ \frac{d}{dx}\left(e^{-x}\right) = - e^{-x}$$
$$ 
\begin{align*}
\frac{d}{dx}\left(4e^{x}+3e^{-x}\right) &=   4e^{x} + 3\times(-1)e^{-x}\\
&= 4e^{x}-3e^{-x}
\end{align*}
$$
A: The derivative of $e^{-x}$ is $-e^{-x}$ and that is where the minus sign comes from.  To see that, if you know the chain rule $e^{-x}=f(g(x))$ where $f(x)=e^x, g(x)=-1$.  Then $f'(g(x))g'(x)=e^{-x}(-1)$
A: The chain rule states that the derivative of $e^u$ is $e^u \cdot \frac{du}{dx}$, where $u$ is a function of $x$. In this case $u = -x$, and $\frac{du}{dx} = -1$.
A: In general, whenever we see exponents in a problem involving derivatives it is always a good idea to use logs to simplify the problem. We need to simplify the function: $y=e^{-x}$. Taking ln of both sides gives us $\space$ 
$\ln(y)=-x$. What this does is it simplifies a problem that dealt with exponentiation into a problem that deals with multiplication. From here we take the derivative giving us: $\space$ $\frac{y'}{y}=-1$, $\space$ $y'=-y$ $\space$ And thus, $\frac{d}{dx}e^{-x}=-e^{-x}$. It may be helpful to keep this strategy in mind when solving problems in the future.
