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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 $-3$?

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2  
I suppose $f(x)$ should be $4e^x+3e^{-x}$. –  azarel Apr 12 '12 at 20:24
4  
You have $f(x)$ wrong. If $f(x)$ was really what you write, then it would be equal to $7e^x$. I suspect you miscopied it/misread it. –  Arturo Magidin Apr 12 '12 at 20:24
    
YES indeed it is $3e^{-x}$ –  Sean87 Apr 12 '12 at 20:27
    
I edited the question, still why the answer has -3 instead of +3? –  Sean87 Apr 12 '12 at 20:28
1  
You can also notice that $e^{-3x} = \frac 1 {e^{3x}}$ and use the rule for fractions. It is comforting to note that you get the same answer. –  Mark Bennet Apr 12 '12 at 21:04

7 Answers 7

up vote 7 down vote accepted

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, $$\frac{d}{dx}(4e^x + 3e^{-x}) = 4\frac{d}{dx}e^x +3\frac{d}{dx}e^{-x} = 4e^x + 3(-e^{-x}) = 4e^x-3e^{-x}.$$

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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 $a=-1$

$$ \frac{d}{dx}\left(e^{-x}\right) = - e^{-x}$$

$$ \begin{align*} \frac{d}{dx}\left(4e^{x}+3e^{-x}\right) &= 4e^{x} + 3\times(-1)e^{-x}\\ &= 4e^{x}-3e^{-x} \end{align*} $$

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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.

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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} $

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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

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

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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 $\frac{du}{dx} = -1$.

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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 $f'(g(x))g'(x)=e^{-x}(-1)$

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