If $\frac{d}{dx}e{^x} = e{^x}$, then why does $\frac{d}{dx}e^{-14}$ = 0? If $\frac{d}{dx}e{^x} = e{^x}$, then why does $\frac{d}{dx}e^{-14}$ = 0?
Why doesn't $\frac{d}{dx}e^{-14}$ = $e^{-14}$?
I don't understand.
 A: Because $e^{-14}$ is a constant value. The derivative of any constant value is $0$.
A: If $f(x)=e^x$ there is a difference between take the derivative of $f(x)$ at the point $x=-14$ and derivate $f(-14)$, the first is:
$$\lim_{h\to 0}\left(\frac{f(-14+h)-f(-14)}{h}\right) = e^{-14}$$
And the second being a constant function $g(x)=f(-14)$:
$$\lim_{h\to 0}\left(\frac{g(-14+h)-g(-14)}{h}\right)=\lim_{h\to 0}\left(\frac{f(-14)-f(-14)}{h}\right) = 0$$
A: If you look at the graphs of the functions $f(x)=e^x$ and $g(x)=e^{-14}$ then you will notice that the slope of each one is different. The slope at any point $x$ on the function $f(x)=e^x$ is given by $e^x$, since $f'(x)=e^x$. The slope of $g(x)=e^{-14}$ is always zero, since $g(x)=e^{-14}$ for all $x$. 
Remember that $e$ is just a number, $e=2.71828 \dots$ and therefore $e^{-14}=(2.7128 \dots)^{-14}=0.00000083152 \dots$
A: Consider the equation:
$$(\forall x)\, x + 1 > x$$
Obviously in this equation, you can substitute $x = -14$  or whatever you want and get an equally valid equation: $-14 + 1 > -14$.
One the other hand, consider the equation:
$$\sum_{x = 1}^4 x + n = 10 + 4n$$
What would it signify to substitute $x=-14$ into the above equation?  It would be nonsense.  For comparison, what if $n = -14$ were substituted into the above equation?  You would get the valid result $\sum_{x = 1}^{4} x + -14 = 1 - 14 + 2 - 14 + 3 - 14 + 4 - 14 = 10 - 4\cdot 14$.  That's because, even though we didn't write it, it's understood that the above equation is:
$$(\forall n)\, \sum_{x = 1}^4 x + n = 10 + 4n$$
So you can substitute any $n = \text{ whatever}$ into the above equation because that is what the forall $\forall$ indicates.
Although it's a pain, the equation
$$\frac{de^x}{dx} = e^x$$
is not a shorthand for 
$$(\forall x)\frac{de^x}{dx} = e^x$$
It is more exactly a shorthand for "the derivative of the exponential function is the exponential function", which if you want to see how ugly that actually looks:
$$\text{diff}(x \rightarrow e^x) = x \rightarrow e^x$$
That's long and difficult so we don't write that.  But the point is that you can't assume an arbitrary substitution like $x = \text{whatever}$ unless the $x$ comes from a $\forall x$.
A: $e^x$ is a function that depends on $x$. $e^{-14}$ is a constant.
