TITLE: If $\frac{d}{dx}e{^x} = e{^x}$, then why does $\frac{d}{dx}e^{-14}$ = 0?
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: 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 $\forall x$.