Why is the derivative of e^x=e^x? Just out of curiosity.

1 Answer
Nov 22, 2017

Because e is related to growth.

Explanation:

In loose terms, e^x has a very special derivative because it is strongly linked to growth. The number e is constructed so that it can measure the amount of growth over time. And the derivative is exactly that, how fast a function grows over time. So if you try and measure how fast the function e^x increases, you will just find that it increases at a rate of e^x, because e^x is already measuring growth.

We can of course prove this using the derivative definition too.

I want to start however by looking at the definition of e, since it will help us to prove the derivative:

e=lim_(n->oo)(1+1/n)^n
Using a bit of limit cleverness (and seeing that 1/n is just approaching 0), we can rewrite it like this:
e=lim_(n->0)(1+n)^(1/n)
(this version of the limit will be useful later)

So, let's start with our derivative. The derivative definition looks like this:
lim_(h->0)(f(x+h)-f(x))/h

So if we plug in e^x, we get:
lim_(h->0)(e^(x+h)-e^x)/h=lim_(h->0)(e^x(e^h-1))/h

By that last factoring, we have e^x in the front of the numerator, and since e^x doesn't change as h->0, we can bring it out of the limit:
e^x*lim_(h->0)(e^h-1)/h

Now, we need to do something quite clever. If we make a substitution that n=e^h-1, and solve for h to get h=ln(n+1), we can think about what happens to the limit.

As h->0, n is going to be e^0-1=0, and by that knowledge, our n-definition of h, is also going to approach 0 since ln(0+1)=0. Using this knowledge, we can rewrite the limit in terms of n, like this:
e^x*lim_(n->0)n/ln(n+1)=e^x*lim_(n->0)(n*1/n)/(1/n*ln(n+1))

On the right there, we multiplied by 1/n on the top and bottom, which might seem quite arbitrary, but you'll soon see why we did it. By logarithm properties, we have the fact that aln(b)=ln(b^a). If we use this on our limit, we get this:
e^x*lim_(n->0)1/ln((1+n)^(1/n))

We can now recognize that the piece inside the ln function is actually the same as in the limit for the definition of e. We just need to do some manipulation to isolate it as the limit.

Since the only things that actually change as n->0 are inside the ln function, we can bring the limit function inside, like this:
e^x*1/ln(lim_(n->0)(1+n)^(1/n))

And now, we can use the definition of e we had above to replace that limit with e:
e^x*1/(ln(e))

If we solve this expression, we get:
e^x*1/1=e^x

And there. We have just proven that the derivative of e^x is equal to e^x.