# Derivatives of Exponentials (why e?)

Why is it that the constant e is defined as being the unique case where the limit as h goes to 0 of (e^h - 1)/h = 1? I mean every exponential function like a^x equals 1 when x equals 0, right? So would it be fair to say that (a^h) approaches 1 as h approaches zero? And that (a^h - 1) approaches zero as h approaches zero? And finally that the limit as h goes to 0 of (a^h - 1)/h also equals one?

I am confused as to why is it that only e is such that d/dx(e^x)=e^x

pwsnafu
Why is it that the constant e is defined as being the unique case where the limit as h goes to 0 of (e^h - 1)/h = 1?
We can give you mathematical proofs, but it seems like in your case, you should see what's going on. Get out a calculator and calculate the left hand side by taking smaller and smaller h. Then repeat with something not e. It's one thing to know the theory, its another to see the raw data infront of you.

I mean every exponential function like a^x equals 1 when x equals 0, right?
a=0

So would it be fair to say that (a^h) approaches 1 as h approaches zero? And that (a^h - 1) approaches zero as h approaches zero? And finally that the limit as h goes to 0 of (a^h - 1)/h also equals one?
But do they approach at the same rate?

I mean every exponential function like a^x equals 1 when x equals 0, right?
a=0
ax, when x=0 is taken to be 1, even for a=0. Read under exponentiation : https://www.physicsforums.com/showthread.php?t=530207 [Broken].

To the OP, try taking y = ax as a general function and find its derivative, by taking logarithm on both sides. See where this leads...Also, as pwsnafu said, raw data will help you get an initial 'feel' of it.

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Do you mean to say that (e^h - 1) approaches zero at the same rate as h does, but (a^h - 1) doesn't ? Why that happens exclusively to e?
Also, is rate the only thing that matters? So if I had two things approaching zero at the same rate, but one of them would reach zero earlier because it starts lower, would the ratio of those two things still be equal to one?
And if that is not the case, does that mean that (e^h - 1) = h as h approaches zero?

HallsofIvy
Yes, it is true that $a^h- 1$ approaches 0, as h goes to 0, for different a at different rates. It should be no surprise to you that different functions can converge to the same number at different rates- that's the whole point of the "slope" or derivative. To verify that, for example, look at graphs of $y= a^x- 1$ for different a. And it happens that the rate is "1" for a= e, essentially because e is defined to be that number.
For example, if we take h= .1, .01, .001, successively, $2^h- 1$ becomes
But $3^h- 1$ becomes 0.11612, 0.01104, and 0.00109. Again dividing by h they give 1.1612, 1.104, and 1.09. Each of those is larger than 1 and converges to a number larger than 1.
We could do the same with, say, $(2.7^h- 1)/h$ and $(2.75^h- 1)/h$ showing that the first converges to a number slightly less than 1, the second to a number just larger than 1. There exist a between 2 and 3, between 2.7 and 2.75, such that $(a^h- 1)/h$ converges to 1. We call that value, "e".