# Derivative and integral of the exponential e^t

• B

## Summary:

Is the integral of e to the t the same as the derivative?
If I take the exponential function e^t and take the derivative, I think I get the same e^t. Even if I keep doing it over and over, second, third derivative, etc. My admittedly naive question, though, is this symmetric? Meaning...if I take the the integral of e^t, do I just get the reverse or do I have t deal with an infinity of constants because it is an indefinite integral?

Delta2

PeroK
Homework Helper
Gold Member
2020 Award
Summary: Is the integral of e to the t the same as the derivative?

If I take the exponential function e^t and take the derivative, I think I get the same e^t. Even if I keep doing it over and over, second, third derivative, etc. My admittedly naive question, though, is this symmetric? Meaning...if I take the the integral of e^t, do I just get the reverse or do I have t deal with an infinity of constants because it is an indefinite integral?
Yes, an indefinite integral produces an equivalence class of functions, not a unique function.

This is important as you can see from the following example.

We want to find functions, ##f(t)## that satisfy ##f''(t) = e^t##. Clearly ##f(t) = e^t## is one such function. But, ##f(t) = e^t + At + B## is also a solution to this differential equation, where ##A, B## are any two constants.

Another way to look at this is to say that if you integrate ##f''(t) = e^t## you do not get a unique result.

Differentiation, on the other hand, always gives a unique function as a result.

DiracPool, Antarres, Stephen Tashi and 1 other person
HallsofIvy
Yes, the derivative of $e^x$ is again $e^x$ and so the nth derivative, for all n, of $e^x$ is $e^x$ and all integrals of $e^x$ are again $e^x$. In fact, that is why "e" is defined as it is. One can show that the derivative or $a^x$, for a any positive number, is $C_aa^x$ where "$C_a$" is a constant (independent of x) that depends upon a. "e" is the unique number such that $C_e= 1$.