D/dx[int(0,x) e^(-t^2) dt] : two methods, two answers

1. Dec 24, 2011

There are two methods to take d/dx [ ∫t=0x exp(-t^2) dt].

First method: using the relationship of integral and antiderivative, one gets
(exp(-t^2) , from t = 0 to x, so exp(-x^2) - 1.

Second method: the integral is (1/2)sqrt(pi)*erf(t) from 0 to x, which is (1/2)sqrt(pi)*erf(x), and the derivative of this is exp(-x^2).

So, which answer is correct, and what is wrong with the other method?

(Wolfram alpha favors the second answer, but I have another source that favors the first answer.)

2. Dec 24, 2011

jgens

The second answer is right. Can you show how you got this so we can help point out your mistake?

3. Dec 24, 2011

Thank you, gladly. I am using the idea that d(∫abf(t)dt)/dx = f(x)|ab; here a=0 , b = x, and f(t)= exp(-t^2).

4. Dec 24, 2011

jgens

5. Dec 24, 2011