Question

How do you find the derivative of `G(x)=int (e^(t^2))dt` from `[1,x]`?

Answer 1

You should get `e^(x^2)`.

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You can use Leibniz's integral rule.

`(dG(x))/(dx) -= d/(dx) [int_(g(x))^(h(x)) f(x,t)dt]`

`= int_(g(x))^(h(x)) (delf)/(delx)dt + f(x,h(x))(dh)/(dx) - f(x,g(x))(dg)/(dx)`

In this case, `g(x) = 1` and `h(x) = x`, so:

• `(dg)/(dx) = 0`
• `(dh)/(dx) = 1`
• `f(x,h(x)) = f(x,t = x) = e^(x^2)`
• `f(x,g(x)) = f(x,t = 1) = e^(1^2) = e`
• `(delf)/(delx) = (del)/(delx)[e^(t^2)]_x = 0`

Therefore:

`color(blue)((dG)/(dx)) = cancel(int_(1)^(x) 0dt)^(0) + e^(x^2)cdot1 - cancel(ecdot0)`

`= color(blue)(e^(x^2))`

Answer 2

`d/dx \ int_1^x e^(t^2) \ dt = e^(x^2)`

Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

`d/dx \ int_a^x \ f(t) \ dt = f(x)` for any constant `a`

(ie the derivative of an integral gives us the original function back).

We are asked to find:

`d/dx \ int_1^x e^(t^2) \ dt`

And this integral is in the correct form for the FTOC to be applied, giving:

`d/dx \ int_1^x e^(t^2) \ dt = e^(x^2)`