Question

Let `F(x)=int_(0)^(x^2)e^(x^2)dx` how do you find `F'(x)` ?

Answer 1

`F'(x)=2xe^(x^4)`

Explanation:

The best way to approach this is to rewrite this integral. It canbe better re-written to the following:
`F(x)=\int_0^{x^2}e^{t^2}dt`
So that no ambiguity is given about which variable is being integrated
Suppose `G(t)=\int e^{t^2}dt`
Then `F(x)=G(x^2)-G(0)` by the fundamental theorem of calculus
Since you can differentiate now, notice that G(0) is a constant so it dissapears. And you are left with
`F'(x)=d/{dx}G(x^2)`
Now if you substitute u=x^2, and use the chain rule then
`F'(x)={du}/{dx}\cdot d/{du}G(u)`
Now since you know `u=x^2` and so therefore `{du}/{dx}=2x`
And you also know that if `G(t)=\int e^{t^2}dt`, then `{dG}/{du}=e^{u^2}`
Therefore putting it all together and substituting the value of `u`
`F'(x)=2xe^{u^2}=2xe^{(x^2)^2}=2xe^(x^4)`