Question

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

Answer 1

The Fundamental Theorem of Calculus states that, if `y(x)=int_a^xf(t)\ dt`, then `dy/dx=f(x)`.

The problem gives a function `G(x)` defined as `int_1^(x^2)e^(t^2)\ dt`.

Since the bound is `x^2`, if we apply the Fundamental Theorem, we get `(dG)/(d(x^2))=e^(x^2)`. However, we want to know what `(dG)/dx` is.

To do this, we apply the chain rule. Given `(dG)/(d(x^2))`, if we multiply this by `(d(x^2))/dx`, we get `(dG)/cancel(d(x^2))*cancel(d(x^2))/dx=(dG)/dx`.

Using the power rule, `(d(x^2))/dx=2x`.

Thus, `(dG)/dx=(dG)/(d(x^2))*(d(x^2))/dx=2xe^(x^2)`