Question

Let `G(x)=int_x^{-2}cos(t^2)dt`. Then what is G'(x) ?

Let `G(x)=int_x^{-2}cos(t^2)dt`. Then what is G'(x) ?

I got two different answers and I'm confused which one is right. The first one I got, G'(x)=2xcos(x^2), by evaluating the integral and then finding the derivative. The 2nd one is G'(x)=cos(4) - cos(x^2) because I think that calculating the derivative of an integral of a function will be equal to the function itself.So which one is right?

Answer 1

`G^'(x) = -cos(x^2)`

Explanation:

Let us try to evaluate the derivative from first principles, i.e. using

`G^'(x) = lim_{h to 0} (G(x+h)-G(x))/h`

Now

`G(x)=int_x^{-2}cos(t^2)dt implies`

`G(x+h-G(x))=int_{x+h}^{-2}cos(t^2)dt - int_x^{-2}cos(t^2)dt`
`qquad = int_{x+h}^{x}cos(t^2)dt =-int_{x}^{x+h}cos(t^2)dt = -h cos(x^2)+O(h^2)`

Thus

`(G(x+h)-G(x))/h = -cos(x^2)+O(h)`

and thus the derivative is

`G^'(x) = -cos(x^2)`

This, of course, is a special case of the general result

`color(red)(d/dx(int_{f(x)}^{g(x)} F(t)dt ) = g^'(x)F(g(x))-f^'(x)F(f(x)))`