How Does FTC1 Help in Finding the Derivative of G(x)?

[aeonsky]

I need to find the derivative of the function below...

Homework Statement

$G(x) = \int_{x}^{1} cos(\sqrt{t}) dt$

Homework Equations

FTC1

If f is continuous on [a,b], then the function g defined by

$g(x) = \int_{a}^{x} f(t) dt$ $a \leq x \leq b$

is continuous on [a,b] and differentiable on (a,b) and $g'(x) = f(x)$

The Attempt at a Solution

Would it be $-cos(sqrt(t))$

Thanks for the time!

 

[HallsofIvy]

It would be $-cos(\sqrt{x})$, not t.

 

[lurflurf]

First you might reverse the limits, which reveses the sighn. FTC1 says the derivative of the integral of a function is the function. Differendiation cancels integration.