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

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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!
 
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It would be -cos(\sqrt{x}), not t.
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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