# How do you use part 1 of the fundamental theorem of calculus to find the derivative of #g(x)= int cos(sqrt( t))dt# 6 to x?

The theorem says:

Given the function:

then:

So:

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To find the derivative of ( g(x) = \int_{6}^{x} \cos(\sqrt{t}) , dt ), you use Part 1 of the Fundamental Theorem of Calculus, which states that if ( f ) is continuous on ([a, b]) and ( F ) is an antiderivative of ( f ) on ([a, b]), then

[ \int_{a}^{x} f(t) , dt = F(x) - F(a) ]

for all ( x ) in ([a, b]).

In this case, ( f(t) = \cos(\sqrt{t}) ) and ( a = 6 ). We need to find an antiderivative of ( f(t) ) to apply the Fundamental Theorem of Calculus.

Let ( F(t) ) be an antiderivative of ( \cos(\sqrt{t}) ). Then, by applying the Fundamental Theorem of Calculus, the derivative of ( g(x) ) is ( F(x) - F(6) ).

So, to find ( F(x) ), you need to find an antiderivative of ( \cos(\sqrt{t}) ), and then evaluate it at ( x ).

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To use Part 1 of the Fundamental Theorem of Calculus to find the derivative of ( g(x) = \int_{6}^{x} \cos(\sqrt{t}) , dt ), you differentiate the integral with respect to ( x ). The derivative of the integral function is equal to the integrand evaluated at the upper limit of integration, which in this case is ( x ). Therefore, the derivative of ( g(x) ) with respect to ( x ) is ( \cos(\sqrt{x}) ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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