# How do you use the second fundamental theorem of Calculus to find the derivative of given #int cos(sqrt( t))dt# from #[6, x]#?

See the explanation below.

The second Fundamental Theorem of Calculus (a.k.a The Fundamental Theorem of Calculus, Part 2) tells us that

Therefore,

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To find the derivative of the integral (\int_{6}^{x} \cos(\sqrt{t}) , dt) with respect to (x), you can use the Second Fundamental Theorem of Calculus, which states that if (f) is continuous on an interval containing (a) and (F(x)) is any antiderivative of (f(x)) on that interval, then

[\frac{d}{dx}\left(\int_{a}^{x} f(t) , dt\right) = f(x).]

In your case, the function (f(t) = \cos(\sqrt{t})) is continuous on the interval ([6, x]).

First, find an antiderivative of (f(t)), denoted by (F(t)). Let (u = \sqrt{t}), then (du = \frac{1}{2\sqrt{t}}dt), or (2\sqrt{t}du = dt). Then, (F(t) = \int \cos(u) \cdot 2u , du = 2\sin(u) + C = 2\sin(\sqrt{t}) + C), where (C) is the constant of integration.

Now apply the Second Fundamental Theorem of Calculus:

[\frac{d}{dx}\left(\int_{6}^{x} \cos(\sqrt{t}) , dt\right) = \frac{d}{dx}(F(x) - F(6)) = F'(x) = 2\sin(\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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