# How do you calculate the derivative of #intcos(t)/t dt# from #[x,3]#?

Use the Fundamental Theorem of Calculus Part 1.

This question asks about the function:

Notice that FTC 1 requires the constant to be the lower limit of integration, so we use the properties of definite integral to write:

That's it. Stop! We are finished. Move on to the next question. (On exams, the trickiest thing about this kind of question is often understanding how easy it is. We expect more work to be needed.)

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To calculate the derivative of the integral of cos(t)/t from [x,3] with respect to x, you can use the Leibniz Rule, which states:

d/dx ∫[a(x), b(x)] f(t,x) dt = ∫[a(x), b(x)] ∂f/∂x dt + f(b(x), x) * db(x)/dx - f(a(x), x) * da(x)/dx

In this case, a(x) = x and b(x) = 3. So, applying the Leibniz Rule:

d/dx ∫[x,3] cos(t)/t dt = ∫[x,3] (∂/∂x (cos(t)/t)) dt + (cos(3)/3) * d(3)/dx - (cos(x)/x) * d(x)/dx

Now, compute the partial derivative of cos(t)/t with respect to x:

∂/∂x (cos(t)/t) = -sin(t)/t

Substitute back into the expression:

d/dx ∫[x,3] cos(t)/t dt = ∫[x,3] (-sin(t)/t) dt + (cos(3)/3) * 0 - (cos(x)/x) * 1

Integrate -sin(t)/t from x to 3:

= ∫[x,3] (-sin(t)/t) dt = -Ci(3) + Ci(x) - π/2

Where Ci is the cosine integral function.

Therefore, the derivative of the integral of cos(t)/t from [x,3] with respect to x is:

= -Ci(3) + Ci(x) - π/2 - (cos(x)/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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