# How do you calculate the derivative of #intarctan t dt # from #[2,1/x]#?

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Use the Fundamental Theorem of Calculus, Part 1 and the chain rule.

In this problem we have a composition, so we need the chain rule:

So the chain rule gives us:

In this case:

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To calculate the derivative of ∫arctan(t) dt from [2, 1/x], where x is a variable:

- First, find the antiderivative of arctan(t), which is ∫arctan(t) dt = t * arctan(t) - (1/2) * ln(1 + t^2) + C, where C is the constant of integration.
- Evaluate the antiderivative at the upper bound (1/x) and subtract from it the value of the antiderivative at the lower bound (2).

So, the derivative of ∫arctan(t) dt from [2, 1/x] is:

d/dx [∫arctan(t) dt from 2 to 1/x] = [1/x * arctan(1/x) - (1/2) * ln(1 + (1/x)^2)] - [2 * arctan(2) - (1/2) * ln(1 + 2^2)]

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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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