Question

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

Answer 1

`d/dxint_2^(1/x) arctan t dt = -(arctan (1/x))/x^2`

Explanation:

`d/dxint_2^(1/x) arctan t dt = arctan (1/x) d/dx(1/x)`
`= -(arctan (1/x))/x^2`

Answer 2

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

Explanation:

Fundamental Theorem of Calculus, Part 1 tells us that if `f` is continuous on interval `[a,b]`, if `g` is defined by

`g(x) = int_a^x f(t) dt" "` for `x in [a,b]`

then (`g` is continuous on `[a,b]`)
and (`g` is differentiable on `(a,b)`)

and `g'(x) = f(x)` (which is what we need here).

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

`g(x) = int_2^u arctan t dt` with `u = 1/x`

So the chain rule gives us:

`g'(x) = arctanu (du)/dx`

In this case:

`g'(x) = arctan(1/x) (-1)/x^2`

`= - arctan(1/x)/x^2`