Question

`d/dx[int_1^tanx sqrt(tan^-1t)dt]`?

Answer 1

By the FTC, Part 1:

`d/dx (int_a^x f(t) dt) = f(x)`

And by the chain rule:

`d/dx (int_a^(g(x)) f(t) dt) = f(g(x)) (dg)/(dx)`

So:

`d/dx(int_1^tanx sqrt(arctan t ) \ dt)`

`= sqrt(arctan (tan x)) d/dx(tan x)`

`= sqrtx sec^2 x`

Answer 2

`d/dx \ int_1^tanx sqrt(tan^-1t) \ dt = sec^2xsqrt(x)`

Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

`d/dx \ int_a^x \ f(t) \ dt = f(x)` for any constant `a`

(ie the derivative of an integral gives us the original function back).

We are asked to find:

`d/dx \ int_1^tanx sqrt(tan^-1t) \ dt`

(notice the upper and lower bounds are not in the correct format for the FTOC to be applied, directly).

We can manipulate the definite integral using a substitution and the chain rule. Let:

`u=tanx => (du)/dx = sec^2x`

The substituting into the integral we get:

`d/dx \ int_1^tanx sqrt(tan^-1t) \ dt = d/dx \ int_1^u sqrt(tan^-1t) \ dt`
`" " = (du)/dx * d/(du) \ int_1^u sqrt(tan^-1t) \ dt`
`" " = sec^2x* d/(du) \ int_1^u sqrt(tan^-1t) \ dt`

And now the derivative of the integral is in the correct form for the FTOC to be applied, giving:

`d/(du) \ int_1^u sqrt(tan^-1t) \ dt = sqrt(tan^-1u)`

And so:

`d/dx \ int_1^tanx sqrt(tan^-1t) \ dt = sec^2x* sqrt(tan^-1u)`
`" " = sec^2x* sqrt(tan^-1(tanx))`
`" " = sec^2xsqrt(x)`