Question

What is the derivative of `arctan sqrt [ (1-x)/(1+x)]`?

Answer 1

I got:

`-(1)/(2sqrt(1-x^2))`

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The derivative of `arctanu` is `(1/(1+(u(x))^2))((du(x))/(dx))`.

So since `u(x) = sqrt((1-x)/(1+x))`:

`d/(dx)(arctansqrt((1-x)/(1+x)))`

`= 1/(1+(1-x)/(1+x))*(1/(2sqrt((1-x)/(1+x))))*[((1+x)*(-1) - (1-x)*(1))/(1+x)^2]`

You can see there are multiple chain rules here.

`= [(-1cancel(-x)-1cancel(+x))/(1+x)^2][1/(2sqrt((1-x)/(1+x))(1+(1-x)/(1+x)))]`

Multiply in the `sqrt((1-x)/(1+x))` and cancel out the `2`:

`= [(-cancel(2))/(1+x)^2][1/(cancel(2)(sqrt((1-x)/(1+x))+((1-x)/(1+x))^"3/2"))]`

Merge the fractions:

`= -(1)/((1+x)^"4/2"(sqrt((1-x)/(1+x))+((1-x)/(1+x))^"3/2"))`

Multiply in the `(1+x)^"4/2"`:

`= -(1)/(sqrt(1-x)*(1+x)^"3/2"+(1-x)^"3/2"*sqrt(1+x)`

Simplify by turning `sqrt(1-x)` and `sqrt(1+x)` into `sqrt(1-x^2)`:

`= -(1)/((1-x)^"1/2"(1+x)^"1/2"(1+x)+(1-x)(1-x)^"1/2"(1+x)^"1/2"`

Factor out the `sqrt(1-x^2)` and cancel what's now inside:

`= -(1)/(sqrt(1-x^2)*(1+x)+(1-x)*sqrt(1-x^2))`

`= -(1)/(sqrt(1-x^2)(1cancel(+x)+1cancel(-x)))`

`= color(blue)(-(1)/(2sqrt(1-x^2)))`

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Wolfram Alpha gives the derivative as (assuming `x > 0`)

`= -(1)/(2sqrt(1-x^2))`,

and in general, gives

`sqrt((1-x)/(1+x))/(2(1-x))`.