Question

What is the derivative of `arctan [(1-x)/(1+x)]^(1/2)`?

Answer 1

`dy/dx = -1/(2(1+x)sqrt((1-x)/(1+x)))`

Explanation:

When tackling the derivative of inverse trig functions. I prefer to rearrange and use Implicit differentiation as I always get the inverse derivatives muddled up, and this way I do not need to remember the inverse derivatives. If you can remember the inverse derivatives then you use the chain rule.

Let `y=arctan sqrt((1-x)/(1+x)) <=> tany=sqrt((1-x)/(1+x))`
`:. tan^2y=(1-x)/(1+x)`

Differentiate Implicitly, and applying product rule:

`2tanysec^2ydy/dx = ((1+x)(-1) - (1-x)(1))/(1+x)^2`
`:. 2tanysec^2ydy/dx = (-1-x-1+x)/(1+x)^2`
`:. 2tanysec^2ydy/dx = -2/(1+x)^2`
`:. tanysec^2ydy/dx = -1/(1+x)^2`
`:. sqrt((1-x)/(1+x))sec^2ydy/dx = -1/(1+x)^2` ..... [1]

Using the `sec"/"tan` identity;

`(1-x)/(1+x)+1=sec^2y`
`:. sec^2y = ((1-x)+(1+x))/(1+x)`
`:. sec^2y = 2/(1+x)`

Substituting into [1]
`sqrt((1-x)/(1+x))(2/(1+x))dy/dx = -1/(1+x)^2`
`:. 2sqrt((1-x)/(1+x))dy/dx = -1/(1+x)`
`:. dy/dx = -1/(2(1+x)sqrt((1-x)/(1+x)))`