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

Jun 5, 2015

I got:

$- \frac{1}{2 \sqrt{1 - {x}^{2}}}$

The derivative of $\arctan u$ is $\left(\frac{1}{1 + {\left(u \left(x\right)\right)}^{2}}\right) \left(\frac{\mathrm{du} \left(x\right)}{\mathrm{dx}}\right)$.

So since $u \left(x\right) = \sqrt{\frac{1 - x}{1 + x}}$:

$\frac{d}{\mathrm{dx}} \left(\arctan \sqrt{\frac{1 - x}{1 + x}}\right)$

$= \frac{1}{1 + \frac{1 - x}{1 + x}} \cdot \left(\frac{1}{2 \sqrt{\frac{1 - x}{1 + x}}}\right) \cdot \left[\frac{\left(1 + x\right) \cdot \left(- 1\right) - \left(1 - x\right) \cdot \left(1\right)}{1 + x} ^ 2\right]$

You can see there are multiple chain rules here.

$= \left[\frac{- 1 \cancel{- x} - 1 \cancel{+ x}}{1 + x} ^ 2\right] \left[\frac{1}{2 \sqrt{\frac{1 - x}{1 + x}} \left(1 + \frac{1 - x}{1 + x}\right)}\right]$

Multiply in the $\sqrt{\frac{1 - x}{1 + x}}$ and cancel out the $2$:

$= \left[\frac{- \cancel{2}}{1 + x} ^ 2\right] \left[\frac{1}{\cancel{2} \left(\sqrt{\frac{1 - x}{1 + x}} + {\left(\frac{1 - x}{1 + x}\right)}^{\text{3/2}}\right)}\right]$

Merge the fractions:

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

Multiply in the ${\left(1 + x\right)}^{\text{4/2}}$:

 = -(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:

$= - \frac{1}{\sqrt{1 - {x}^{2}} \cdot \left(1 + x\right) + \left(1 - x\right) \cdot \sqrt{1 - {x}^{2}}}$

$= - \frac{1}{\sqrt{1 - {x}^{2}} \left(1 \cancel{+ x} + 1 \cancel{- x}\right)}$

$= \textcolor{b l u e}{- \frac{1}{2 \sqrt{1 - {x}^{2}}}}$

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

$= - \frac{1}{2 \sqrt{1 - {x}^{2}}}$,

and in general, gives

$\frac{\sqrt{\frac{1 - x}{1 + x}}}{2 \left(1 - x\right)}$.