# Arctan Derivative

1. Jul 3, 2014

### emmaerin

1. The problem statement, all variables and given/known data
Find the derivative of the function. Simplify where possible.
y= arctan ( (1+x)/(1-x))^1/2

2. Relevant equations

d/dx (arctan x) = 1/(1+x^2)

3. The attempt at a solution
I'm really not sure where to even begin, so any help would be greatly appreciated!!

2. Jul 3, 2014

### SammyS

Staff Emeritus
Hello emmaerin. Welcome to PF !

You will need to use the Chain Rule.

Also, the Quotient Rule will likely come into play.

To remove any ambiguity regarding the function you are differentiating, is it

$\displaystyle y= \arctan\left(\left(\frac{1+x}{1-x}\right)^{1/2}\right)\ ?$

3. Jul 3, 2014

### emmaerin

Thank you! And yes, that is the correct function.

4. Jul 3, 2014

### Pranav-Arora

Use the substitution $x=\cos(2\theta)$ to simplify the function and then find the derivative.

5. Jul 3, 2014

### emmaerin

Could you please explain further? Where did the cos come from? (I apologize if this is basic; I've missed a week's worth of classes due to a lung infection and I'm desperately trying to catch up.)

6. Jul 3, 2014

### Pranav-Arora

Notice that the domain of the function is $[-1,1)$, this allows us to use the substitution $x=\cos(2\theta)$. Algebra simplifies greatly with this.
$$y=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)=\arctan\left(\sqrt{\frac{1+\cos(2\theta)}{1-\cos(2\theta)}}\right)$$
Using the identity: $\cos(2\theta)=2\cos^2\theta-1=1-2\sin^2\theta$,
$$y=\arctan\left(\sqrt{\frac{\cos^2\theta}{\sin^2\theta}}\right) = \arctan \left(\tan\left(\frac{\pi}{2}-\theta\right)\right)=\frac{\pi}{2}-\theta$$
Substitute back $\theta=\frac{1}{2}\cos^{-1}x$ and finding the derivative of this is easier than the original function.

Or you can simply use chain rule find the derivative.

7. Jul 3, 2014

### ehild

It would simplify the problem, but you do not need it. Just apply Chain Rule. Do you know what it is? You might see

ehild

Last edited by a moderator: Sep 25, 2014
8. Jul 3, 2014

### emmaerin

This is where I am so far, using the chain rule. Could someone help me figure out where I'm going wrong/what comes next? Thanks!!

y = arctan ( (1+x)/(1-x) )^1/2

1. y' = 1/ (1 + (1+x)/(1-x)) * 1/2 ( (1+x)/(1-x) )^-1/2 * ( (1-x)(1) - (1+x)(-1) ) / (1-x)^2

2. y' = (1-x)/2 * 1/2 ( (1-x)/(1+x) )^1/2 * 2 / (1-x)^2

3. y'=1/2 ( (1-x)/(1+x) ) ^1/2 * 1 / (1-x)

9. Jul 3, 2014

### D H

Staff Emeritus
The next step is to simplify. Hint: For x<1, 1-x>0 so 1-x = sqrt((1-x)^2) if x<1.

10. Jul 3, 2014

### emmaerin

1 / (2 * (1 - x^2)^1/2 ) ?

11. Jul 4, 2014

### D H

Staff Emeritus
Very good.