Derivative of Arctan Function: Get Help!

[emmaerin]

Homework Statement

Find the derivative of the function. Simplify where possible.
y= arctan ( (1+x)/(1-x))^1/2

Homework Equations

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

The Attempt at a Solution

I'm really not sure where to even begin, so any help would be greatly appreciated!

 

[SammyS]

Homework Statement

Find the derivative of the function. Simplify where possible.
y= arctan ( (1+x)/(1-x))^1/2

Homework Equations

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

The Attempt at a Solution

I'm really not sure where to even begin, so any help would be greatly appreciated!

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

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

 

[emmaerin]

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

 

[Saitama]

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

 

[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.)

 

[Saitama]

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.)

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.

 

[ehild]

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.)

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

 

[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)

 

[D H]

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

 

[emmaerin]

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

 

[D H]

Very good.