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Homework Help: Arctan Derivative

  1. Jul 3, 2014 #1
    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. jcsd
  3. Jul 3, 2014 #2


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    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]
  4. Jul 3, 2014 #3
    Thank you! And yes, that is the correct function.
  5. Jul 3, 2014 #4
    Use the substitution ##x=\cos(2\theta)## to simplify the function and then find the derivative.
  6. Jul 3, 2014 #5
    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.)
  7. Jul 3, 2014 #6
    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.
    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.
  8. Jul 3, 2014 #7


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

    Last edited by a moderator: Sep 25, 2014
  9. Jul 3, 2014 #8
    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)
  10. Jul 3, 2014 #9

    D H

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    The next step is to simplify. Hint: For x<1, 1-x>0 so 1-x = sqrt((1-x)^2) if x<1.
  11. Jul 3, 2014 #10
    1 / (2 * (1 - x^2)^1/2 ) ?
  12. Jul 4, 2014 #11

    D H

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    Very good.
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