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Derivative of an inverse trig

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data
    tan^-1(x/(1-x^2)^1/2) find the derivative

    the problem comes from 3g from MIT's PDF I found
    http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-implicit-differentiation-and-inverse-functions/problem-set-2/MIT18_01SC_pset5sol.pdf [Broken]

    here is the solution key
    http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-implicit-differentiation-and-inverse-functions/problem-set-2/MIT18_01SC_pset5prb.pdf [Broken]

    2. Relevant equations

    The solution key says if y = x/(1-x^2)^1/2 then y' = (1-x^2)^-3/2 When I do it, y' comes out differently. This is how I attempt to solve it.

    3. The attempt at a solution

    Original problem: tan^-1((x/(1-x^2)^1/2)) what is the derivative with respect to x?

    let y=x/(1-x^2)^1/2 y=x*(1-x^2)^-1/2 using the product rule I get:
    y'=(1)(1-x^2)^-1/2 + x(-1/2)((1-x^2)^-3/2)(-2)
    y'=(1-x^2)^-1/2 + (x^2)(1-x^2)^-3/2
    This looks a lot different than the y' stated in the solution manual: y' = (1-x^2)^-3/2

    Am I on the right track? Is the solution manual wrong? or am I missing a step? Thank you ahead.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 16, 2011 #2

    hunt_mat

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    Homework Helper

    You need to use the chain rule, you have:
    [tex]
    f(x)=\tan^{-1}\left[ \frac{x}{\sqrt{1-x^{2}}} \right]
    [/tex]
    write (as you did) y=x/(1-x^2)^1/2 and use the chain rule:
    [tex]
    \frac{df}{dx}=\frac{df}{dy}\frac{dy}{dx}
    [/tex]
    I believe you have already computed the second factor.
     
  4. Aug 16, 2011 #3
    Thank you for the reply. However my question is why is my computed derivative of the inside term different than the answer key?
     
  5. Aug 16, 2011 #4

    hunt_mat

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    Homework Helper

    It's not different at all, take out a factor of [itex]1/\sqrt{1-x^{2}}[/itex] and you will see (with a little work) why they are the same.
     
  6. Aug 16, 2011 #5
    Oh nice. I was wondering what I could do. Ty.
     
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