1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inverse trig

  1. Sep 27, 2009 #1
    x^2+xsin^-1y=ye^x

    implict dif

    2x+xy'/(1-x^2)^(1/2)+sin^(-1)y=ye^x+y'e^x
    xy'/(1-x^2)^(1/2)-y'e^x=ye^x-sin^(-1)y-2x
    y'=(ye^x-sin^(-1)y-2x)/(x/(1-x^2)^(1/2)-e^x)
    can somone please verify
     
  2. jcsd
  3. Sep 27, 2009 #2

    ideasrule

    User Avatar
    Homework Helper

    You're differentiating sin^-1y with respect to x, so you use the chain rule and first differentiate with respect to y, then derive y with respect to x, then multiply the two derivatives together. That would give you 1/sqrt(1-y^2) * dy/dx, not 1/sqrt(1-x^2) *dy/dx
     
  4. Sep 28, 2009 #3
    ah yes that was a typeo on my part my solution is

    (ye^x-sin^-1y-2x)/(x/(1-y^2)^(1/2)-e^x)
    my book is showing
    (ye^x-2x-sin^-1y)/(x/((1-y^2)^(1/2)-e^x))

    which divides (1-y^2)^(1/2)-e^x into x not just x/(1-y^2)^(1/2) i dont believe these solutions are equal
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Inverse trig
  1. Inverse Trig (Replies: 2)

  2. Inverse trig. (Replies: 3)

  3. Inverse trig. (Replies: 5)

  4. Inverse trig value (Replies: 1)

Loading...