Results 1 to 2 of 2

Math Help - Derivations of Inverse Trig Functions

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    32

    Derivations of Inverse Trig Functions

    My professor did not explain how to find the derivation of an inverse trig function very well, so I need some basic help in understanding what steps to take. I understand the derivations, and the need to use the chain rule and sometimes the product/quotient rule, but I don't understand these problems:

    y = arcsin(2x +1)

    Would this be the next step? But then what?: y' = 1/ (1 - ((2x+1)^2)

    y = arccos(e^2x)

    Other than plugging this into the formula for the derivative of arccos, I don't know what to do...


    Also, is this implicit derivation problem? If so, why?:

    ysin(x^2) = xsin(y^2)

    product rule

    ycosx^2 + y'sinx^2 = xcosy^2(2yy') +siny^2

    y'sinx^2 - 2xyy'cosy^2 = -ycosx^2 +siny^2

    ect, take out y' and solve...
    Last edited by Maziana; September 30th 2009 at 03:04 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,888
    Thanks
    682
    Quote Originally Posted by Maziana View Post
    My professor did not explain how to find the derivation of an inverse trig function very well, so I need some basic help in understanding what steps to take. I understand the derivations, and the need to use the chain rule and sometimes the product/quotient rule, but I don't understand these problems:

    y = arcsin(2x +1)

    Would this be the next step? But then what?: y' = 1/ (1 - ((2x+1)^2)

    y = arccos(e^2x)

    Other than plugging this into the formula for the derivative of arccos, I don't know what to do...


    Also, is this implicit derivation problem? If so, why?:

    ysin(x^2) = xsin(y^2)

    product rule

    ycosx^2 + y'sinx^2 = xcosy^2(2yy') +siny^2

    y'sinx^2 - 2xyy'cosy^2 = -ycosx^2 +siny^2

    ect, take out y' and solve...
    y = \arcsin(2x+1)

    y' = \frac{2}{\sqrt{1 - (2x+1)^2}}


    y = \arccos(e^x)

    y' = -\frac{e^x}{\sqrt{1 - e^{2x}}}


    yes on implicit ...

    y\sin(x^2) = x\sin(y^2)

    y \cdot 2x\cos(x^2) + \sin(x^2) \cdot y' = x \cdot 2y \cdot y' \cdot \cos(y^2) + \sin(y^2)

    solve for y'
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inverse of Trig functions
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: September 5th 2011, 04:03 AM
  2. Inverse Trig Functions
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: September 20th 2009, 10:08 PM
  3. Inverse Trig functions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 15th 2009, 08:23 PM
  4. Inverse Trig Functions
    Posted in the Trigonometry Forum
    Replies: 0
    Last Post: December 7th 2008, 09:24 AM
  5. Inverse Trig Functions and Advance Trig
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 24th 2008, 03:13 PM

Search Tags


/mathhelpforum @mathhelpforum