Results 1 to 2 of 2

Math Help - Derivative of Inverse Trigonometric Function

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    42

    Derivative of Inverse Trigonometric Function



    Find the derivative...

    I know d/dx(arctan u)= u'/1+u^2

    Can anyone help with this?
    Last edited by tradar; March 28th 2009 at 07:18 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    You are correct that \frac{d}{dx}\arctan u=\frac{u'}{1+u^2}. Applying this formula to y, we obtain

    \frac{dy}{dx}=\frac{d}{dx}\left(\arctan\left(\frac  {x}{5}\right)-\frac{1}{2(x^2+25)}\right)
    =\frac{d}{dx}\arctan\left(\frac{x}{5}\right)-\frac{d}{dx}\left(\frac{1}{2(x^2+25)}\right)
    =\frac{\frac{d}{dx}\left(\frac{x}{5}\right)}{1+\le  ft(\frac{x}{5}\right)^2}+\frac{1}{(2(x^2+25))^2}\c  dot(4x)
    =\frac{\frac{1}{5}}{1+\frac{x^2}{25}}+\frac{x}{(x^  2+25)^2}
    =\frac{5}{25+x^2}+\frac{x}{(x^2+25)^2}
    =\frac{5(x^2+25)}{(x^2+25)^2}+\frac{x}{(x^2+25)^2}
    =\frac{5x^2+125}{(x^2+25)^2}+\frac{x}{(x^2+25)^2}
    =\frac{5x^2+x+125}{(x^2+25)^2}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: November 20th 2009, 02:16 AM
  2. Replies: 8
    Last Post: September 13th 2009, 04:34 AM
  3. Inverse trigonometric function
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 28th 2009, 11:31 AM
  4. Replies: 4
    Last Post: November 3rd 2008, 07:09 PM
  5. Inverse trigonometric function
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: October 15th 2008, 06:08 AM

Search Tags


/mathhelpforum @mathhelpforum