Results 1 to 5 of 5

Math Help - product of inverse trig function

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    162

    product of inverse trig function

    I need help with taking the derivative with y=\frac{arctanx}{x}. I cannot seem to even begin. In the solution manual it does the product derivative which I try doing, but I cannot figure out how to do it with the inverse of the tan function. I don't know what to do and the solution manual just jumps to the last step before the solution to "show" its work. Any hints?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,959
    Thanks
    1783
    Awards
    1

    Re: product of inverse trig function

    Quote Originally Posted by Barthayn View Post
    I need help with taking the derivative with y=\frac{arctanx}{x}.
    y=\frac{\arctan(x)}{x}=(x^{-1})\arctan(x)} use product rule.

    \frac{d[\arctan(x)]}{dx}=\frac{1}{1+x^2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    162

    Re: product of inverse trig function

    I understand I have to use the product rule, but I don't see how \frac{d[\arctan(x)]}{dx}=\frac{1}{1+x^2}.

    I get {tany}=x
    sec^2y\frac{d}{dx}=1
    \frac{d}{dx}=cos^2y

    Don't see how cos^2y = \frac{1}{1+x^2}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,959
    Thanks
    1783
    Awards
    1

    Re: product of inverse trig function

    Quote Originally Posted by Barthayn View Post
    I understand I have to use the product rule, but I don't see how \frac{d[\arctan(x)]}{dx}=\frac{1}{1+x^2}.
    I get {tany}=x
    sec^2y\frac{d}{dx}=1
    \frac{d}{dx}=cos^2y
    Don't see how cos^2y = \frac{1}{1+x^2}
    \sec^2(y)=1+\tan^2(y)=1+x^2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2009
    Posts
    162

    Re: product of inverse trig function

    Quote Originally Posted by Plato View Post
    \sec^2(y)=1+\tan^2(y)=1+x^2
    Thank you so very much! I was on this problem for a long time yesterday and today and I never thought to use secy where it was I always used cosy and got stuck so many times! Thank you!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: April 26th 2011, 06:08 AM
  2. inverse trig function
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: December 7th 2009, 09:25 AM
  3. inverse trig function help,
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 10th 2009, 05:12 AM
  4. How to do this Inverse Trig Function?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 26th 2008, 05:30 PM
  5. inverse trig function
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: June 1st 2007, 02:52 PM

Search Tags


/mathhelpforum @mathhelpforum