Results 1 to 4 of 4

Math Help - chain rule help!

  1. #1
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517

    chain rule help!

    differentiate

    <br /> <br />
f(x) = tan(\frac{x}{73})<br /> <br />

    any ideas? i know i have to use the chain rule but confused on the sec^2 and x/73
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by jvignacio View Post
    differentiate

    <br /> <br />
f(x) = tan(\frac{x}{73})<br /> <br />


    any ideas? i know i have to use the chain rule but confused on the sec^2 and x/73
    well, chain rule says that if h(x)=f(g(x)), then h'(x)=f'(g(x))\cdot g'(x) of course in the assumption that f and g are differentiable..

    so

    your f'(x) = \left[\sec^2{\left(\frac{x}{73}\right)}\right] \cdot \frac{1}{73}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    Quote Originally Posted by kalagota View Post
    well, chain rule says that if h(x)=f(g(x)), then h'(x)=f'(g(x))\cdot g'(x) of course in the assumption that f and g are differentiable..

    so

    your f'(x) = \left[\sec^2{\left(\frac{x}{73}\right)}\right] \cdot \frac{1}{73}
    so the differentiation of x/73 is 1/73?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Yes, \frac{1}{73} is just a constant multiplied to the variable x.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: November 9th 2010, 01:40 AM
  2. Chain Rule Inside of Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 22nd 2009, 08:50 PM
  3. Replies: 5
    Last Post: October 19th 2009, 01:04 PM
  4. Replies: 3
    Last Post: May 25th 2009, 06:15 AM
  5. Replies: 2
    Last Post: December 13th 2007, 05:14 AM

Search Tags


/mathhelpforum @mathhelpforum