Results 1 to 8 of 8

Math Help - tricky derivative?

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    98

    tricky derivative?

    Find the derivative of
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    We use the rule that

    a^r=e^{r\ln a}

    for a>0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    Posts
    98
    Quote Originally Posted by Scott H View Post
    We use the rule that

    a^r=e^{r\ln a}

    for a>0.
    yea I've been doing that but I'm not getting the right answer.. Must be doing something wrong.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Feb 2010
    Posts
    148
    Thanks
    2
    Quote Originally Posted by tbenne3 View Post
    Find the derivative of
    \ln{y}=4^x\ln{x}

    \frac{1}{y}\,\frac{dy}{dx}=4^x\,\frac{1}{x}+\ln{x}  \,(4^x\,\ln{4})

    Are you able to finish from there?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    I imagined that that differential equation would be a bit tricky. We can rewrite the function as

    y=x^{4^x}=e^{4^x\ln x}

    and differentiate using the Chain Rule.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Feb 2010
    Posts
    98
    Quote Originally Posted by ione View Post
    \ln{y}=4^x\ln{x}

    \frac{1}{y}\,\frac{dy}{dx}=4^x\,\frac{1}{x}+\ln{x}  \,(4^x\,\ln{4})

    Are you able to finish from there?
    honestly.. no.. our teacher doesn't really know how to teach and I haven't had time to sit down and teach it to myself yet
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    651
    Thanks
    1
    Awards
    1
    I recommend Scott's method.

    Change the function to:

    y = e^{4^x\ln{x}}

    Then you differentiate:

    y' = e^{4^x\ln{x}}\frac{d}{dx}[4^x\ln{x}]

    All you use now is the product rule.

    y' = e^{4^x\ln{x}}\left(4^x\log{4}\ln{x} + \frac{4^x}{x}\right)

    We remember that: e^{4^x\ln{x}} = x^{4^x}

    y' = 4^xx^{4^x}\left(\ln{x}\log{4} + \frac{1}{x}\right)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Feb 2010
    Posts
    98
    Quote Originally Posted by Aryth View Post
    I recommend Scott's method.

    Change the function to:

    y = e^{4^x\ln{x}}

    Then you differentiate:

    y' = e^{4^x\ln{x}}\frac{d}{dx}[4^x\ln{x}]

    All you use now is the product rule.

    y' = e^{4^x\ln{x}}\left(4^x\log{4}\ln{x} + \frac{4^x}{x}\right)

    We remember that: e^{4^x\ln{x}} = x^{4^x}

    y' = 4^xx^{4^x}\left(\ln{x}\log{4} + \frac{1}{x}\right)
    thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tricky Derivative Question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 22nd 2011, 09:40 PM
  2. Tricky Derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 15th 2009, 04:23 AM
  3. Tricky Derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 12th 2009, 10:31 AM
  4. Derivative, really tricky one!! Am I right???
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 12th 2008, 08:52 PM
  5. Tricky derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 12th 2007, 03:57 AM

Search Tags


/mathhelpforum @mathhelpforum