Results 1 to 5 of 5

Math Help - log differntial

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    29

    log differntial

    y= ln(3x+4)^4

    i got
    4 ln (3x+4) then = 4 * 1/ (3x+4) * 3
    4 * 3/ (3x+4) y' = 12/(3x+4)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Ah, \log(x^2) = 2\log(x) \neq (\log(x))^2 which is where you've gone wrong.

    Just differentiate it normally...

    y = (\log(3x + 4))^4.

    then \frac{dy}{dx} = 4(\log(3x+4))^3 \cdot (\log(3x+4))'

    = 4(\log(3x+4))^3 \cdot \frac{3}{3x+4} = \frac{12(\log(3x+4))^3}{3x+4}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    29
    another one

    ln 2/x

    = ln2 - ln x
    ln 2= 1/2 *0 = 0
    ln x = 1/x *1

    so ln 2/x = 1/x ?
    the question above can you do it in the form that i did much easy for me to understand.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Quote Originally Posted by matlondon View Post
    another one

    ln 2/x

    = ln2 - ln x
    ln 2= 1/2 *0 = 0
    ln x = 1/x *1

    so ln 2/x = 1/x ?
    the question above can you do it in the form that i did much easy for me to understand.
    You missed a minus sign in the above, otherwise correct.

    ln 2/x = -1/x

    For your original question... Fine, but learn how to properly write it out as you'll get probably close to 0 for the above type of working. It makes no mathematical sense although I know what you are saying with it.

    y = ln(3x+4)^4

    = 4 * ln(3x+4)^3 * 3/(3x+4)

    = 12 ln(3x+4)/(3x+4)


    I'm sure you are aware that the derivative of x^n = nx^(n-1) * 1 since 1 is the derivative of x.

    Same thing applies for functions.

    Derivative of (some function)^n = n((some function)^(n-1) * derivative of (some function)


    Or in maths notation.

    y = (f(x))^n

    \frac{dy}{dx} = n(f(x))^{n-1} \cdot f'(x)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2009
    Posts
    29
    thx for the time
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. strange differntial equation
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: February 5th 2010, 11:33 PM
  2. with this differntial equation problem
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 9th 2009, 04:08 PM
  3. Help With differntial equation
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: April 30th 2009, 02:27 PM
  4. Differntial Equation Help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 9th 2008, 03:22 AM
  5. Differntial Equations
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 27th 2006, 09:43 AM

Search Tags


/mathhelpforum @mathhelpforum