Results 1 to 7 of 7

Math Help - Derivative of Natural Logarithmic Funtion

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    11

    Derivative of Natural Logarithmic Funtion

    Find the equation of the tangent to the curve defined by y=ln(1+2^(-x)) at the point where x=0.

    Help will be appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,408
    Take the derivative of y and evaluate it at x = 0. This will be your slope value. To find the equation to the tangent, you use the formula:
    y - y_{1} = m(x - x_{1})
    where (x_{1}, y_{1}) is the point you're concerned with.

    As for taking the derivative, make sure you're careful about using your chain rule! Show us what you've gotten down and we'll help you from there
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by rhhs11 View Post
    Find the equation of the tangent to the curve defined by y=ln(1+2^(-x)) at the point where x=0.

    Help will be appreciated

    Note that 2^{-x}=e^{-xln(2)} if we take the derivative we get.

    \frac{d}{dx}2^{-x}=-ln(2)e^{-xln(2)}=-ln(2)2^{-x}<br />

    So now back to your derivative

    \frac{dy}{dx}=\frac{-ln(2)2^{-x}}{1+2^{-x}}

    evaluating at zero we get...

    \frac{dy}{dx}|_{x=0}=\frac{-ln(2)2^{-0}}{1+2^{-0}}=\frac{-ln(2)}{2}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2008
    Posts
    11
    So i got
    y` = 1 / (1+e^-x) * e^-x * (-1)
    = - e^(-x) / (1+ e^-x)

    mtangent @ x=0 = -1 / (1+1)
    = -1/2

    y - ln2 = 1/2(x-0)
    2y - 2ln2 = x
    x - 2y + 2ln2 = 0

    Im not sure if that's right but looks fair enough. I would like any advice though.

    I also have another question where it says: Find the equation of the tangent to the curve defined by y = e^x that is perpendicular to the line 3x + y = 1.

    Thanks a lot in advance!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2008
    Posts
    11
    Im sorry but the question is y = ln(1+e^-x)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by rhhs11 View Post
    So i got
    y` = 1 / (1+e^-x) * e^-x * (-1)
    = - e^(-x) / (1+ e^-x)

    mtangent @ x=0 = -1 / (1+1)
    = -1/2

    y - ln2 = 1/2(x-0)
    2y - 2ln2 = x
    x - 2y + 2ln2 = 0

    Im not sure if that's right but looks fair enough. I would like any advice though.

    I also have another question where it says: Find the equation of the tangent to the curve defined by y = e^x that is perpendicular to the line 3x + y = 1.

    Thanks a lot in advance!

    How did your function change from

    y=\ln(1+2^{-x}) to y=\ln(1+e^{-x})
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2008
    Posts
    11
    It was a mistake when i was entering the question
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: August 28th 2011, 10:16 AM
  2. Replies: 6
    Last Post: July 11th 2010, 09:11 AM
  3. the natural logarithmic function?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: November 7th 2009, 04:00 PM
  4. Replies: 1
    Last Post: April 15th 2009, 03:51 AM
  5. Finding the derivative of a funtion
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 8th 2007, 07:29 PM

Search Tags


/mathhelpforum @mathhelpforum