Results 1 to 4 of 4

Math Help - Partial Derivatives

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    19

    Arrow Partial Derivatives

    Can someone show me the steps to solving these...

    f(x,y) = x ln y + y ln x

    df/dx =

    df/dy =

    df/dx^2 =

    df/dy^2 =

    df/dydx =

    df/dxdy =

    I believe df/dx = ln y + y/x

    and df/dy = x/y + lnx

    how would I find the critical points from that?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by vodka View Post
    Can someone show me the steps to solving these...

    f(x,y) = x ln y + y ln x

    df/dx =
    df/dx is the partial derivative of f(x, y) with respect to x; every other variable is considered a constant, so let's derive:

    \frac{df}{dx} = \ln{y}*x'+ y*(\ln{x})'

    Derivative w.r.t x of x is just 1. The derivative of ln(x) is 1/x

    \frac{df}{dx} = \ln{y} + \frac{y}{x}

    That simple? Yep. I believe you can solve the rest of the problem.
    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 vodka View Post
    Can someone show me the steps to solving these...

    I believe df/dx = ln y + y/x

    and df/dy = x/y + lnx

    how would I find the critical points from that?
    Finding the critical points is not easy.

    I don't think it can be solved analytically(I could be wrong)

    I drew a graph to see what it looked like and here it is.

    Partial Derivatives-ln.jpg

    So we want to solve the system of equations

    \ln(y)+\frac{y}{x}=0 \iff \frac{x}{y}=-\frac{1}{\ln(y)}

     \frac{x}{y}+\ln(x)=0 subbing the first into the 2nd we get

    -\frac{1}{\ln(y)}+\ln(x)=0 \iff -1 +\ln(x)\ln(y)=0 \iff ln(x)\ln(y)=1

    finally we get \ln(y)=\frac{1}{\ln(x)} \iff y=e^{\frac{1}{\ln(x)}}

    If we sub this back into the first equation we get

    \frac{1}{\ln(x)}+\frac{e^{\frac{1}{\ln(x)}}}{x}=0

    Finally we can use newtons method on the last equation

    x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

    Based on the graph from above I used x_0=0.5

    After applying newtons method three times I got the values(on a calculator)

    0.5,0.3628837,0.36784455,0.36787943

    Oddly enough when I repeated the process but eliminted the x's and used newtons method again I got the same value for y.

    so the critical numbers are (\approx 0.36787943,\approx 0.36787943 )

    I hope this helps.

    P.S if anyone knows another way I would love to see it

    Thanks, The Empty Set.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    I have been thinking about this and just noticed that

    \ln(y)+\frac{y}{x}=0 \iff x=-\frac{y}{\ln(y)} and

    \frac{x}{y}+\ln(x)=0 \iff y=-\frac{x}{\ln(x)}

    Are inverses of each other. So they are symmetric over the line y=x

    This means that any of their intersections would have to be on the line y=x.

    So if I sub this into the first equation I get

    x=-\frac{x}{\ln(x)} \iff \ln(x)=-1 \iff x=e^{-1}=\frac{1}{e} \approx 0.36787

    So the exact value is x=y=\frac{1}{e}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with Partial Derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 15th 2011, 05:07 PM
  2. Partial Derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 22nd 2011, 01:18 PM
  3. partial derivatives
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 24th 2011, 05:22 AM
  4. Partial derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 20th 2010, 04:04 AM
  5. partial derivatives
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 7th 2008, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum