Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By hollywood

Math Help - Advanced Proof of Chain Rule

  1. #1
    Member
    Joined
    Aug 2011
    Posts
    84

    Advanced Proof of Chain Rule

    assuming differentiability of functionsthe substitution t=g(x,y) converts F(t) into f(x,y) where f(x,y)=F[(g(x,y))]
    Show that

    \frac{\partial f}{\partial x}=F'[(g(x,y))]\frac{\partial g}{\partial x}, \frac{\partial f}{\partial y}=F'[(g(x,y))]\frac{\partial g}{\partial y}

    am I supposed to use a first order Taylor function with error term?
    for instance h(a+y)-h(a)=f[g(a+y)]-f[g(a)]=f(b+v)-f(b) can be used to prove that h is differentiable at a (I guess it means that the limit exists) and the total derivative h'(a) is equal to the composition f'(b)=g'(a)
    Last edited by mathnerd15; February 28th 2014 at 04:23 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    Re: Advanced Proof of Chain Rule

    I think you need to go back to the definition of the derivative (as a limit of the difference quotient). For the first equation, start with:

    \frac{f(x,y_0)-f(x_0,y_0)}{x-x_0} = \frac{F(g(x,y_0))-F(g(x_0,y_0))}{g(x,y_0)-g(x_0,y_0)}\cdot\frac{g(x,y_0)-g(x_0,y_0)}{x-x_0}

    The left-hand side and the second fraction on the right-hand side are good. There's a problem with the first fraction on the right-hand side, though, since g(x,y_0) could be equal to g(x_0,y_0) for any or even all x in an interval around x_0. The trick is to define a function Q(y) which is equal to \frac{F(y)-F(g(x_0,y_0))}{y-g(x_0,y_0)} when y \ne g(x_0,y_0) and f'(g(x_0,y_0)) when y=g(x_0,y_0). So you have:

    \frac{f(x,y_0)-f(x_0,y_0)}{x-x_0} = Q(g(x,y_0))\cdot\frac{g(x,y_0)-g(x_0,y_0)}{x-x_0}

    because when g(x,y_0) \ne g(x_0,y_0), the g(x,y_0)-g(x_0,y_0) terms cancel, and when g(x,y_0) = g(x_0,y_0), both sides are zero. Now you can take the limit as x goes to x_0.

    - Hollywood
    Thanks from mathnerd15
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2011
    Posts
    84

    Re: Advanced Proof of Chain Rule

    Thanks so much! this is p.275 in vol. 2 Apostol, I'm not sure if I should try analysis instead. My lungs are failing and don't work so well.... so I guess it's adequate to say that when y=g(x_0,y_0), both sides are 0 since you multiply the left by the denominator of the right, therefore the limit is 0. And when g(x,y_0) \ne g(x_0,y_0) then you have holding y as constant
    \lim_{x\rightarrow x_0}\frac{f(x,y_0)-f(x_0,y_0)}{x-x_0}=\frac{\partial f}{\partial x}= \lim_{x\rightarrow x_0} \frac{F(g(x,y_0))-F(g(x_0,y_0))}{x-x_0}=F'[g(x,y)]\frac{\partial g}{\partial x}

    and also holding x constant we can define a function Q(x) similarly as well so that

    \lim_{y\rightarrow y_0}\frac{f(x_0,y_)-f(x_0,y_0)}{y-y_0}=\frac{\partial f}{\partial y}= \lim_{y\rightarrow y_0} \frac{F(g(x_0,y))-F(g(x_0,y_0))}{y-y_0}=F'[g(x,y)]\frac{\partial g}{\partial y}

    and for the special case where
    F(t)=e^{sint}, g(x,y)=cos(x^{2}+y^{2}), \frac{\partial f}{\partial x}=-e^{sin(cos(x^{2}+y^{2}))}cos(cos(x^{2}+y^{2}))sin(  x^{2}+y^{2}))(2x)
    which has level curves of concentric circles about the origin
    Last edited by mathnerd15; April 10th 2014 at 12:38 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Chain rule proof
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 15th 2011, 07:26 AM
  2. [SOLVED] A small part of a chain rule proof I don't get
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: January 11th 2011, 04:03 AM
  3. multivariable chain rule proof
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 11th 2010, 05:28 PM
  4. Proof Involving Partial Derivatives Chain Rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 26th 2010, 01:03 AM
  5. Chain Rule - Proof Using...
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 25th 2009, 04:59 PM

Search Tags


/mathhelpforum @mathhelpforum