Results 1 to 2 of 2

Math Help - Evaluation of a partial derivative using the chain rule

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    29

    [SOLVED] Evaluation of a partial derivative using the chain rule

    I was solving a practice test for the GRE math subject test and I came across a problem which I thought was unsolvable. The solution provided in the back makes no sense to me, so I hope someone can shed some light on this situation.

    Let f, g, and h be functions of two variables that are differentiable everywhere such that z=f(x,y) where x and y in turn are functions of u and v. We know that
    x=g(u,v), y=h(u,v) and that g(0,1)=2 and that h(0,1)=1. Let P=(0,1) and Q=(2,1). We are also given that

    \frac{\partial f}{\partial x}\bigg|_Q = 11, \;\; \frac{\partial f}{\partial y}\bigg|_Q = -3, \;\; \frac{\partial g}{\partial u}\bigg|_P = 1, \;\; \frac{\partial h}{\partial u}\bigg|_P = -3, \;\; \frac{\partial g}{\partial v}\bigg|_P = \frac{\partial h}{\partial v}\bigg|_P = 2

    The object is to find \frac{\partial z}{\partial v} \bigg|_P.

    Now note that the partial derivatives given are evaluated at different points, but the solution is not afraid to say that

    \frac{\partial z}{\partial v} \bigg|_P = \frac{\partial f}{\partial x} \bigg|_Q \frac{\partial x}{\partial v} \bigg|_P + \frac{\partial f}{\partial y} \bigg|_Q \frac{\partial y}{\partial v} \bigg|_P = (11)(2) + (-3)(2) = 16

    Now, it makes no sense to me that you can just pick and choose where you evaluate the partials involved in the chain rule... but that's what the solution does. Shouldn't they all be evaluated at P? Why is it allowed to evaluate some at Q and some at P? Does the solution somehow implicitly make use of g(0,1) and h(0,1) which are given?

    Thank you for taking the time to read my question!
    Last edited by eeyore; September 25th 2009 at 12:35 PM. Reason: Changed title to say the problem is solved.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Sep 2009
    Posts
    29
    I answered my own question after some research...

    When evaluating the partial derivatives the formula is

    \frac{\partial{z}}{\partial{v}} \bigg|_P = \frac{\partial{z}}{\partial{x}} \bigg|_{(x(P), y(P))} \frac{\partial{x}}{\partial{v}} \bigg|_P + \frac{\partial{z}}{\partial{y}} \bigg|_{(x(P),y(P))} \frac{\partial{y}}{\partial{v}} \bigg|_P

    so the solution did use the fact that (x(P), y(P)) = Q. Ugh.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Partial Derivative - Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 1st 2011, 07:41 AM
  2. partial derivative and the chain rule
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 8th 2010, 09:22 AM
  3. Chain Rule in Partial Derivative
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 31st 2010, 02:45 PM
  4. Partial Derivative and chain rule...
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 3rd 2009, 07:49 AM
  5. Partial derivative chain rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 11th 2008, 03:02 PM

Search Tags


/mathhelpforum @mathhelpforum