Results 1 to 3 of 3

Math Help - Help with a second partial derivative, please

  1. #1
    Junior Member
    Joined
    Apr 2006
    Posts
    25

    Help with a second partial derivative, please

    I am doing a chapter on bivariate calculus and so far have done about 30 problems without any mistakes. Yet I've come across one where I can't match the author's answer. Given that I've found a few typos before, I'd like to know if I'm right or wrong, and if wrong, why.

    The relevant part of the problem asks:

    If z = f(x,y) = x^3 + y^2 - 8xy + (2x^3)(y^2) + 11, find

    a) the partial derivative of f with respect to y, and

    b) the partial of f [subscript] y (i.e. the 2nd derivative) with respect to y.

    The method I am using is to treat all "x"s as constants and to take the derivative of all "y"s.

    Thus, when y' means "derivative of y", and when

    f(x,y) = x^3 + y^2 - 8xy + (2x^3)(y^2) + 11,

    For a) I get

    (0) + (y^2)' - (8x)(y)' + (2x^3)(y^2)' + (0)
    = 2y - 8x(1) + (2x^3)(2y)
    = 2y - 8x + (4x^3)(y),

    which matches the book's answer.

    For b) I use the same approach.

    If the partial derivative of f with respect to y

    = 2y - 8x + (4x^3)(y),

    then

    the partial of f [subscript] y with respect to y should be

    = 2(y)' - (0) + (4x^3)(y)'
    = 2 (1) + (4x^3)(1)
    = 2 + 4x^3.

    The book, however, gives the answer as 2 + (12x^2)(y), without showing any work.

    Am I missing something?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,739
    Thanks
    645
    Hello, lingyai!

    Your work is correct!

    If z \:= \:f(x,y) \:= \:x^3 + y^2 - 8xy + 2x^3y^2 + 11, find

    a) the partial derivative of f with respect to y

    b) the partial of f_y with respect to y


    The method I am using is to treat all x's as constants and to take the derivative of all y's.

    For a) I get: f_y \:= \:0 + (y^2)' - (8x)(y)' + (2x^3)(y^2)' + 0
    . . = \:2y - 8x(1) + (2x^3)(2y) \:= \:2y - 8x + 4x^3y . . . which matches the book's answer.

    For b) I use the same approach.

    If f_y \:=\: 2y - 8x + 4x^3y, then f_{yy} should be:

    . . f_{yy} \:= \:2(y)' - (0) + (4x^3)(y)' \:= \:2 (1) + (4x^3)(1)\:=\:2 + 4x^3 . . . Correct!

    The book, however, gives the answer as: 2 + 12x^2y ??

    Am I missing something? . . . Not a thing!
    The 12x^2y looks like they differentiated with respect to x . . . but even that is wrong.

    If f_y\:=\:2y - 8x + 4x^3y, then: f_{yx}\:=\:-8 + 12x^2y

    Since f_x\:=\:3x^2 - 8y + 6x^2y^2, then f_{xx} \:=\:6x + 12xy^2

    Their answer doesn't match anything in the problem!
    . . It must be just one more typo on their part.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2006
    Posts
    25
    Thanks very much for that feedback. Having only started with multivariate calculus today, I was inclined to assume I'd missed something. Not being able to figure it out was troubling.

    I'm all in favor of students learning from "spot the deliberate mistake" type answers, but typos in what is supposed to be a self-contained, self-teaching text are very annoying!

    Thanks again,

    Ken (lingyai)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivative of arctan in a partial derivative
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 12th 2010, 01:52 PM
  2. Partial Derivative?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2010, 05:56 PM
  3. how do you take this partial derivative
    Posted in the Calculus Forum
    Replies: 10
    Last Post: July 26th 2009, 07:09 PM
  4. Partial derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 16th 2009, 11:21 AM
  5. partial derivative...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 9th 2009, 05:06 PM

Search Tags


/mathhelpforum @mathhelpforum