Results 1 to 6 of 6

Math Help - Basic multivariable differentiation

  1. #1
    Junior Member Spimon's Avatar
    Joined
    Sep 2007
    Posts
    41

    Basic multivariable differentiation

    Hi, this is my first post so I hope I get it right.
    I'm a 3rd year engineering student from Sydney, Australia.

    I'm just starting multivariable calculus and am a little unsure. Here's my question and solution.
    Could someone please confirm where/if I'm going wrong?


    (Should read 'Let u = (4-x^2-y^2)^1/2). I forgot to put in the power of 1/2.
    EDIT: Oh no! I just saw another mistake. I shouldn't have a 4 in the last line of either function as it's constant. Sorry guys...



    What I'm basically 'trying' to do is apply the chain rule as with a single variable problem but when substituting (4-x^2-y^2) into u (last 2 lines) I'm letting the variable I'm not dealing with become a constant and thus = 0.
    Ie for f'(x), y = 0 and for f'(y) x=0.
    (This is why I have (4-0-y^2) in the last line of f'(y). I didn't put the zero into the f'(x) but maybe should have to make my thinking more clear.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    It looks to me as if you really have not made the leap to multiple variables. You are going to an awful lot of effort to separate x from y. You don't get to do that. The 2D definition of a derivative gives away the significant generalization that there are no longer only two ways to approach a limit, left or right. There are now infinitely many ways to approach the limit. For this reason, one must consider both simultaneously.

    You must do some reading on:

    Partial Derivative - Clearly, you have seen some of this. Find some more.
    Directional Derivative
    Gradient
    Parametric Representation of Curves
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Spimon's Avatar
    Joined
    Sep 2007
    Posts
    41
    Ah ok. Thanks for the reply. I was trying to separate the 2 variables as per the diagram below. On each axis, the other is constant at any arbitary point. I'll go read through my notes form the beginning again and see if I can make more sense out of it.

    Last edited by Spimon; September 14th 2007 at 11:41 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member Spimon's Avatar
    Joined
    Sep 2007
    Posts
    41
    I think I may have been getting a little ahead of myself. I think I was actually trying to find the partial derivative but thought I was finding the true derivative.
    Thanks for the pointers anyway. Hopefully I've found what I want to do and stumbled across a few examples to help me out.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Spimon View Post
    Hi, this is my first post so I hope I get it right.
    I'm a 3rd year engineering student from Sydney, Australia.

    I'm just starting multivariable calculus and am a little unsure. Here's my question and solution.
    Could someone please confirm where/if I'm going wrong?


    (Should read 'Let u = (4-x^2-y^2)^1/2). I forgot to put in the power of 1/2.
    EDIT: Oh no! I just saw another mistake. I shouldn't have a 4 in the last line of either function as it's constant. Sorry guys...



    What I'm basically 'trying' to do is apply the chain rule as with a single variable problem but when substituting (4-x^2-y^2) into u (last 2 lines) I'm letting the variable I'm not dealing with become a constant and thus = 0.
    Ie for f'(x), y = 0 and for f'(y) x=0.
    (This is why I have (4-0-y^2) in the last line of f'(y). I didn't put the zero into the f'(x) but maybe should have to make my thinking more clear.
    yes, i think you want the partial derivatives.

    Notation: for a differentiable function of two variables, f(x,y), we use the notation:

    \frac {\partial f}{\partial x} or f_x(x,y) (or simply f_x) to mean the partial derivative with respect to x,

    \frac {\partial f}{\partial y} or f_y (or simply f_y) to mean the partial derivative with respect to y, and

    \mbox {grad}f or \nabla f = <f_x,f_y> to mean the directional derivative of f(x,y), which you can think of as the "general derivative." I just had a discussion with my colleague as to whether we can represent the "general" derivative in 3-dimensional space without a vector function, but neither of us think we can, and i think it makes sense that we can't

    when we are finding the partial derivative with respect to one variable, we treat the other variable as if it were a constant.

    so, given f(x,y) = \sqrt {4 - x^2 - y^2} = \left( 4 - x^2 - y^2 \right)^{\frac 12}

    \Rightarrow f_x = \frac 12 \left( 4 - x^2 - {\color {red} y^2} \right)^{- \frac 12} \cdot (-2x) = -x \left( 4 - x^2 - y^2 \right)^{- \frac 12}

    and

    \Rightarrow f_y = \frac 12 \left( 4 - {\color {red} x^2} - y^2 \right)^{- \frac 12} \cdot (-2y) = -y \left( 4 - x^2 - y^2 \right)^{- \frac 12}


    the variables in red are the ones you left out. recall that for the chain rule, we do not trouble what's inside the brackets until we are ready to multiply by the derivative of what's in the brackets. also, you had the powers at the end wrong, they should be -1/2
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member Spimon's Avatar
    Joined
    Sep 2007
    Posts
    41
    Thanks so much for the reply!
    I gosh I think I have it!

    Give that man a New!! (Aussie expression of gratitude)

    (BTW, I realise I had the powers at the end wrong. Oops!)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Multivariable Differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 14th 2010, 09:55 PM
  2. Multivariable Differentiation
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 2nd 2010, 02:06 PM
  3. Differentiation of Multivariable Functions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 6th 2010, 04:05 AM
  4. Multivariable Differentiation Problem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: October 30th 2009, 12:32 PM
  5. basic multivariable calculus question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 16th 2008, 03:54 PM

Search Tags


/mathhelpforum @mathhelpforum