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Homework Help: Using the chain rule with 2 variables ?

  1. Oct 23, 2005 #1
    Hey, im a bit confused oh how to use the chain rule when i have 2 variables in an equation...
    Example : f(x,y) = (Squareroot(x)).(cosh(x+y^2)) x(s,t)=st y(s,t)=s/t
    When i have 2 variables, im not sure how to split it up and use the chain rule, all the examples i found only have 1 variable :shy: Thanks :(
     
    Last edited: Oct 23, 2005
  2. jcsd
  3. Oct 23, 2005 #2

    benorin

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    When differentiating a multivariate function with respect to one variable, you hold the other variables constant, e.g.
    [tex]\frac{d}{dy}\left(x\sin(xy^{2})\right)=x\frac{d}{dy}\sin(xy^{2})=x\cos(xy^{2})\frac{d}{dy}(xy^{2})=2x^{2}y\cos(xy^{2})[/tex]

    in the above example x is a constant. Whereas in the example below, y is a constant:

    [tex]\frac{d}{dx}\left(x\sin(xy^{2})\right)=\frac{d}{dx}\left(x\right)\cdot\sin(xy^{2})+x\frac{d}{dx}\left(\sin(xy^{2})\right)=1\cdot\sin(xy^{2})+x\cos(xy^{2})\frac{d}{dx}(xy^{2})=\sin(xy^{2})+xy^{2}\cos(xy^{2})[/tex]
     
    Last edited: Oct 23, 2005
  4. Oct 23, 2005 #3
    Ok, so say when i derive cosh(x + y^2) with respect to x, i get (1+y^2)sinh(x+y^2) correct ?? basically i dont touch y^2 and treat it as a whole number ?? :shy: But i dont get it, since if i had like cosh(x+3), then the derivative would just be sinh(x+3) using the chain rule correct ?? so why is the y^2 out the front in the other one ?
     
    Last edited: Oct 23, 2005
  5. Oct 23, 2005 #4

    Galileo

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    When you differentiate with respect to one variable, say x, you treat the other, y, as constant. Specifically:

    [tex]\frac{\partial}{\partial x}f(x,y)=\lim_{h\to 0}\frac{f(x
    +h,y)-f(x,y)}{h}[/tex]
    So it's just the rate of change of the function as you vary x you're interested in.
    By the way, this alone has nothing to do with the chain rule, which tells you how to differentiate composite functions.

    Now, differentiating cosh(x+y^2) with respect to x just means treating y constant. (So for this purpose we can treat the function as just dependent on x alone):
    [tex]\frac{d}{dx}\cosh(x+y^2)=\sinh(x+y^2)\frac{d}{dx}(x+y^2)=\sinh(x+y^2)[/tex]
    Since the derivative of (x+y^2) wrt x is 1.
     
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