1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Using the chain rule with 2 variables ?
Loading...