Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivatives of functions of products of variables

  1. Aug 3, 2013 #1
    Is there a general formula for (total) derivatives of functions of the form [itex]f(xy(x)+z(x)[/itex]?

    I tried the most simple function of that form [itex]f(xy(x)+z(x))=xy(x)+z(x)[/itex] and the formula I got was [itex]\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y} \frac{dy}{dx}+\frac{\partial f}{\partial z}\frac{dz}{dx}[/itex] though I'm unsure if it's even close to what I'm after. The formula Mathematica gave is [itex]y\frac{\partial f}{\partial x}+x\frac{\partial f}{\partial y} \frac{dy}{dx}+\frac{\partial f}{\partial z}\frac{dz}{dx}[/itex]

  2. jcsd
  3. Aug 3, 2013 #2


    User Avatar
    Science Advisor

    You have a missing right parenthesis here. Do you mean "f(xy(x)+ z(x))"? And do you mean y(x) and z(x) as functions of x rather that y times x and z times x? Assuming that, the usual "chain rule" for functions of several variables applies:
    [tex]\frac{df}{dx}= \frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}\frac{dy}{dx}+ \frac{\partial f}{\partial z}\frac{dz}{dx}[/tex]
    In this case, writing f' as the derivative of f(u) with respec to u, that becomes
    [tex]\frac{df}{dx}= f'(xy+ z)y+ f'(xy+ z)x\frac{dy}{dx}+ f'(xy+z)\frac{dz}{dx}[/tex].

    Well, that's what I wrote before but with f(u)= u, as here, f'= 1 so that is
    [tex]\frac{df}{dx}= y+ x\frac{dy}{dx}+ \frac{dz}{dx}[/tex].

    As a check you can take [tex]y= 2x[/tex] and [tex]z= x^2[/tex]
    Then [tex]f(xy+ z)= xy+ z= 2x^2+ x^2= 3x^2[/tex] which has derivative 6x.

    In this example, dy/dx= 2 and dz/dx= 2x so the formula above would give
    [tex]\frac{df}{dx}= y+ x\frac{dy}{dx}+ \frac{dz}{dx}= 2x+ x(2)+ 2x= 6x[/tex]

    Yes, that is precisely the formula I gave!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook