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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]

    Thanks
     
  2. jcsd
  3. Aug 3, 2013 #2

    HallsofIvy

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    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!
     
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