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Thread: multivariate: chain rule

  1. #1
    Nov 2008

    multivariate: chain rule

    Suppose that you are given an equation of the form:
    for example, something like x^3z+ycos(z)+(siny)/z=0. Then we may consider z to be defined implicitly as a function z(x,y).

    a) Use the chain rule to show that if F and z(x,y) are both assumed to be differentiable, then

    dz/dx = - Fx(x,y,z)/Fz(x,y,z) and dz/dy = -Fy(x,y,z)/Fz(x,y,z)

    note: the letters after the F's are subscripts
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    Differentiate the equation $\displaystyle F(x,y,z)=0$ partially with respect to x, using the chain rule, and you get $\displaystyle \textstyle\frac{\partial F}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} = 0$. In this equation, x and y are the independent variables, and z is a function of x and y. When we differentiate partially with respect to x, we keep y constant. So $\displaystyle \textstyle\frac{\partial x}{\partial x} = 1$ and $\displaystyle \textstyle\frac{\partial y}{\partial x}=0$. Thus the equation becomes $\displaystyle \textstyle F_x + F_z\frac{\partial z}{\partial x} = 0$, or $\displaystyle \textstyle\frac{\partial z}{\partial x} = -F_x/F_z$.
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