Implicit differentiation of a two variable function
Let $f(u,v)$ be a differentiable function of two variables, and let z be a differentiable function of x and y defined implicitly by $f(xz,yz)$ = 0. Prove that $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -z$. I've got no idea how to start this. Any guidance would be appreciated.
Thanks in advance!
$\endgroup$ 11 Answer
$\begingroup$Define $g(x,y,z)=(xz,yz)$ and $h(x,y) = (x,y,z(x,y)).$ The equation $f(xz,yz)=0$ is equivalent to $(f\circ g \circ h)(x,y) = 0,$ where "$\circ$" denotes function composition. Taking the total derivative of both sides of the composition equation yields $$ (Df)(Dg)(Dh) = 0 $$ where $Df,$ $Dg,$ and $Dh$ are the matrices of partial derivatives of the vector fields $f,$ $g,$ and $h.$
In matrix form, this is $$ (f_x \ f_y) \left( \begin{array}{lll} z & 0 & x\\ 0 & z & y \end{array} \right) \left( \begin{array}{ll} 1 & 0 \\ 0 & 1\\ z_x & z_y \end{array} \right) = 0. $$ Multiplying the last two matrices and transposing gives $$ \left( \begin{array}{ll} z+xz_x & yz_x \\ xz_y & z+yz_y \end{array} \right) \left( \begin{array}{l} f_x \\ f_y \end{array} \right) =0. $$
If the matrix in this 2x2 linear system is non-singular, the solution $(f_x, f_y)$ must be 0, meaning $f$ is constant, which is obviously not valid because then $f$ cannot be used to define $z.$ So the matrix must be singular, meaning the determinant $z^2 + z(xz_x + yz_y)=0,$ or $z+xz_x+yz_y=0.$
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