The partial derivative as a quotient

Posted on 11 March, 2015 by jesperg

Let z=f(x,y). Then we have

\displaystyle\frac{dy}{dx}=-\frac{\partial z/\partial x}{\partial z/\partial y},

which of course is not possible to interpret as quotients. But if we change the notation to

\displaystyle\frac{dy}{dx}=-\frac{\partial z/\partial x}{\delta z/\delta y},

then this is possible to interpret as quotients.

At a point (x,y,z) on the function graph, the tangent plane defines a plane through the origin in the tangent space. Let (dx,dy,0), (\partial x,0,\partial z) and (0,\delta y,\delta z) be three vectors in the tangent space that all belong to the tangent plane of the function. Then they are linearly dependent and we have that

\displaystyle\begin{vmatrix}dx & dy & 0\\\partial x & 0 & \partial z\\ 0 & \delta y & \delta z\end{vmatrix}=0,

which gives that

dx\delta y\partial z+\partial xdy\delta z=0

and it follows that

\displaystyle\frac{dy}{dx}=-\frac{\partial z/\partial x}{\delta z/\delta y}.

A purely algebraic proof can be given by starting with

\displaystyle df=f_xdx+f_ydy=\frac{\partial z}{\partial x}dx+\frac{\delta z}{\delta y}dy

Setting df=0 gives that

\displaystyle \frac{\partial z}{\partial x}dx+\frac{\delta z}{\delta y}dy=0

dx\delta y\partial z+\partial xdy\delta z=0

\displaystyle\frac{dy}{dx}=-\frac{\partial z/\partial x}{\delta z/\delta y}.

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