I Derive the formula for gradient using chain rule

[Happiness]

Consider a surface defined by the equation $g(x, y, z)=0$. The intersection between this surface and the plane $z=c$ produces a curve that can be plotted on an x-y plane. Find the gradient of this curve.

By chain rule,

$\frac{\partial y}{\partial x}=\frac{\partial y}{\partial g}\frac{\partial g}{\partial x}$

Using the reciprocity relation $\frac{\partial y}{\partial g}=\Big(\frac{\partial g}{\partial y}\Big)^{-1}$, we have

$\frac{\partial y}{\partial x}=\frac{(\frac{\partial g}{\partial x})}{(\frac{\partial g}{\partial y})}$

This differs from the correct answer by a negative sign. What's wrong with this method?

 

[andrewkirk]

Greater care is needed when handling partial derivatives. They cannot be treated in general the same as total derivatives. In particular it is not necessarily the case that

$\frac{\partial y}{\partial x}=\frac{\partial y}{\partial g}\frac{\partial g}{\partial x}$

Indeed, in this case the $\frac{\partial y}{\partial x}$ that is being calculated is with the values of $g$ and $z$ held constant, so $\frac{\partial y}{\partial g}$ is undefined or infinite, while $\frac{\partial g}{\partial x}$ is zero.

 

[Happiness]

The following website calculates the gradient as follows:

Differentiate $g$ wrt $x$ while holding $z$ constant.

Then we will have $\frac{\partial y}{\partial x}=-\frac{(\frac{\partial g}{\partial x})}{(\frac{\partial g}{\partial y})}$. Why ain't $\frac{\partial g}{\partial x}$ and $\frac{\partial g}{\partial y}$ zero in this case?

In what situations must the two partials on the RHS of the chain rule have the same variables held constant? Like you mentioned for post #1, $(\frac{\partial y}{\partial g})_{gz}(\frac{\partial g}{\partial x})_{gz}$. But why is it not required in the attachment above, where $(\frac{\partial g}{\partial y})_{xz}(\frac{\partial y}{\partial x})_{gz}$?

Source:
http://www.sjsu.edu/faculty/watkins/envelopetheo.htm

 

[andrewkirk]

I find the presentation on that web page unnecessarily confusing. It attempts to use the formula for a total derivative in a case where all derivatives are partial. Sometimes one can get away with this, but sometimes one can't, so it is best avoided.

A more correct derivation would be as follows:

Say the point $(x_0,y_0)$ is on the curve. Let us parametrise the curve through that point by function $\gamma:\mathbb R\to\mathbb R^2$ and wlog set $\gamma(0)=(x_0,y_0)$. Then we have $g(\gamma_1(t).\gamma_2(t),c)=0$, where $\gamma_1$ and $\gamma_2$ are the component functions of $\gamma$.

We then apply $\frac d{dt}$ to this - ie taking a total derivative - to get:

$$0=\frac d{dt}0=\frac d{dt}g(\gamma_1(t).\gamma_2(t),c)=\frac{\partial g}{\partial x}\frac {dx}{dt}+
\frac{\partial g}{\partial y}\frac {dy}{dt}+\frac{\partial g}{\partial c}\frac {dc}{dt}
=\frac{\partial g}{\partial x}\frac {dx}{dt}+
\frac{\partial g}{\partial y}\frac {dy}{dt}+\frac{\partial g}{\partial c}\cdot 0$$

Rearranging, we get

$$-\frac{\partial g}{\partial x}/\frac{\partial g}{\partial y}=\frac{dy}{dt}/\frac{dx}{dt}=\frac{dy}{dx}$$
as required. Note that the transformation in the final step is allowed because we are dealing with total derivatives, not partial derivatives.