Partial derivatives with respect to two functions while keeping one of them constant

I have a function of $f(x_1, x_2)$, and two other functions of $g_1(x_1, x_2)$ and $g_2(x_1, x_2)$ in such a way that the original function can be written as $f(g_1, g_2)$.

Now I need to fine partial derivative of $f$ with respect to $g_1$ or $g_2$ while keeping the other constant. That is

$$\Big( \dfrac{\partial f}{\partial g_1} \Big)_{g_2 = const}$$

and

$$\Big( \dfrac{\partial f}{\partial g_2} \Big)_{g_1 = const}$$

What is the correct formula to do this?

• What are the domains and codomains of your functions? – user258700 Jan 3 '16 at 23:00
• Don't really know. I am more of a structural engineer trying to sort out some form of equation for simulation... I think your question goes way beyond my scope... – Quang Thinh Ha Jan 3 '16 at 23:03
• Not really. I am just asking where do $f$ and $g$ get their inputs from, and what do they output. For instance, is $f:\Bbb R^2 \to \Bbb R$? – user258700 Jan 3 '16 at 23:05
• Ah I see. Yes they do. – Quang Thinh Ha Jan 3 '16 at 23:07

This is probably not the most rigorous or direct way of doing it, but it should explain what's going on:

By the chain rule, we can write \begin{align} \mathrm{d}f &= f_x\ \mathrm{d}x + f_y\ \mathrm{d}y \\ \mathrm{d}g_1 &= g_{1x}\ \mathrm{d}x + g_{1y}\ \mathrm{d}y \\ \mathrm{d}g_2 &= g_{2x}\ \mathrm{d}x + g_{2y}\ \mathrm{d}y \\ \end{align}

Let us think about $(\partial f/\partial g_1)_{g_2}$. The condition $g_2 = \text{constant}$ means that we are confined to move along a contour of $g_2$, so let us first work out what those contours are, and then work out the rates of change of $f$ and $g_1$ if we move along such a contour.

Since $g_2 = \text{constant}$ along a contour, we must have $$g_{2x}\ \mathrm{d}x + g_{2y}\ \mathrm{d}y = 0$$ and so $$\mathrm{d}y = - \frac{g_{2x}}{g_{2y}}\ \mathrm{d}x$$ on contours of $g_2$ (but see Note 1). The contours are therefore described by the differential equation $$\frac{\mathrm{d}y}{\mathrm{d}x} = - \frac{g_{2x}}{g_{2y}}.$$

Now we have a relationship between $\mathrm{d}y$ and $\mathrm{d}x$. We shall substitute these into the expressions for $\mathrm{d}f$ and $\mathrm{d}g_1$ to work out the rates of change of $f$ and $g_1$ along the contours.

For $f$, we have $$\mathrm{d}f = \left(f_x - \frac{g_{2x}}{g_{2y}} f_y \right)\ \mathrm{d}x$$ and a corresponding equation for $\mathrm{d}g_1$. Dividing the two (see Note 2) then gives the desired expression: $$\left(\frac{\partial f}{\partial g_1}\right)_{g_2} = \frac{ f_x g_{2y} - f_y g_{2x} }{ g_{1x} g_{2y} - g_{1y} g_{2x} },$$ or, in terms of 'Jacobian determinants': $$\left(\frac{\partial f}{\partial g_1}\right)_{g_2} = \frac{ \partial(f,g_2)/\partial(x,y) }{ \partial(g_1,g_2)/\partial(x,y) }.$$

Notes

1. If $g_{2y}=0$, then we can proceed instead by writing $\mathrm{d}x$ in terms of $\mathrm{d}y$ instead, and proceed accordingly to get the same result.

If both $g_{2x}$ and $g_{2y}$ are zero at the same point, then $g_2$ is stationary there, and $(\partial f/\partial g_1)_{g_2}$ is not defined at that point.

2. The partial derivative $(\partial f/\partial g_1)_{g_2}$ is undefined at points where $\partial(g_1,g_2)/\partial(x,y) = 0$. At such points, the gradients of $g_1$ and $g_2$ are not linearly dependent, which means that the contour of $g_2$ is also a contour of $g_1$; the rate of change of $g_1$ is therefore zero there.