$g(x,y,z(x,y))$ - how to compute $\frac{\partial g}{\partial x}$ using chain rule?

Question

I was thinking about this:

Let's say $g(x,y,z)$ is some function of $x,y,z$, but $z$ is a function of $x,y$. How would you compute $\frac{\partial g}{\partial x}$?

Here's my thoughts intuitively but it doesn't seem to make sense mathematically.

I would think that this would be true:

$$\frac{\partial g}{\partial x} = \frac{\partial g}{\partial x} + \frac{\partial g}{\partial z} \frac{\partial z}{\partial x}$$

This is what I think because $g$ will be changed by $x$ directly and $z$ will change because of $x$ also, which results in an additional change to $g$.

But this would imply:

$$0 = \frac{\partial g}{\partial z} \frac{\partial z}{\partial x}$$

So then one of these terms has to be zero. I don't believe this could be correct but I can't figure out the flaw in my reasoning.

Answer 1

Partial derivative notation is terrible; this example is one of the reasons why. In the equation

$$\frac{\partial g}{\partial x} = \frac{\partial g}{\partial x} + \frac{\partial g}{\partial z} \frac{\partial z}{\partial x}$$

the appearance of $\frac{\partial g}{\partial x}$ on the left hand side means something completely different from the appearance of $\frac{\partial g}{\partial x}$ on the right hand side.

When working with functions (in my example, $f$ is a univariate function), I find the best practice is:

• Don't write $f$ in a place where a scalar is expected.
• Don't write $f(x)$ in a place where a function is expected
• Write $f'$ for the derivative of $f$. With a function, like your $g$, that has multiple places, I like the notation $g_1$, $g_2$, and $g_3$ for the derivatives in the three places.

Then, the correct formula is

$$ \frac{\partial g(x, y, z(x,y))}{\partial x} = g_1(x, y, z(x,y)) + g_3(x, y, z(x,y)) z_1(x, y) $$

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That said, I prefer to compute with differentials than with partial derivatives when possible. I will also make the following change, since it's probably what you actually want:

• $z$ is not a function, but a scalar, related to $x$ and $y$ through an auxiliary function $z = h(x,y)$

Then, the differential is given by

$$ \mathrm{d}g(x,y,z) = g_1(x,y,z) \mathrm{d}x + g_2(x,y,z) \mathrm{d}y + g_3(x,y,z) \mathrm{d}z $$

(note that this equation is true even if $z$ is independent from $x$ and $y$)

Furthermore, we have

$$ \mathrm{d}z = h_1(x,y) \mathrm{d}x + h_2(x,y) \mathrm{d}y$$

which we can substitute into the above.

The premise asks for the derivative under the restriction that $y$ is held constant — that is, $\mathrm{d}y = 0$. Under that restriction we have

$$ \mathrm{d}g(x,y,z) = \left( g_1(x,y,z) + g_3(x,y,z) h_1(x,y) \right) \mathrm{d}x $$

Answer 2

Really you have:

$$g(u(x,y),v(x,y),z(x,y))$$

where $u(x,y)=x$, $v(x,y)=y$. So then:

$$g_x=g_u u_x+g_v v_x+g_z z_x$$

Simplifying:

$$g_x(x,y,z)=g_u(x,y,z)+g_z(x,y,z)z_x(x,y)$$