# Is there a chain rule for partial derivatives?

##### 1 Answer
Oct 28, 2015

Yes, there is.

Suppose you have three functions: $y \left(x , z\right)$, $x \left(t\right)$, and $z \left(t\right)$.

$y = x + z$
$x = 2 t$
$z = {t}^{2}$

The function $y$ varies according to $x$ and $z$, while $x$ and $z$ are function themselves that vary with $t$. When you take the partial derivative with respect to $x$, for instance:

$\frac{\partial y}{\partial x} = 1 \cdot \textcolor{g r e e n}{\frac{\partial x}{\partial x}} + 1 \cdot \textcolor{g r e e n}{\frac{\partial z}{\partial x}}$

But since $z$ is not a function of $x$, nothing happens to it.

$= 1 + z$

So you still see the chain rule, but it may not be obvious, and here it looks a bit redundant. $\frac{\partial x}{\partial x}$ merely becomes a $1$ multiplier in the equation.

This is similar to the chain rule you see when doing related rates, for instance.

However, if you take the exact differential with respect to $t$, which is like a variable "indirectly associated" with $y$ or "embedded" within $y$, it is easier to see the effects of the chain rule here, where you have embedded partial derivatives (inexact differentials) due to the fact you have multiple variables:

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left[x + z\right]$

$= \frac{\partial y}{\partial x} \cdot \textcolor{g r e e n}{\frac{\partial x}{\partial t}} + \frac{\partial y}{\partial z} \cdot \textcolor{g r e e n}{\frac{\partial z}{\partial t}}$

You may see that this is a convenient notation that allows you to see the rationale for formulating the way you take these partial derivatives:

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{\partial y}{\cancel{\partial x}} \cdot \textcolor{g r e e n}{\frac{\cancel{\partial x}}{\partial t}} + \frac{\partial y}{\cancel{\partial z}} \cdot \textcolor{g r e e n}{\frac{\cancel{\partial z}}{\partial t}}$

That aside, you get, from $\frac{\partial y}{\partial x} = 1$ and $\frac{\partial y}{\partial z} = 1$:

$= 1 \cdot \textcolor{g r e e n}{\frac{\partial x}{\partial t}} + 1 \cdot \textcolor{g r e e n}{\frac{\partial z}{\partial t}}$

$= \textcolor{b l u e}{2 + 2 t}$