Multivariate Chain Rule

The (univariate) chain rule can be extended for multivariate functions, however, with each new argument, the complexity of the computation increases.

Definition §

Chain Rule (Bivariate) ^definition-bivariate

Let $f(x,y)$ be a bivariate function: ${\mathbb{R}}^2\to \mathbb{R}$

• Univariate dependency: If $x=g(t)$ and $y=h(t)$, i.e. $f$ implicitly depends only on $t$. Then the chain rule is:

$$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$

• Multivariate dependency: However, if $x$ and $y$ depend on more than one variable, for example $x=g(r,s,t,\dots )$ and $y=h(r,s,t,\dots )$ then $f$ now depends on multiple variables. Then $f$ can be rewritten as $f(g(r,s,t,\dots ),h(r,s,t,\dots ))$. Now the chain rule is, essentially, using a monovarietal chain rule, twice:

$$\begin{array}{r}\frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}\\ \frac{\partial f}{\partial s}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}\\ \dots \end{array}$$

Chain Rule (Multivariate) ^definition-multivariate

Let $f(x_1,x_2,\dots x_n)$ be a multivariate function: ${\mathbb{R}}^n\to \mathbb{R}$

• Univariate dependency: If $x_1=g_1(t),x_2=g_2(t),\dots ,x_n=g_n(t)$, i.e. $f$ implicitly depends only on $t$. Then the chain rule is:

$$\frac{df}{dt}=\frac{\partial f}{\partial x_1}\frac{d{x}_1}{dt}+\frac{\partial f}{\partial x_2}\frac{d{x}_2}{dt}+\cdots +\frac{\partial f}{\partial x_n}\frac{d{x}_n}{dt}$$

• Multivariate dependency: However, if $x_i$ depend on more than one variable, i.e. $x_i=g_i(t_1,t_2,t_3,\dots t_n)$ then $f$ now depends on multiple variables. Then $f$ can be rewritten as $f(g_1(t_1,t_2,\dots ,t_m),g_2(t_1,t_2,\dots ,t_m),\dots g_n(\dots t_m))$. The chain rule has a matrix-like structure:

$$\begin{array}{r}\frac{\partial f}{\partial t_1}=\frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial t_1}+\frac{\partial f}{\partial x_2}\frac{\partial x_2}{\partial t_2}+\cdots +\frac{\partial f}{\partial x_n}\frac{\partial x_n}{\partial t_m}\\ \frac{\partial f}{\partial t_2}=\frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial t_2}+\frac{\partial f}{\partial x_2}\frac{\partial x_2}{\partial t_2}+\cdots +\frac{\partial f}{\partial x_n}\frac{\partial x_n}{\partial t_m}\\ \\ \dots \end{array}$$

In matrix notation:

$$\underset{1\times m}{\underbrace{\left[\begin{array}{cccc}\frac{\partial f}{\partial t_1}& \frac{\partial f}{\partial t_2}& \dots & \frac{\partial f}{\partial t_m}\end{array}\right]}}=\underset{1\times n}{\underbrace{\left[\begin{array}{cccc}\frac{\partial f}{\partial x_1}& \frac{\partial f}{\partial x_2}& \dots & \frac{\partial f}{\partial x_n}\end{array}\right]}}\underset{n\times m}{\underbrace{\left[\begin{array}{cccc}\frac{\partial x_1}{\partial t_1}& \frac{\partial x_1}{\partial t_2}& \dots & \frac{\partial x_1}{\partial t_m}\\ \frac{\partial x_2}{\partial t_1}& \frac{\partial x_2}{\partial t_2}& \dots & \frac{\partial x_2}{\partial t_m}\\ ⋮& ⋮& \ddots & ⋮\\ \frac{\partial x_n}{\partial t_1}& \frac{\partial x_n}{\partial t_2}& \dots & \frac{\partial x_n}{\partial t_m}\end{array}\right]}}$$

The last matrix is also known as the Jacobi Matrix

Functional Chain Rule (Multivariate) ^definition-functional-multivariate

Dependency Graphs §

One way to visualise the multivariate chain rule is using (what I like to call) dependency graphs, where dependencies of variables are represented as lines. Then we need to simply trace all paths that end with that dependency.

For example, take $f(u,v,w)$ to be multivariate function mapping ${\mathbb{R}}^3\to \mathbb{R}$ and let $u=u(x,y,z)$ (where $u$ is a function, not multiplying!), $v=v(x,y,z)$ and $w=w(x,y,z)$. The dependency graph would look like

%%🖋 Edit in Excalidraw, and the dark exported image%%

Then to find, say $\frac{\partial f}{\partial x}$, we simply need to trace all paths that start at $f$ and end at $x$. Any type we traverse an edge $(a,b)$ we multiply by $\frac{\partial a}{\partial b}$ and any time we take a new path, we add: %%🖋 Edit in Excalidraw, and the dark exported image%%

Examples §

1: 1-variable Dependency Chain Rule

If $z=x^2-y^2$, $x=\sin (t)$,$y=\cos (t)$. Find $\frac{dz}{dt}$ at $t=\frac{\pi }{6}$

Solution

• $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$
• $\frac{\partial z}{\partial x}=2x=2\sin (t)$
• $\frac{\partial z}{\partial y}=-2y=-2\cos (t)$
• $\frac{dx}{dt}=\cos (t)$
• $\frac{dy}{dt}=-\sin (t)$

Hence,

$$\begin{array}{rl}\frac{dz}{dt}& =\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\\ & =2\sin (t)\cos (t)+-2\cos (t)\times -\sin (t)\\ & =4\sin (t)\cos (t)\end{array}$$

Now, to find it at $t=\frac{\pi }{6}$

$$\begin{array}{rl}{\frac{dz}{dt}|}_{t=\pi /6}& =4\sin \left(\frac{\pi }{6}\right)\cos \left(\frac{\pi }{6}\right)\\ & =4\times \frac{1}{2}\times \frac{\sqrt{3}}{2}=\sqrt{3}\end{array}$$

2: 2-variable dependancy Chain Rule

If $z=e^x\sinh (y)$, $x=s{t}^2$, $y=s^{2}t$. Find $\frac{\partial z}{\partial s}$

Solution

• $\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$
• $\frac{\partial z}{\partial x}=e^x\sinh (y)=e^{s{t}^2}\sinh (s^{2}t)$
• $\frac{\partial z}{\partial y}=e^x\cosh (y)=e^{s{t}^2}\cosh (s^{2}t)$
• $\frac{\partial x}{\partial s}=t^2$
• $\frac{\partial y}{\partial s}=2st$

Hence, $\frac{\partial z}{\partial s}=e^{s{t}^2}(\sinh (s^{2}t)t^2+2st\cosh (s^{2}t))$