Chain rules for partial differentiation

Multivariate functions: Partial derivatives

Chain rules for partial differentiation

The chain rule for differentiation describes what the derivative of a composition of two functions looks like. The composition of two functions $f$ and $g$ is the function $f\circ g$ given by $f\circ g(t) = f(g(t))$ .

Chain Rule of a function of a single variable

Before we discuss chain rules of multivariate functions, we briefly look back at the chain rule of functions of a single variable. Let $f$ and $g$ be two functions of one variable. The chain rule is $(f\circ g)' = (f'\circ g)\cdot g'$ . In terms of functional rules, this reads $\frac{\dd f(g(t))}{\dd t}=\frac{\dd f(x)}{\dd x}\left(g(t)\right)\cdot\frac{\dd g(t)}{\dd t}\tiny.$ The term $\frac{\dd f(x)}{\dd x}\left(g(t)\right)$ represents the value of the function $\frac{\dd f(x)}{\dd x}$ at $g(t)$ and can also be written as $\left. \frac{\dd f(x)}{\dd x}\right|_{x=g(t)}$ .

Often, a shorter notation is used. Write $x=g(t)$ , and $y=f(x)$ . Then $\frac{\dd y}{\dd x}$ represents both the functional rule of $\frac{\dd f(x)}{\dd x}$ and the value of $\frac{\dd f(x)}{\dd x}$ in $x=g(t)$ , and we can write the chain rule simply as $\frac{\dd y}{\dd t}=\frac{\dd y}{\dd x}\cdot\frac{\dd x}{\dd t}\tiny.$ We will continue to work with this notation when we discuss chain rules for multivariate functions.

The function $(3t^2-1)^5$ of $t$ is composed of the functions $f$ and $g$ with functional rules $f(x)=x^5\quad\text{and}\quad g(t)=3t^2-1\tiny.$ Indeed, $f\circ g(t)=f(g(t))=(g(t))^5=(3t^2-1)^5$ .

The derivatives of $f$ and $g$ are $\frac{\dd f(x)}{\dd x}=5x^4$ and $\frac{\dd g(t)}{\dd t}=6t$ so, because of the chain rule, $\begin{array}{rcl}\frac{\dd(3t^2-1)^5}{\dd t}&=&\frac{\dd f(g(t))}{\dd t}\\&=&\left.\frac{\dd f(x)}{\dd x}\right|_{x=g(t)}\cdot\frac{\dd g(t)}{\dd t}\\ &=&\left.5x^4\right|_{x=3t^2-1}\cdot6t\\&=&5(3t^2-1)^4\cdot6t\\&=&30t\cdot(3t^2-1)^4\end {array}$

In the short notation we write $y=f(x)=x^5$ and $x=g(t)=3t^2-1$ . Using the chain rule in the short form, we find:

$\begin{array}{rcl}\frac{\dd y}{\dd t}&=&\frac{\dd y}{\dd x}\cdot\frac{\dd x}{\dd t}\\ &=&\frac{\dd x^5}{\dd x}\cdot\frac{\dd (3t^2-1)}{\dd t}\\ &=&5x^4\cdot6t\\&=&5(3t^2-1)^4\cdot6t\\&=&30t\cdot(3t^2-1)^4\end {array}$

Note that in $\frac{\dd y}{\dd t}$ we use the variable $y$ as a function of $t$ and in $\frac{\dd y}{\dd x}$ as a function of $x$ .

We now discuss chain rules for functions of two variables.

Chain Rules for partial differentiation of bivariate functions

Let $f(x,y)$ be a function of two variables $x$ and $y$ , so that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are continuous functions.

If $x$ and $y$ are differentiable functions of $t$ , then for $f$ as a function of $t$ holds: $\frac{\dd f}{\dd t}=\frac{\partial f}{\partial x}\cdot\frac{\dd x}{\dd t}+ \frac{\partial f}{\partial y}\cdot\frac{\dd y}{\dd t}$
If $x$ and $y$ are differentiable functions of two variables $s$ and $t$ , then for $f$ as a function of $s$ and $t$ holds: $\begin{array}{rcl}\frac{\partial f}{\partial s} &=&\frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial s}+ \frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial s} \\ \\ \frac{\partial f}{\partial t} &=&\frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial t}\end {array}$

If the concept of continuity of functions like $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ is not known, then view it as a mild condition, which in our examples is almost always fulfilled.

When we talk about $f$ as a function of $s$ and $t$ , we mean the composite function $f\circ \varphi$ , where $\varphi(s,t)=\rv{x(s,t),y(s,t)}$ is a mapping from a subset of ${\mathbb R}^2$ to ${\mathbb R}^2$ .

This double meaning of $f$ (as a function of $x$ and $y$ and as a function of $s$ and $t$ ) is also reflected in $x$ and $y$ , which are seen both as independent variables (e.g. in the denominators of expressions such as $\frac{\partial f}{\partial y}$ ), and as functions of $s$ and $t$ (e.g. in the numerator of $\frac{\partial y}{\partial t}$ ).

For functions of more than two variables similar chain rules exist:

If $w=w(x_1,x_2,\ldots,x_m)$ is a function of $m$ variables $x_i$ $(i=1,2,\ldots,m)$ with continuous partial derivatives, and each $x_i$ is a differentiable function of $t_1,t_2,\ldots,t_n$ , then, for each $j$ $(1\le j\le n)$ , the function $w$ of $t_1,t_2,\ldots,t_n$ , has partial derivative $\frac{\partial w}{\partial t_j}=\frac{\partial w}{\partial x_1}\cdot\frac{\partial x_1}{\partial t_j}+ \frac{\partial w}{\partial x_2}\cdot\frac{\partial x_2}{\partial t_j}+\cdots+\frac{\partial w}{\partial x_m}\cdot\frac{\partial x_m}{\partial t_j}\tiny.$

Two more special cases:

If $w=w(x)$ is a function of a single variable $x$ with continuous partial derivative, and $x$ is a differentiable function of $s$ and $t$ , then $w$ , as a function of $s$ and $t$ , has partial derivatives $\frac{\partial w}{\partial s}=\frac{\dd w}{\dd x}\cdot\frac{\partial x}{\partial s}\quad\text{and}\quad\frac{\partial w}{\partial t}=\frac{\dd w}{\dd x}\cdot\frac{\partial x}{\partial t}\tiny.$

If $w=w(x,y,z)$ is a function of three variables with continuous partial derivatives, and $x$ , $y$ , and $z$ are differentiable functions of $t$ , then $w$ , as a function of $t$ , has the derivative $\frac{\dd w}{\dd t}=\frac{\partial w}{\partial x}\cdot\frac{\dd x}{\dd t}+ \frac{\partial w}{\partial y}\cdot\frac{\dd y}{\dd t}+\frac{\partial w}{\partial z}\cdot\frac{\dd z}{\dd t}\tiny.$

Suppose $z(x,y)=x\cdot y, \quad x=t^{3},\quad \text{and} \quad y=t^{4}\tiny.$ Then $z$ , viewed as a function of $t$ , is equal to $t^{3}\cdot t^{4}=t^{7}$ , so $\frac{\dd z}{\dd t}=7 t^{6}$ . This result also follows from the Chain Rule for partial differentiation. How?

$\begin{array}{rcl} \frac{\dd z}{\dd t} &=&\frac{\partial z}{\partial x}\cdot\frac{\dd x}{\dd t}+ \frac{\partial z}{\partial y}\cdot\frac{\dd y}{\dd t} \\ &=&\frac{\partial \left(x\cdot y\right)}{\partial x}\cdot\frac{\partial\left( t^{3}\right)}{\partial t}+ \frac{\partial \left(x\cdot y\right)}{\partial y}\cdot\frac{\partial \left(t^{4}\right)}{\partial t} \\&=& y\cdot 3 t^{2}+x\cdot 4 t^{3} \\ &=& t^{4}\cdot 3 t^{2}+t^{3}\cdot 4 t^{3} \\&=& 7 t^{6}\end{array}$

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