Chain rules

Functions of several variables: Partial derivatives

Chain rules

First, we look at the chain rule for differentiating functions of one variable.

Chain rule for differentiating a function of one variable We consider the function

$y(t)=(3t^2-1)^5$ as being composed of the functions

$y=x^5\quad\text{and}\quad x=3t^2-1$ Then:

$\frac{\dd y}{\dd x}=5x^4\quad\text{and}\quad \frac{\dd x}{\dd t}=6t$ In terms of differentials we have

$\dd y = 5x^4\,\dd x\quad\text{and}\quad \dd x=6t\,\dd t$ Because

$x=3t^2-1\quad\text{and}\quad \dd x=6t\,\dd t$ we have

$\dd y=5(3t^2-1)^4 6t\,\dd t$ and thus

$\frac{\dd y}{\dd t}=\frac{\dd y}{\dd x}\cdot \frac{\dd x}{\dd t}$ where we can write the derivative $\frac{\dd y}{\dd x}$ , which initially is a function of $x$ , via $x=x(t)$ again in terms of the variable $t$ . This last rule applies more generally, for any function $y\circ x$ , that is, for the composition of the functions $y$ and $x$ .

We now discuss chain rules for differentiating functions of two variables.

If $z=z(x,y)$ is a differentiable function of two variables, and $x$ and $y$ are differentiable functions of $t$ , then $z$ is a function of $t$ and

$\frac{\dd z}{\dd t}=\frac{\partial z}{\partial x}\cdot\frac{\dd x}{\dd t}+ \frac{\partial z}{\partial y}\cdot\frac{\dd y}{\dd t}$

Example

Consider

$z(x,y)=x\cdot y, \quad x=\cos(t)\quad \text{and} \quad y=\sin(t)$ Then $z$ is, as function of $t$ , equal to $\cos(t)\sin(t)=\tfrac{1}{2}\sin(2t)$ and thus $z'(t)=\cos(2t)$ . But this result also follows from the above chain rule:

$\begin{aligned} \frac{\dd z}{\dd t}&=\frac{\partial z}{\partial x}\cdot\frac{\dd x}{\dd t}+ \frac{\partial z}{\partial y}\cdot\frac{\dd y}{\dd t}\\ \\ &= \sin(t)\cdot -\sin(t) + \cos(t)\cdot\cos(t)\\ \\ &= \cos^2(t)-\sin^2(t)=\cos(2t)\end{aligned}$

More generally, the following rule also is true.

If $z=z(x,y)$ is a differentiable function of two variables, and $x$ and $y$ are differentiable functions of two variables $s$ and $t$ , then $z$ is a function of $s$ and $t$ , and

$\begin{aligned}\frac{\partial z}{\partial s} &=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial s}+ \frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial s} \\ \\ \frac{\partial z}{\partial t} &=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial t}+ \frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial t}\end{aligned}$

Example Considee

$z(x,y)=x\cdot y, \quad x=s^2+t^2\quad \text{and} \quad y=s^2-t^2$ Then $z$ is equal to $(s^2+t^2)(s^2-t^2)=s^4-t^4$ as a function of $s$ and $t$ , and thus $\displaystyle \frac{\partial z}{\partial s}=4s^3$ and $\displaystyle \frac{\partial z}{\partial s}=-4t^3$ . But this result also follows from the above chain rule:

$\begin{aligned} \frac{\partial z}{\partial s} &=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial s}+ \frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial s}\\ \\ &= (s^2-t^2)2s+(s^2+t^2)2s = 4s^3\\ \\ \frac{\partial z}{\partial t} &=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial t}+ \frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial t}\\ \\ &= (s^2-t^2)2t+(s^2+t^2)\cdot -2t=-4t^3\end{aligned}$

An important application of the second chain rule is the conversion from Cartesian coordinates $x, y$ to polar coordinates $r,\varphi$ via the definitions $x=r\cos\varphi, y=r\sin \varphi$ .

Example of polar coordinates Consider the function

$z(x,y)=\frac{2xy}{x^2-y^2}$ where

$x=r\cos(\varphi)\quad\text{and}\quad y=r\sin(\varphi)$
The definition of the function is simpler in polar coordinates $r, \varphi$ :

$z(r,\varphi) = \frac{2\cdot\bigl(r\cos(\varphi)\bigr)\cdot\bigl(r\sin(\varphi)\bigr)}{(r\cos\bigl(\varphi)\bigr)^2+\bigl(r\sin(\varphi)\bigr)^2}=\frac{2\sin(\varphi)\cos(\varphi)}{\cos^2(\varphi)-\sin^2(\varphi)}=\frac{\sin(2\varphi)}{\cos(2\varphi)}=\tan(2\varphi)$ After this rewriting of the function in polar coordinates you come to the conclusion that

$\frac{\partial z}{\partial r} = 0\quad\text{and}\quad \frac{\partial z}{\partial \varphi}=\frac{2}{\cos^2(2\varphi)}$ But this result also follows from the above chain rule. First, we just calculate the required partial derivatives:

$\begin{aligned} \frac{\partial z}{\partial x} &= \frac{2y\cdot (x^2-y^2)-2xy\cdot 2x}{(x^2-y^2)^2}=\frac{-2y(x^2+y^2)}{(x^2-y^2)^2}\\ \\ \frac{\partial z}{\partial y} &= \frac{2x\cdot (x^2-y^2)-2xy\cdot -2y}{(x^2-y^2)^2}=\frac{2x (x^2+y^2}{(x^2-y^2)^2}\\ \\ \frac{\partial x}{\partial r} &= \cos(\varphi)\qquad\qquad \frac{\partial x}{\partial \varphi}= -r\sin(\varphi)\\ \\ \frac{\partial y}{\partial r}&= \sin(\varphi)\qquad\qquad \frac{\partial y}{\partial \varphi}= r\cos(\varphi)\end{aligned}$ Now we can apply the chain rule:

$\begin{aligned} \frac{\partial z}{\partial r} &=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial r}+ \frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial r}\\ \\ &= \frac{-2y\,(x^2+y^2)}{(x^2-y^2)^2}\cdot \cos(\varphi)+ \frac{2x\,(x^2+y^2)}{(x^2-y^2)^2}\cdot \sin(\varphi) \\ \\ &= \frac{\bigl(2x\sin(\varphi)-2y\cos(\varphi)\bigr)\cdot\bigl(x^2+y^2\bigr)}{(x^2-y^2)^2} = 0 \\ \\ \frac{\partial z}{\partial \varphi} &=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial \varphi}+ \frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial \varphi}\\ \\ &=\frac{-2y(x^2+y^2)}{(x^2-y^2)^2}\cdot -r\sin\varphi+\frac{2x(x^2+y^2)}{(x^2-y^2)^2}\cdot r\cos(\varphi)\\ \\ &= \frac{\bigl(2ry\sin(\varphi)+2rx\cos(\varphi)\bigr)\bigl(x^2+y^2\bigr)}{(x^2-y^2)^2} \\ \\ &= \frac{\bigl(2r^2(\sin^2(\varphi)+\cos^2(\varphi)\bigr)r^2 }{(x^2-y^2)^2} = \frac{2r^4}{(x^2-y^2)^2} \\ \\ &= \frac{2r^4}{(r^2\bigl(\cos^2\varphi-\sin^2\varphi\bigr)^2} = \frac{2}{\cos^2(2\varphi)} \end{aligned}$ Much formula manipulation, but the example also shows that some functions are simpler in a another coordinate system than they are defined and that calculations are simpler.

You can formulate the chain rule when you use polar coordinates in the follwoing way:

If $f(x,y)=f\bigl(r\cos(\varphi), r\sin(\varphi)\bigr)$ then

$\begin{aligned} \frac{\partial f}{\partial r} &= \frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial r}+ \frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial r} = \phantom{-}\,\,\frac{\partial f}{\partial x}\cdot\cos(\varphi)\;\,+\frac{\partial f}{\partial y}\cdot\sin(\varphi) \\ \\ \frac{\partial z}{\partial \varphi} &= \frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial\varphi}+ \frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial \varphi} = -\frac{\partial f}{\partial x}\cdot r\sin(\varphi)+\frac{\partial f}{\partial y}\cdot r\cos(\varphi) \end{aligned}$

For functions of more than two variables one can derive similar chain rules for differentiation. For example, the following rule holds for functions of three variables.

If $w=w(x,y,z)$ is a differentiable function of three variables, and if $x$ , $y$ and $z$ are differentiable functions $t$ , then $w$ is a function of $t$ and

$\frac{\dd w}{\dd t}=\frac{\partial w}{\partial x}\cdot\frac{\dd x}{\dd t}+ \frac{\partial w}{\partial y}\cdot\frac{\dd y}{\dd t}+\frac{\partial w}{\partial z}\cdot\frac{\dd z}{\dd t}$