Chapter 6 Chain Rule

6.1 Definition

Consider univariate functions $y=y(x)$ and $x=x(t)$. Note a change in $t$ will result in a change in $x$, which subsequently results in a change in $y$. How does $y$ change with $t$? One has the well known formula: $$\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}.$$ How does this translate to the multivariate case? Consider a function $z=z(x,y)$, where $x=x(u,v)$ and $y=y(u,v)$. As either $u$ or $v$ change both $x$ and $y$ will change, and subsequently $z$ will change. How does $z$ change with $u$, and how does $z$ change with $v$? It would be theoretically possible to calculate this by a substitution of $x=x(u,v)$ and $y=y(u,v)$ into $z=z(x,y)$, but for most functions this will a tedious computation.

More generally consider the function $z = z(x_1,x_2 \ldots , x_n)$, which is dependent on $n$ variables. Each of the $x_j$ is a function of $k$ variables, that is, $x_j = x_j (t_1, t_2, \ldots, t_k )$. A variation in any of the $t_i$, results in a change in each of the $x_i$, which subsequently changes the value of $z$. How does $z$ change with respect to change in each of the $t_i$.

Chain Rule. Given $z = z(x_1,x_2 \ldots , x_n)$ where $x_i = x_i (t_1, t_2, \ldots, t_k )$, it follows that $$\frac{\partial z}{\partial t_i} = \sum\limits_{j=1}^{n} \frac{\partial z}{\partial x_j} \frac{\partial x_j}{\partial t_i}.$$

Note that the derivatives that appear on the right hand side of the Chain Rule will be routine to compute from the expressions $z = z(x_1,x_2 \ldots , x_n)$ and $x_i = x_i (t_1, t_2, \ldots, t_k )$.

Consider $z=z(x,y)$ where $x=x(u,v)$ and $y=y(u,v)$ as above. By the chain rule $$\begin{align*} \frac{\partial z}{\partial u} &= \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u}, \\ \frac{\partial z}{\partial v} &= \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v}. \end{align*}$$

Consider $u=u(x,y,z)$ where $x=x(\alpha,\beta), y=y(\alpha,\beta)$ and $z=z(\alpha,\beta)$. By the chain rule $$\frac{\partial u}{\partial \alpha} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial \alpha} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial \alpha} + \frac{\partial u}{\partial z} \frac{\partial z}{\partial \alpha}.$$

Consider $u=u(x,y)$ where $x=x(\alpha,\beta, \gamma)$ and $y=y(\alpha,\beta, \gamma)$. By the chain rule $$\frac{\partial u}{\partial \beta} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial \beta} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial \beta}.$$

Calculate $\frac{\partial u}{\partial \beta}$ in Example 6.1.3, and $\frac{\partial u}{\partial \gamma}$ in Example 6.1.4.

Find the derivative of $w = \frac{1}{2} xy+ 2$ with respect to $t$ along the path $x= \cos t$, $y = \sin t$. What is the value of the derivative at $t = \frac{\pi}{2}$?

The question has a nice geometrical interpretation:

Note that $w$ is a function of $x$ and $y$, while $x$ and $y$ are both functions of $t$. Therefore by the Chain rule: $$\begin{align*} \frac{\partial w}{\partial t} &= \frac{\partial w}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial t} \\ &= \frac{1}{2} y \cdot \left( - \sin t \right) + \frac{1}{2} x \cdot \cos t \\ &= - \frac{1}{2} \sin^{2} t + \frac{1}{2} \cos^{2} t \\ &= \frac{1}{2} \cos (2t). \end{align*}$$ At $t = \frac{\pi}{2}$, evaluate: $$\frac{\partial w}{\partial t} \left( \frac{\pi}{2} \right) = - \frac{1}{2}.$$

6.2 Implicit Differentiation

Suppose a function $y(x)$ is given implicitly, that is, $y(x)$ satisfies some equation $F(x,y) =0$. One can deduce a simple formula for $\frac{d y}{d x}$ via the Chain Rule.

Given $y(x)$ satisfying $F(x,y) =0$, then $$\frac{d y}{d x} = - \frac{F_x}{F_y},$$ if $F_y \neq 0$.

Let $w = F(x,y)$, so $w$ is a function of $x$ and $y$. Now $y$ is a function of $x$, and trivially so is $x$ itself. Therefore by the Chain Rule: $$\begin{align*} \frac{dw}{dx} &= \frac{\partial w}{\partial x} \frac{\partial x}{\partial x} + \frac{\partial w}{\partial y} \frac{d y}{d x} \\ &= \frac{\partial F}{\partial x} \cdot 1 + \frac{\partial F}{\partial y} \frac{d y}{d x} \\ &= \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \frac{d y}{d x}. \end{align*}$$

Since $w = F(x,y)=0$ by definition, it follows that $w$ is constant and so $\frac{dw}{dx}=0$. By substitution $$\begin{align*} 0 &= \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \frac{d y}{d x} \\[6pt] \implies \qquad \frac{d y}{d x} &= - \frac{\frac{\partial F}{\partial x}}{ \frac{\partial F}{\partial y}} \\[3pt] &= - \frac{F_x}{F_y} \end{align*}$$

provided $F_y \neq 0$.

Find $\frac{dy}{dx}$ where $y^2 -x^2 - \sin (xy) =0$.

Define $F(x,y) = y^2 -x^2 - \sin (xy)$. Calculate that:

$$\begin{align*} F_x &= -2x - y \cos (xy), \\ F_y &= 2y - x \cos (xy). \end{align*}$$

Therefore by Lemma 6.2.1, provided $F_y \neq 0$:

$$\frac{dy}{dx} = \frac{2x+y \cos (xy)}{2y - x \cos (xy)}.$$

6.3 Transforming Partial Differential Equations

In the Autumn term, we saw how to solve certain collections of ordinary differential equations (ODEs), that is equations that impose conditions on the derivatives of a univariate function. A partial differential equation (PDE), is an equation that imposes a condition on the partial derivatives of a multivariate function. The study of PDEs and their solutions is a deep and active area of research which can be further learnt about in the module MATH2012: Modelling with Differential Equations.

The one-dimensional wave equation for $u(x,t)$ is the PDE

$$\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2},$$

where $c>0$ is some constant.

The wave equation occurs frequently in natural phenonmenon:

• Consider a horizontal vibrating string that is fixed at both ends. Let $t$ represent time, $x$ represent a distance along the string, and $u(x,t)$ be the vertical displacement of the point $x$ at time $t$. Then $u(x,t)$ satisfies the wave equation.

• Consider a sound wave emanating from a point source along a line. Let $t$ represent time, $x$ represent a point at displacement $x$ from the source, and $u(x,t)$ be the perturbation in air density at the point $x$ at time $t$. Then $u(x,t)$ satisfies the wave equation.

• Consider a wave of water travelling along a cross section of a body of water. Let $t$ represent time, $x$ represent a point along the cross section, and $u(x,t)$ be the sea level at the point $x$ at time $t$. Then $u(x,t)$ satisfies the wave equation.

The solution $u(x,t)$ to the wave equation $u_{tt} = c^2 u_{xx}$ is of the form $$u(x,t)=f(x-ct) + g(x+ct)$$ where $f$ and $g$ are some general functions.

Set $$\alpha = \alpha(x,t) = x-ct, \qquad \text{and} \qquad \beta = \beta(x,t) = x+ct.$$ The plan is to interpret the wave equation in terms of $\alpha, \beta$ rather than $x,t$. The differential equation will then be solved in terms of $\alpha,\beta$ before the solution is converted back to in terms of $x,t$.

Calculate that $$x = \frac{\alpha+\beta}{2}, \qquad \text{and} \qquad t = \frac{\beta - \alpha}{2c}.$$ By substitution, one could think of $u$ as a function of $\alpha$ and $\beta$. Therefore by the Chain Rule $$\begin{align*} u_x &= \frac{\partial u}{\partial x} \\ &= \frac{\partial u}{\partial \alpha} \frac{\partial \alpha}{\partial x} + \frac{\partial u}{\partial \beta} \frac{\partial \beta}{\partial x} \\ &= \frac{\partial u}{\partial \alpha} \cdot 1 + \frac{\partial u}{\partial \beta} \cdot 1 \\ &= \frac{\partial u}{\partial \alpha} + \frac{\partial u}{\partial \beta} \\ &= u_\alpha + u_\beta. \end{align*}$$

Note that this equation holds for any function $u$. Therefore it this setting it is correct to state there is an equivalence of operators: $$\frac{\partial}{\partial x} = \frac{\partial}{\partial \alpha} + \frac{\partial}{\partial \beta},$$ that is, differentiating a function with respect to $x$ is the same as the sum of the derivative with respect to $\alpha$ and the derivative with respect to $\beta$.

Similarly the Chain Rule gives $$u_t = \frac{\partial u}{\partial t} = -c \frac{\partial u}{\partial \alpha} + c \frac{\partial u}{\partial \beta} = -c u_\alpha + c u_\beta,$$ and in terms of operators $$\frac{\partial}{\partial t} = -c \frac{\partial}{\partial \alpha} + c \frac{\partial}{\partial \beta}$$ Now calculating the second order derivatives of $u$ that appear in the PDE, one has

$$\begin{align*} u_{xx} &= \frac{\partial}{\partial x} u_x \\ &= \left( \frac{\partial}{\partial \alpha} + \frac{\partial}{\partial \beta} \right) \left( u_\alpha + u_\beta \right) \\ &= u_{\alpha \alpha} + u_{\beta \alpha} + u_{\alpha \beta} + u_{\beta \beta} \\ &= u_{\alpha \alpha} + 2 u_{\beta \alpha} + u_{\beta \beta} \\[5pt] u_{tt} &= \frac{\partial}{\partial t} u_t \\ &= \left( -c \frac{\partial}{\partial \alpha} + c \frac{\partial}{\partial \beta} \right) \left( -c u_\alpha + c u_\beta \right) \\ &= c^2 u_{\alpha \alpha} -c^2 u_{\beta \alpha} -c^2 u_{\alpha \beta} + c^2 u_{\beta \beta} \\ &= c^2 u_{\alpha \alpha} - 2 c^2 u_{\alpha \beta} + c^2 u_{\beta \beta} \end{align*}$$

By substitution into the wave equation

$$\begin{align*} 0 &= u_{tt} - c^2 u_{xx} \\ &= \Big( c^2 u_{\alpha \alpha} - 2 c^2 u_{\alpha \beta} + c^2 u_{\beta \beta} \Big) - c^2 \Big( u_{\alpha \alpha} + 2 u_{\beta \alpha} + u_{\beta \beta} \Big) \\ &= -4 c^2 u_{\alpha \beta} \\ \iff \qquad u_{\alpha \beta} &= 0 \end{align*}$$

Solving this new differential equation:

$$\begin{align*} u_{\alpha \beta} &= 0 \\ \int u_{\alpha \beta} \,d\beta &= \int 0 \,d\beta \\ u_{\alpha} &= c(\alpha) \\ \int u_{\alpha} \,d\alpha &= \int c(\alpha) \,d\alpha \\ u &= f(\alpha) + g(\beta), \qquad \text{where } \frac{df}{d \alpha} = c. \end{align*}$$

Therefore

$$u(x,t) = f(x-ct) + g(x+ct)$$

is the general solution to the wave equation.

Physically the variable $t$ represents time, and the variable $x$ represents some sort of spatial displacement in a one-dimensional space. It follows that the solution to the wave equation is a combination of two general waves moving in opposite directions along the $x$-axis with speed $c$.