Partial Differentiation

The usual rules for differentiation apply when dealing with several variables, but we now require to treat the variables one at a time, keeping the others constant. It is for this reason that a new symbol for differentiation is introduced. Consider the function

f (x, y) = $${\frac{{2y}}{{y + \cos x}}}$$

We can consider y fixed, and so treat it as a constant, to get a partial derivative

$${\frac{{\partial f}}{{\partial x}}}$$ = $${\frac{{2y \sin x}}{{(2y + \cos x)^2}}}$$

where we have differentiated with respect to x as usual. Or we can treat x as a constant, and differentiate with respect to y, to get

$${\frac{{\partial f}}{{\partial y}}}$$ = $${\frac{{(2y + \cos x).2 - 2y.2}}{{(2y + \cos x)^2}}}$$ = $${\frac{{2\cos x}}{{(2y + \cos x)^2}}}$$.

Although a partial derivative is itself a function of several variables, we often want to evaluate it at some fixed point, such as (x₀, y₀). We thus often write the partial derivative as

$${\frac{{\partial f}}{{\partial x}}}$$(x₀, y₀).

There are a number of different notations in use to try to help understanding in different situations. All of the following mean the same thing:-

$${\frac{{\partial f}}{{\partial x}}}$$(x₀, y₀),        f₁(x₀, y₀),        f_x(x₀, y₀)    and    D₁F(x₀, y₀).

Note also that there is a simple definition of the derivative in terms of a Newton quotient:-

$${\frac{{\partial f}}{{\partial x}}}$$ = $$\lim_{{\delta x \to 0}}^{}$$$${\frac{{f(x_0 + \delta x,y_0) - f(x_0,y_0)}}{{\delta x}}}$$

provided of course that the limit exists.

Example 8.5   Let z = sin(x/y). Compute x$${\frac{{\partial z}}{{\partial x}}}$$ + y$${\frac{{\partial z}}{{\partial y}}}$$.

Solution. Treating first y and then x as constants, we have

$${\frac{{\partial z}}{{\partial x}}}$$ = $${\frac{{1}}{{y}}}$$cos$$\left(\vphantom{\frac{x}{y}}\right.$$$${\frac{{x}}{{y}}}$$$$\left.\vphantom{\frac{x}{y}}\right)$$    and    $${\frac{{\partial z}}{{\partial y}}}$$ = $${\frac{{-x}}{{y^2}}}$$cos$$\left(\vphantom{\frac{x}{y}}\right.$$$${\frac{{x}}{{y}}}$$$$\left.\vphantom{\frac{x}{y}}\right)$$.

thus

x$${\frac{{\partial z}}{{\partial x}}}$$ + y$${\frac{{\partial z}}{{\partial y}}}$$ = $${\frac{{x}}{{y}}}$$cos$$\left(\vphantom{\frac{x}{y}}\right.$$$${\frac{{x}}{{y}}}$$$$\left.\vphantom{\frac{x}{y}}\right)$$ - $${\frac{{x}}{{y}}}$$cos$$\left(\vphantom{\frac{x}{y}}\right.$$$${\frac{{x}}{{y}}}$$$$\left.\vphantom{\frac{x}{y}}\right)$$ = 0.

Note:This equation is an equation satisfied by the function we started with, which involves both the function, and its partial derivatives. We shall meet a number of examples of such a partial differential equation later.

Exercise 8.6   Let z = log(x/y). Show that x$${\frac{{\partial z}}{{\partial x}}}$$ + y$${\frac{{\partial z}}{{\partial y}}}$$ = 0.

The fact that the last two function satisfy the same differential equation is not a co-incidence. With our next result, we can see that for any suitably differentiable function f, the function z(x, y) = f (x/y) satisfies this partial differential equation.

Exercise 8.7   Let z = f (x/y), where f is suitably differentiable. Show that x$${\frac{{\partial z}}{{\partial x}}}$$ + y$${\frac{{\partial z}}{{\partial y}}}$$ = 0.

Because the definitions are really just version of the 1-variable result, these examples are quite typical; most of the usual rules for differentiation apply in the obvious way to partial derivatives exactly as you would expect. But there are variants. Here is how we differentiate compositions.

Theorem 8.8   Assume that f and all its partial derivatives f_x and f_y are continuous, and that x = x(t) and y = y(t) are themselves differentiable functions of t. Let

F(t) = f (x(t), y(t)).

Then F is differentiable and

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

Proof. Write x = x(t), x₀ = x(t₀) etc. Then we calculate the Newton quotient for F.

F(t) - F(t₀) = f (x, y) - f (x₀, y₀)

= f (x, y) - f (x₀, y) + f (x₀, y) - f (x₀, y₀)

$${\frac{{\partial f}}{{\partial x}}} \xi {\frac{{\partial f}}{{\partial y}}} \eta$$ =

Here we have used the Mean Value Theorem (5.18) to write

f (x, y) - f (x₀, y) = $${\frac{{\partial f}}{{\partial x}}}$$($$\xi$$, y)(x - x₀)

for some point $\xi$ between x and x₀, and have argued similarly for the other part. Note that $\xi$, pronounced ``Xi'' is the Greek letter ``x''; in the same way $\eta$, pronounced ``Eta'' is the Greek letter ``y''. Thus

$${\frac{{F(t) - F(t_0)}}{{t - t_0}}}$$ = $${\frac{{\partial f}}{{\partial x}}}$$($$\xi$$, y)$${\frac{{(x - x_0)}}{{t - t_0}}}$$ + $${\frac{{\partial f}}{{\partial y}}}$$(x₀,$$\eta$$)$${\frac{{(y - y_0)}}{{t - t_0}}}$$

Now let t$\to$t₀, and note that in this case, x$\to$x₀ and y$\to$y₀; and since $\xi$ and $\eta$ are trapped between x and x₀, and y and y₀ respectively, then also $\xi$$\to$x₀ and $\eta$$\to$y₀. The result then follows from the continuity of the partial derivatives. $\qedsymbol$

Example 8.9   Let f (x, y) = xy, and let x = cos t, y = sin t. Compute $${\frac{{df}}{{dt}}}$$ when t = $\pi$/2.

Solution. From the chain rule,

$${\frac{{df}}{{dt}}}$$ = $${\frac{{\partial f}}{{\partial x}}}$$$${\frac{{dx}}{{dt}}}$$ + $${\frac{{\partial f}}{{\partial y}}}$$$${\frac{{dy}}{{dt}}}$$ = - y(t)sin t + x(t)cos t = - 1.sin($$\pi$$/2) = - 1.

The chain rule easily extends to the several variable case; only the notation is complicated. We state a typical example

Proposition 8.10   Let x = x(u, v), y = y(u, v) and z = z(u, v), and let f be a function defined on a subset U $\in$ $\mathbb {R}$³, and suppose that all the partial derivatives of f are continuous. Write

F(u, v) = f (x(u, v), y(u, v), z(u, v)).

Then

$${\frac{{\partial F}}{{\partial u}}}$$ = $${\frac{{\partial f}}{{\partial x}}}$$$${\frac{{\partial x}}{{\partial u}}}$$ + $${\frac{{\partial f}}{{\partial y}}}$$$${\frac{{\partial y}}{{\partial u}}}$$ + $${\frac{{\partial f}}{{\partial z}}}$$$${\frac{{\partial z}}{{\partial u}}}$$    and    $${\frac{{\partial F}}{{\partial v}}}$$ = = $${\frac{{\partial f}}{{\partial x}}}$$$${\frac{{\partial x}}{{\partial v}}}$$ + $${\frac{{\partial f}}{{\partial y}}}$$$${\frac{{\partial y}}{{\partial v}}}$$ + $${\frac{{\partial f}}{{\partial z}}}$$$${\frac{{\partial z}}{{\partial v}}}$$.

The introduction of the domain of f above, simply to show it was a function of three variables is clumsy. We often do it more quickly by saying

Let f (x, y, z) have continuous partial derivatives

This has the advantage that you are reminded of the names of the variables on which f acts, although strictly speaking, these names are not bound to the corresponding places. This is an example where we adopt the notation which is very common in engineering maths. But note the confusion if you ever want to talk about the value f (y, z, x), perhaps to define a new function g(x, y, z).

Example 8.11   Assume that f (u, v, w) has continuous partial derivatives, and that

u = x - y,        v = y - z        w = z - x.

Let

F(x, y, z) = f (u(x, y, z), v(x, y, z), w(x, y, z)).

Show that

$${\frac{{\partial F}}{{\partial x}}}$$ + $${\frac{{\partial F}}{{\partial y}}}$$ + $${\frac{{\partial F}}{{\partial z}}}$$ = 0.

Solution. We apply the chain rule, noting first that from the change of variable formulae, we have

$${\frac{{\partial u}}{{\partial x}}}$$ = 1,

$${\frac{{\partial u}}{{\partial y}}}$$ = - 1,

$${\frac{{\partial u}}{{\partial z}}}$$ = 0,

        $$\begin{array}{rl} \displaystyle \frac{\partial v}{\partial x} &=... ...&= 1,\ [2ex] \displaystyle \frac{\partial v}{\partial z} &= -1, \end{array}$$        $$\begin{array}{rl} \displaystyle \frac{\partial w}{\partial x} &=... ... &= 0,\ [2ex] \displaystyle \frac{\partial w}{\partial z} &= 1. \end{array}$$

Then

$${\frac{{\partial F}}{{\partial x}}} {\frac{{\partial f}}{{\partial u}}} {\frac{{\partial f}}{{\partial v}}} {\frac{{\partial f}}{{\partial w}}} {\frac{{\partial f}}{{\partial u}}} {\frac{{\partial f}}{{\partial w}}}$$ =

$${\frac{{\partial F}}{{\partial y}}} {\frac{{\partial f}}{{\partial u}}} {\frac{{\partial f}}{{\partial v}}} {\frac{{\partial f}}{{\partial w}}} {\frac{{\partial f}}{{\partial u}}} {\frac{{\partial f}}{{\partial v}}}$$ =

$${\frac{{\partial F}}{{\partial z}}} {\frac{{\partial f}}{{\partial u}}} {\frac{{\partial f}}{{\partial v}}} {\frac{{\partial f}}{{\partial w}}} {\frac{{\partial f}}{{\partial v}}} {\frac{{\partial f}}{{\partial w}}}$$ =

Adding then gives the result claimed.