Supplement 2.6: Differentials and derivatives (1/4)

Functions of one variable

In analysis in schools, functions as

y = f (x)

are dealt with, which depend on only one variable namely $x$ . Examples:

y = x^{2}

y = sin x

y = e^{a x}

The derivatives of $x$ are symbolically labelled with $y'$ , with $f' (x)$ or with $\frac{d y}{d x}$ .

In accordance with the third option, the derivative is the differential quotient of two quantities $d y$ and $d x$ . These quantities are called differentials, because they become infinitely small differences from finitely large differences $Δ y = y_{2} - y_{1}$ and $Δ x = x_{2} - x_{1}$ through passing a limit

lim_{Δ x \to 0} \frac{Δ y}{Δ x} = \frac{d y}{d x}

They become what is named infinitesimal differences.

Differentials do not necessarily have to be parts of quotients but may stand alone. Example:

y = x^{2} \Rightarrow \frac{d y}{d x} = 2 x \Leftrightarrow d y = 2 x d x

Function of several variables

Natural processes in most cases depend on several environmental variables and this is what occurs in mathematical modelling of such processes as well. One example: the temperature at a particular location takes on different values at other locations and therefore is a function of the room coordinates $x$ , $y$ and $z$ . Moreover, it takes on different values for different times $t$ . The temperature thus is a function of four variables: $T = f (x, y, z, t)$ .

This example of changes in temperature in space and time will be discussed in the following in order to concretely show the importance of differentials and derivatives. But first, the use of these terms shall be developed strictly formally, for what the named quantities are not of descriptive means.

A function $z$ shall be dependent on two variables $x$ and $y$ :

z = f (x, y)

How can the derivative of $z$ be formed? How is it possible to show that $z$ is derived with respect to $x$ or $y$ or even both at the same time? It is clear, that the styles $z'$ or $f' (x, y)$ do not help since they leave the question unanswered. What serves simplification is writing it as a differential quotient.

Equations ↓

Looking at the derivative with respect to $x$ first:

\frac{\partial z}{\partial x} = \frac{\partial f (x, y)}{\partial x}

Through the $\partial$ -symbol at the place of $d$ , it is shown that the function $z$ is derived with respect to $x$ only; $y$ -components are considered to be constant. Therefore, this is called partial derivative of $z$ with respect to $x$ . The partial derivative with respect to $y$ is written in the same way:

\frac{\partial z}{\partial y} = \frac{\partial f (x, y)}{\partial y}

Sometimes, the variables taken as constant are written as an index in brackets on the bottom right, i.e.:

{(\frac{d z}{d x})}_{y} = \frac{\partial z}{\partial x}

{(\frac{d z}{d y})}_{x} = \frac{\partial z}{\partial y}

This convention is frequently used in thermodynamics to show, what variables shall be kept as constants; in most cases - and in this special one - the $\partial$ -symbol is sufficient.

Through multiplication of the partial derivatives with the differentials of the respective variables and summation of both terms, it is possible to write the differential of the function $z$ :

d z = \frac{\partial f (x, y)}{\partial x} d x + \frac{\partial f (x, y)}{\partial y} d y

in short:

d z = \frac{\partial z}{\partial x} d x + \frac{\partial z}{\partial y} d y

Task 1: partial derivatives and differentials ↓

Please calculate the partial derivatives and the total differential of the following functions:

z = x^{3} y + x^{2} y^{2} + x y^{3}

z = x sin y - y sin x

You may find the solutions on the site. But try solving it yourself first!

Above we have written down the total differential of $z$ . It is obviously not possible to show the first derivative of the function $z$ with respect to both variables. The second derivative is easier to specify, it is written $\frac{\partial^{2} z}{\partial x \partial y}$ .

We may find an expression for the first derivative, if the variables $x$ and $y$ depend on a common parameter $t$ . If

x = f_{x} (t)

y = f_{y} (t)

then the derivatives of $x$ and $y$ with respect to $t$ can be calculated. Considering these derivatives in the equation of the differential $d z$ , it follows:

\frac{d z}{d t} = \frac{\partial z}{\partial x} \frac{d x}{d t} + \frac{\partial z}{\partial y} \frac{d y}{d t}

In this equation, $z$ is derived with respect to all variables. Therefore the expression $\frac{d z}{d t}$ is the total derivative of the function $z = f (x, y)$ .

The terms $\frac{d x}{d t}$ and $\frac{d y}{d t}$ do not have to be written with the $\partial$ -symbol since $x$ and $y$ depend on $t$ only and not on other variables; partial derivatives thus are not necessary at this point.

Task 2: total derivatives ↓

Find the total derivative of

z = x^{2} + 3 x y + 5 y^{2}

with $x = sin t$ und $y = cos t$ .

Again, you may find the solutions on this page.

Exercise 3: Partial derivative and total differential of a vector ↓

The calculation rule for the differential applies not only to scalar functions, but also to vectors in the same way. We consider the vector $\vec{A} (x, y, z)$ :