Mathematical Introduction for Physics and Engineering

by Samuel Dagan (Copyright © 2007-2020)

**Previous topic: page 1 Definition and Rules**

**Next topic: page 3 Elementary Functions**

The present page concerns the properties of the derivative and therefore only deals with points of an open differentiable interval of a function (open - in order to exclude the end points).

From the definition, **if at a given point of the independent variable x the derivative is positive(negative), the function must be strictly increasing(decreasing) there**. In this case,

By using the mathematical language **the positive(negative) sign of a derivative at a given point is sufficient (but not necessary) condition for a function to be increasing(decreasing) there**.

If the derivative is zero at a given point, it means that the tangent line is parallel to the *x* axis and locally the function does not change. Such point is called **a stationary point** of the function, and **is defined as a root of the derivative**.

Let's take a closer look at a srictly **increasing(decreasing) function in an interval, but with one stationary point. This means that** the derivative is positive(negative) for all the interval, except for this point. Therefore **the derivative itself has a minimum(maximum) at that particular point**.

Let's take the following ** example**:

$\begin{array}{l}y={x}^{3}\\ \frac{\text{d}y}{\text{d}x}=3{x}^{2}\end{array}\}$ | (1.3.2.1) |
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This function is strictly increasing and its derivative is positive except **at point x = 0** , where

The graphic presentation of this example (1.3.2.1) is at **Fig. Derivative of x³** .

A stationary point is obtained **at a (local) maximum(minimum) of a differentiable function**, since **the derivative is positive(negative) at the left-hand side of the point, and - is negative(positive) at the right-hand side**. In other words the derivative should intersect the *x* axis at that point. By using mathematical language **a stationary point is a necessary, but not sufficient condition for a maximum(minimum)** of a differentiable fuction.

We have learned so far how to look at the sign of a derivative in order to study if a function is increasing or decreasing, but we cannot specify what happens at a stationary point without knowing how the derivative is changing there. In order to learn if a derivative is increasing or decreasing, **we need** to know **the derivative of the derivative, which is called the second order derivative and commonly - the second derivative**.

**A second derivative is obtained by the differentiation of the first derivative with the use of exactly the same rules**. Formally speaking the second order derivative is obtained by the limit:

$\frac{{\text{d}}^{2}y}{\text{d}{x}^{2}}=\underset{\Delta x\to 0}{\mathrm{lim}}\frac{\frac{\text{d}y}{\text{d}x}\left(x+\Delta x\right)-\frac{\text{d}y}{\text{d}x}\left(x\right)}{\Delta x}$ | (1.3.2.2) |
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The **notation
$\frac{{\text{d}}^{2}y}{\text{d}{x}^{2}}$** is not symmetric about the superscript 2 due to the second division by Δ*x* . It **shows** also the physical dimensions: **the dimensions of y divided by the square of the dimensions of x** . For instance if

$\begin{array}{l}v=\frac{\text{d}x}{\text{d}t}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\text{velocity=}\frac{\text{length}}{\text{time}}\\ a=\frac{\text{d}v}{\text{d}t}=\frac{{\text{d}}^{2}x}{\text{d}{t}^{2}}\text{\hspace{1em}}\text{acceleration=}\frac{\text{length}}{{\text{time}}^{\text{2}}}\end{array}\}$ | (1.3.2.3) |
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One defines in the same way, high order derivatives. **The n^{th} derivative is denoted by
$\frac{{\text{d}}^{n}y}{\text{d}{x}^{n}}$** , and its physical dimensions are the ratio of the dimensions of

So far we did not ask the question: **Does the differentiability of a function implement the differentiability of its derivative? The answer is: No! ** Any consecutive derivative could increase the number of the points that are not differentiable any more. But usually such behaviour contains in itself important information about the function, as the case of diverengence to infinity which will be discussed later on.

We are going to see how the properties of the **high order derivatives at a given stationary point can help us in the analysis of a function**.

In case of a maximum(minimum) the first derivative should decrease(increase) since its values are changing from positive(negative) to negative(positive) values. Therefore, if the second derivative is negative(positive), we are sure that there is a maximum(minimum) there. By use of the mathematical language: **a stationary point with negative(positive) second derivative is a sufficient condition for a maximum(minimum) there**.

As an example the function *y* = *x*(*x*-2) has a stationary point at *x* = 1 :

$\begin{array}{l}y=x(x-2)\\ \frac{\text{d}y}{\text{d}x}=2\left(x-1\right)\\ \frac{{\text{d}}^{\text{2}}y}{\text{d}{x}^{2}}=2\end{array}\}$ | (1.3.2.4) |
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Since the second derivative is positive, there is minimum at that point. This example is graphically presented at **Fig. Derivatives of x²** .

We now can go **back** **to a stationary point of an increasing(decreasing) function** and find out how to implement the information from the higher order derivatives. We already know that in such a case the first derivative should have a minimum(maximum) there. This minimum(maximum) point should behave exactly the same way as any other of the same kind:

- Since the
**first derivative**has a minimum(maximum) at the stationary point, then - The
**second derivative**will be also zero at that point because of this minimum(maximum). We know also that a derivative of a function with minimum(maximum) is by itself an increasing(decreasing) function there. As a result the second derivative should be an increasing(decreasing) function and we are back to square one. - If we are lucky the
**next (third) derivative**will be positive(negative), which will indicate that the original function is increasing(decreasing). If we are not lucky and we again obtain zero, we have to continue till we find that the order of the first non-zero derivative is odd, and use its sign for the original function.

Let's go back now to the example (1.3.2.1):

$\begin{array}{l}y={x}^{3}\\ \frac{\text{d}y}{\text{d}x}=3{x}^{2}\\ \frac{{\text{d}}^{2}y}{\text{d}{x}^{2}}=6x\\ \frac{{\text{d}}^{3}y}{\text{d}{x}^{3}}=6\end{array}\}$ | (1.3.2.5) |
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In this example all the terms vanish at *x* = 0 except the third derivative which is positive and indicates that the function *y* = *x*³ is increasing there. The graphic presentation of (1.3.2.5) is at **Fig. Derivatives of x³** .

One can extend the previous discussion to the case of a maximum(minimum). We already know that the sufficient condition for a maximum(minimum) is a stationary point with negative(positive) second derivative as consequence of the decrease(increase) of the first derivative. But we already know that the derivative of a decreasing(increasing) function could vanish. In such a case, following the same pattern of reasoning, **in case of minimum(maximum) we have to continue the differentiation, until we find that the order of the first non-zero derivative is even, and use its sign as the indicator**.

As an example of a minimum with vanishing second derivative we'll use:

$\begin{array}{l}y={\left(x-1\right)}^{4}-1\\ \frac{\text{d}y}{\text{d}x}=4{\left(x-1\right)}^{3}\\ \frac{{\text{d}}^{2}y}{\text{d}{x}^{2}}=12{\left(x-1\right)}^{2}\\ \frac{{\text{d}}^{3}y}{\text{d}{x}^{3}}=24\left(x-1\right)\\ \frac{{\text{d}}^{4}y}{\text{d}{x}^{4}}=24\end{array}\}$ | (1.3.2.6) |
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The function has a minimum at *x* = 1 . The lower order derivative that does not vanish is 4. The graphic presentation of (1.3.2.6) is at **Fig. Derivatives of x^{4}
** .

Before summarising the rules for a stationary point of a function, there is a question to be asked. In the previous discussions, we looked at a row of higher order derivatives for the first one that does not vanish. What happens if there is no such one? **The only function having an infinite row of vanishing derivatives is y=constant** and therefore we do not have to worry.

- If its order is odd, then
- if its sign is positive, the function is increasing
- if its sign is negative, the function is decreasing

- If its order is even, then
- if its sign is positive, the function is at minimum
- if its sign is negative, the function is at maximum

So far we dealt with stationary points of a function, meaning we looked for the roots of the first derivative. In case of inflection points, **we look for the roots of the second derivative** independently of the first.

When a function curves (is not a straight line) and **the slope** (the first derivative) changes, e.g. **diminishes**, graphically this is expressed as a **clockwise** (mathemetically negative) **rotation** if we follow the function in the direction of the independent variable (*x*). If **the slope is growing, the rotation is counter-clockwise** (positive mathematical sense). This graphical presentation is enhanced furthermore if we look at the animated motion of the tangent line. **The point** (if it exists) **where this sense of rotation changes direction, is called an inflection point**.

If the sense of rotation changes from clockwise(anticlockwise) to anticlockwise(clockwise), it means that the first derivative which is decreasing(increasing) starts to increase(decrease), and therefore **the first derivative should have a minimum(maximum) at the inflection point**. Of course the second derivative should vanish and the first consecutive non-zero derivative should be of odd-order. The sign of this derivative is related to the minimum or maximum of the first derivative. A root of **the second derivative is a necessary**, but not sufficient **condition for an inflection point**. **A point of minimum or maximum of the first derivative is a necessary and sufficient condition for an inflection point**.

The stationary point of *y* = *x*³ is also an inflection point. This and more examples are shown graphically at
**Fig. Inflection points**.

There are many practical verbal problems of minimum or maximum where the function to be dealt with is not presented, but has to be worked out. An ** example** of such problem is the following:

The luminosity at a given point is proportional to the light power of a source divided by its distance-squared. Two light sources of different power are situated at a given distance amongst them. One has to find the point of the minimum luminosity situated on a straight line between them. Solution:

First one have to parametrize the problem. Assume that the sources are situated on the *x* axis, the first of power *A* at point *x* = *a* and the second of power *B* at point *x* = *b* .
We know that the luminosity is proportional to power/distance^{2} and for finding its minimum we don't need the factor of proportion. Therefore we can write the luminosity *y* at point *x* by

$y=A{\left(x-a\right)}^{-2}+B{\left(x-b\right)}^{-2}$ |
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After parametrising the problem, we have to find the stationary point for a *x* between the *a* and *b* , and to continue differentiating till we make sure that the point is a minimum. The first derivative is

$\frac{\text{d}y}{\text{d}x}=-2\left[A{\left(x-a\right)}^{-3}+B{\left(x-b\right)}^{-3}\right]$ |
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By requiring $\frac{\text{d}y}{\text{d}x}=0$ one obtains

$A{\left(x-b\right)}^{3}=-B{\left(x-a\right)}^{3}$ |
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After taking the third root of both sides we obtain

$x=\frac{a\text{\hspace{0.17em}}\sqrt[3]{B}+b\text{\hspace{0.17em}}\sqrt[3]{A}}{\sqrt[3]{B}+\sqrt[3]{A}}$ |
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that is obviously situated between *a* and *b* . In order to validate the minimum we look at the second derivative:

$\frac{{\text{d}}^{\text{2}}y}{\text{d}{x}^{2}}=6\left[A{\left(x-a\right)}^{-4}+B{\left(x-b\right)}^{-4}\right]$ |
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which is obviously positive as expected for a minimum.

We'll now have a look at **continuous functions with a point of a derivative diverging to infinity**, which we consider as a point where the function is not differentiable. If the function is *y* = *f*(*x*) , and the point corresponds to *x* = *x*_{0} then the line ** x = x_{0} defines the tangent line** there.

There are **two categories: the divergence from the left and from the right have the same sign or - different signs**.

In **case of the same sign**: The derivative from the left and from the right diverges to positive(negative) infinity, it means that the function from both sides is increasing(decreasing), which **makes the function** also **increasing(decreasing)** at that point. The infinity means that for an increasing(decreasing) function the derivative increases(decreases) at the left and decreases(increases) at the right. In other words this **indicates an inflection point**.

Let's take the following ** example**:

$\begin{array}{l}y=\sqrt[3]{x}\\ \frac{\text{d}y}{\text{d}x}=\frac{1}{3\text{\hspace{0.28em}}\sqrt[3]{{x}^{2}}}\end{array}\}$ | (1.3.2.7) |
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The function is increasing and the derivative is positive at both - the lefthand and the righthand sides of the divergence point *x* = 0 .

In **case of different signs**: If the derivative from the left diverges to positive(negative) infinity means that the function increases(decreases) at the left side, and since the sign is opposite at the right, it decreases(increases) there. As result - **the point is a maximum(minimum)**.

A suitable ** example** is:

$y=\sqrt[3]{\left|x\right|}=\left\{\begin{array}{l}\text{for}\text{\hspace{1em}}x<0\{\begin{array}{l}y=-\sqrt[3]{x}\\ \frac{\text{d}y}{\text{d}x}=-\frac{1}{3\text{\hspace{0.28em}}\sqrt[3]{{x}^{2}}}\end{array}\\ \text{for}\text{\hspace{1em}}x>0\{\begin{array}{l}y=\sqrt[3]{x}\\ \frac{\text{d}y}{\text{d}x}=\frac{1}{3\text{\hspace{0.28em}}\sqrt[3]{{x}^{2}}}\end{array}\end{array}\right\}$ | (1.3.2.8) |
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which corresponds to a minimum at *x* = 0 .

These two examples are shown graphically at **Fig. Infinite derivative**.

Study and do approximate graphic presentations of the following functions. The study should contain:

- Roots and infinities
- Increasing and decreasing intervals
- Stationary points and their characteristics
- Points of infinite derivative and their characteristics
- Inflection points

$y={x}^{3}+5{x}^{2}+4x$ |
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$y={x}^{{\scriptstyle \frac{2}{3}}}\left(x-1\right)$ |
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$y=\frac{2x}{1-{x}^{2}}$ |
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**Exercise 4.** The diagonal of a rectangle is *d*. What is its maximal possible area?

It is possible to obtain a minimal area also. What is it, and why don't you get it from the stationary point?

**Exercise 5.** ${x}_{1\text{\hspace{0.28em}}}{x}_{2\text{\hspace{0.28em}}}\mathrm{.....}\text{\hspace{0.28em}}{x}_{n-1}\text{\hspace{0.28em}}{x}_{n}$ are *n* given points on the *x* axis. Find the *x* that makes

$\sum _{k=1}^{n}}{\left({x}_{k}-x\right)}^{2$ |
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**Previous topic: page 1 Definition and Rules**

**Next topic: page 3 Elementary Functions**