Notes for Calculus

Chapter 3 : Differentiation

Author: Kenneth, S.K. Cheng

Table of Contentss)

• Notes for Calculus
  • Chapter 3 : Differentiation
    • Author: Kenneth, S.K. Cheng
  • Table of Contentss)
    • 3.1 Introduction
    • 3.2 Definition of the Derivative
      • Exercises for this section
    • 3.3 More about Differentiation
      • Exercises for this section
    • 3.4 Curves and Tangents
      • Exercises for this section
    • 3.5 Techniques of Differentiation
      • Exercises for this section
    • 3.6 Differentials. Further Notation
    • 3.7 Implicit Differentiation
      • Exercises for this section
    • 3.8 Higher Derivatives
      • Exercises for this section
    • 3.9 Maxima and Minima
    • 3.10 Rolle's Theorem and Further Applications
      • Exercises for this section
    • 3.11 Applications
      • Exercises for this section
    • 3.12 Antiderivatives
      • Exercises for this section
    • 3.13 Using Python to Differentiate
      • Exercises for this section
    • References

3.1 Introduction

When we face physical problems, they habitually involve the "rate of change" of one quantity with respect to another.

For example, in a first course on physics, you will encounter the rate of change of distance with respect to time, called velocity, the rate of change of velocity with respect to time, called acceleration, the rate of change of length (of a metal rod, say) with respect to temperature, called the coefficient of linear expansion, the rate of change of mass (of a wire of variable cross section, say) with respect to length, called the linear density. These are all special cases of the general mathematical concept of the rate of change or derivative of a function with respect to its argument.

To see just how the derivative should be defined, we now take a closer look at one of the physical quantities just enumerated:

Example 3.1.1 The velocity of a particle moving along a straight line is given by the function $v(t)$, where $t$ is the time. The average velocity of the particle over the time interval $[t, t + \Delta t]$ is given by the formula

$$\frac{v(t+\Delta t) - v(t)}{\Delta t}.$$

the instantaneous velocity of the particle at time $t$ is the limit of the average velocity as $\Delta t$ approaches zero. This limit is called the derivative of $v(t)$ at $t$ and is denoted by $v'(t)$.

3.2 Definition of the Derivative

Definition 3.2.1 Let $f$ be a function defined on an open interval $I$ containing the point $a$. The derivative of $f$ at $a$, denoted by $f'(a)$, is defined by the formula

$$f'(a) = \lim_{\Delta x \to 0} \frac{f(a + \Delta x) - f(a)}{\Delta x},$$

provided that this limit exists. The operation leads $f$ to its derivative is called differentiation. If $f$ fas a finite derivative at $a$, then $f$ is said to be differentiable at $a$. Generally, if $f$ is differentiable at every point of an interval $I$, then $f$ is said to be differentiable on $I$.

Remark: The quotient $\frac{f(a + \Delta x) - f(a)}{\Delta x}$ is called the difference quotient of $f$ at $a$. It is introduced to measure the average rate of change of $f$ over the interval $[a, a + \Delta x]$.

Remark: . The distinction between $f$ having a derivative at $x$ and being differentiable at $x$ is important. In the first case, $f'(x)$ may be infinite, but not in the second case.

Example 3.2.1 The constant function $f(x) = k$ is differentiable at every point of its domain, and its derivative is zero at every point. The reason is that

$$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{k - k}{\Delta x} = 0.$$

Example 3.2.2 The function $f(x) = x$ is also differentiable at every point of its domain, and its derivative is one at every point. The reason is that

$$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{x + \Delta x - x}{\Delta x} = \lim_{\Delta x \to 0} \frac{\Delta x}{\Delta x} = 1.$$

Example 3.2.3 Let $f(x) = x^n$ where $n$ is a positive integer, then differentiating $f$.

Solution: First we have,

$$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{(x + \Delta x)^n - x^n}{\Delta x}.$$

Then one may think of using the Binomial Theorem, which results in the following

$$f'(x) = \lim_{\Delta x \to 0} \frac{x^n + nx^{n-1}\Delta x + \cdots + \Delta x^n - x^n}{\Delta x} = \lim_{\Delta x \to 0} nx^{n-1} + \cdots + \Delta x^{n-1} = nx^{n-1}.$$

Since every term in the limit is zero except the first term, we have $f'(x) = nx^{n-1}$.

Remark: This is the power rule for differentiation. It is a very useful rule for differentiating polynomials. But one may note that the power is only allowed to be a positive integer, for the cases that the power is a negative integer or a fraction, the power rule is not applicable.

Exercises for this section

1. A particle moves along a straight line with equation of motion

$$s(t) = \frac{1}{3}t^3 - 3t^2 + 9t + 1.$$

Find the velocity and acceleration of the particle at time $t$.

2. Differentiate the following functions:
   1. $f(x) = \frac{1}{x}$
   2. $f(x) = x^2$
   3. $f(x) = \sqrt{2x+1}$

3.3 More about Differentiation

We first think of the following examples:

Example 3.3.1: Differentiate the $\sin x$ and $\cos x$.

Solution: We have

$$\begin{aligned} \sin'(x) &= \lim_{\Delta x \to 0} \frac{\sin(x + \Delta x) - \sin x}{\Delta x} \\&= \lim_{\Delta x \to 0} \frac{\sin x \cos \Delta x + \cos x \sin \Delta x - \sin x}{\Delta x} \\ &= \lim_{\Delta x \to 0} \sin x \frac{\cos \Delta x - 1}{\Delta x} + \cos x \frac{\sin \Delta x}{\Delta x} = \cos x. \end{aligned}$$

Similarly, we have

$$\begin{aligned} \cos'(x) &= \lim_{\Delta x \to 0} \frac{\cos(x + \Delta x) - \cos x}{\Delta x} \\&= \lim_{\Delta x \to 0} \frac{\cos x \cos \Delta x - \sin x \sin \Delta x - \cos x}{\Delta x} \\ &= \lim_{\Delta x \to 0} \cos x \frac{\cos \Delta x - 1}{\Delta x} - \sin x \frac{\sin \Delta x}{\Delta x} = -\sin x. \end{aligned}$$

Example 3.3.2: Differentiate the function $f(x) = e^x$.

Solution: We have

$$\begin{aligned} f'(x) &= \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \\&= \lim_{\Delta x \to 0} \frac{e^{x + \Delta x} - e^x}{\Delta x} \\ &= \lim_{\Delta x \to 0} e^x \frac{e^{\Delta x} - 1}{\Delta x} = e^x. \end{aligned}$$

Remark: The derivative of $e^x$ is $e^x$. This is a very important result, as it shows that the exponential function is its own derivative.

Next, we would like to interpret the relationship between the derivative and continuity.

Theorem 3.3.1: If $f$ is differentiable at $a$, then $f$ is continuous at $a$.

Proof: Since $f$ is differentiable at $a$, we have

$$f'(a) = \lim_{\Delta x \to 0} \frac{f(a + \Delta x) - f(a)}{\Delta x}.$$

Then we have

$$\lim_{\Delta x \to 0} f(a + \Delta x) - f(a) = \lim_{\Delta x \to 0} f'(a) \Delta x = 0.$$

It follows that

$$\lim_{x \to a} f(x) = f(a),$$

which means that $f$ is continuous at $a$.

But what about the converse of the theorem? That is, if $f$ is continuous at $a$, is $f$ differentiable at $a$? We would like to make it as your exercise.

Exercises for this section

1. For what choice of $a,b$ is the function $f(x) = \begin{cases} x^2 & \text{if } x < 0, \\ ax + b & \text{if } x \geq 0, \end{cases}$ differentiable at $x = 0$?
2. Find the derivatives of the following functions:
   1. $f(x) = \cos 2x + \sin 2x$
   2. $f(x) = x\sin x$

3.4 Curves and Tangents

Roughly speaking, a curve is a geometric figure which can be drawn without lifting pen from paper. It can be a straight line, a circle, a parabola, or a more complicated figure. The tangent to a curve at a point is a straight line which touches the curve at that point. The slope of the tangent is the derivative of the curve at that point.

Example 3.4.1: The graph of $f(x) = \frac{\sin x}{x}$ is not a curve. The reason is that the function is not defined at $x = 0$.

Example 3.4.2: The graph of $f(x) = \frac{1}{x}$ if $x > 0$ is a curve, but the graph of $f(x) = \frac{1}{x}$ if $x \neq 0$ is not a curve. The reason is that the function is not continuous at $x = 0$.

Now let $P(x_0,y_0)$ be a fixed point and $Q(x,y)$ be a variable point on the curve $y = f(x)$. Now the line passing through $P$ and $Q$ is called the secant to the curve at $Q$. Then let $\theta$ be the inclination of the secant to the $x$-axis. Then the slope of the secant is $\tan \theta = \frac{y - y_0}{x - x_0},$ or

$$\tan \theta = \frac{\Delta y}{\Delta x}$$

in terms of the increment $\Delta x = x - x_0$ and $\Delta y = y - y_0$. The slope of the tangent is the limit of the slope of the secant as $Q$ approaches $P$. That is, the slope of the tangent is

To illustrate, one may think of the following figure:

Now suppose the point $Q$ is moving along the curve $y = f(x)$ and approaches the point $P$. More exactly the distance between $P$ and $Q$ is

$$|PQ| = \sqrt{(\Delta x)^2 + (\Delta y)^2}.$$

approches zero. Then the secant line also varies, rotating about the point $P$. If the curve $y = f(x)$ is "well-bahaved" near the point $P$, then the limit

$$m = \lim_{Q \to P} \frac{\Delta y}{\Delta x}= \lim_{|PQ| \to 0} \frac{\Delta y}{\Delta x}$$

exists (one may note that it might be infinite). The line $T$ through $P$ with slope $m$ is called the tangent to the curve at $P$. It follows from theorem that $T$ is the line

$$y = m(x - x_0) + y_0.$$

if $m$ is finite, and the line $x = x_0$ if $m$ is infinite. Speaking qualitatively, the tangent at $P$ is the "limiting position" of the secant through $P$ and $Q$ as $Q$ approaches $P$.

The following theorem gives the connection between tangents and derivatives.

Theorem 3.4.1: The curve $y = f(x)$ has a tangent $T$ at the point $P(x_0,y_0)$ if and only if the derivative $f'(x_0)$ exists. In this case, the slope of the tangent is $f'(x_0)$.

Proof: Let $Q$ be the point $(x_0+\Delta x, f(x_0+\Delta x))$ on the curve. Since $y = f(x)$ is a curve, the function is continuous at $x_0$ which implies that

$$\lim_{\Delta x \to 0} f(x_0 + \Delta x) = f(x_0).$$

or equivalently

$$\lim_{\Delta x \to 0} f(x_0 + \Delta x) - f(x_0) = \lim_{\Delta x \to 0} \Delta y = 0.$$

Therefore the distance between $P$ and $Q$ is

$$|PQ| = \sqrt{(\Delta x)^2 + (\Delta y)^2}$$

approaches zero as $\Delta x$ approaches zero. Therefore,

$$\lim_{|PQ| \to 0} \frac{\Delta y}{\Delta x}$$

exists. This limit is the slope of the tangent at $P$. This completes the proof.

Now, time for some simple example.

Example 3.4.3: Find the equation of the tangent to the curve $y = x^2$ at the point $P(x_0,y_0)$.

Solution: The derivative of $y = x^2$ is $y' = 2x$. Therefore the slope of the tangent at $P(x_0,y_0)$ is $2x_0$. The equation of the tangent is

$$y - y_0 = 2x_0(x - x_0).$$

or equivalently

$$y = 2x_0(x - x_0) + y_0.$$

Example 3.4.4: For what value of $a$ are the curves $y=f_1(x) = 1-ax^2$ and $y=f_2(x) = x^2$ orthogonal?

Solution: Solving the equation

$$1- ax^2 = x^2,$$

we find that the curves intersect at the point $(\pm x_0, x^2_0)$, where

$$x_0 = \frac{1}{\sqrt{1+a}}.$$

At these points, the slopes of the tangents to the curves are

$$m_1 = f'_1(\pm x_0) = \mp \frac{2a}{\sqrt{1+a}}$$

and

$$m_2 = f'_2(\pm x_0) = \pm \frac{2}{\sqrt{1+a}}$$

Therefore, the tangents and hence the curves are orthogonal if and only if

$$m_1 \cdot m_2 = - \frac{4a}{1+a} = -1,$$

which implies that $a = \frac{1}{3}$.

Exercises for this section

1. Find the tangent to the curves:
   1. $y = \frac{1}{3}x^3$ at the point $(-1,-\frac{1}{3})$.
   2. $y = \frac{8}{x^2+4}$ at the point $(1,1)$.
   3. $y = \sin x$ at the point $(\pi, 0)$.

3.5 Techniques of Differentiation

After we have discussed derivatives in a theoretical way, we now discuss some techniques of differentiation.

Theorem 3.5.1 (Derivative of sum or difference): If $f$ and $g$ are differentiable at $x$, then $f \pm g$ is differentiable at $x$ and

$$(f \pm g)'(x) = f'(x) \pm g'(x).$$

One may note that the proof of this theorem is straightforward from the definition of the derivative. We leave it as an exercise.

Corollary 3.5.1: If $f_1, f_2, \cdots, f_n$ are differentiable at $x$, then $f_1 \pm f_2 \pm \cdots \pm f_n$ is differentiable at $x$ and

$$(f_1 \pm f_2 \pm \cdots \pm f_n)'(x) = f_1'(x) \pm f_2'(x) \pm \cdots \pm f_n'(x).$$

Theorem 3.5.2 (Derivative of a constant multiple): If $f$ is differentiable at $x$ and $k$ is a constant, then $kf$ is differentiable at $x$ and

$$(kf)'(x) = kf'(x).$$

Theorem 3.5.3 (Derivative of a product): If $f$ and $g$ are differentiable at $x$, then $fg$ is differentiable at $x$ and

$$(fg)'(x) = f'(x)g(x) + f(x)g'(x).$$

Proof: By definition, we have

$$\begin{aligned} (fg)'(x) &= \lim_{\Delta x \to 0} \frac{f(x + \Delta x)g(x + \Delta x) - f(x)g(x)}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{f(x + \Delta x)g(x + \Delta x) - f(x)g(x + \Delta x) + f(x)g(x + \Delta x) - f(x)g(x)}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}g(x) + \lim_{\Delta x \to 0}f(x)\frac{g(x + \Delta x) - g(x)}{\Delta x} \\ &= f'(x)g(x) + f(x)g'(x). \end{aligned}$$

It completes the proof.

Corollary 3.5.2: If $f_1, f_2, \cdots, f_n$ are differentiable at $x$, then $f_1f_2\cdots f_n$ is differentiable at $x$ and

$$(f_1f_2\cdots f_n)'(x) = f_1'(x)f_2(x)\cdots f_n(x) + f_1(x)f_2'(x)\cdots f_n(x) + \cdots + f_1(x)f_2(x)\cdots f_n'(x).$$

Example 3.5.1: Differentiate the function $f(x) = x^2 \sin x$.

Solution: We have

$$\begin{aligned} f'(x) &= (x^2)' \sin x + x^2(\sin x)' \\ &= 2x \sin x + x^2 \cos x. \end{aligned}$$

Theorem 3.5.4 (Derivative of a quotient): If $f$ and $g$ are differentiable at $x$ and $g(x) \neq 0$, then $\frac{f}{g}$ is differentiable at $x$ and

$$\left(\frac{f}{g}\right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}.$$

Proof: By definition, we have

$$\begin{aligned} \left(\frac{f}{g}\right)'(x) &= \lim_{\Delta x \to 0} \frac{\frac{f(x + \Delta x)}{g(x + \Delta x)} - \frac{f(x)}{g(x)}}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{f(x + \Delta x)g(x) - f(x)g(x + \Delta x)}{g(x + \Delta x)g(x)\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{f(x + \Delta x)g(x) - f(x)g(x)}{g(x + \Delta x)g(x)\Delta x} + \lim_{\Delta x \to 0} \frac{f(x)g(x) - f(x)g(x + \Delta x)}{g(x + \Delta x)g(x)\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{g(x + \Delta x)\Delta x} + \lim_{\Delta x \to 0} \frac{g(x) - g(x + \Delta x)}{g(x + \Delta x)\Delta x} \\ &= \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}. \end{aligned}$$

Example 3.5.2: Differentiate the function $f(x) = \tan x$.

Solution: One may think of $f(x) = \tan x = \frac{\sin x}{\cos x}$. Then we have

$$f'(x) = \frac{\cos x \cos x - \sin x (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x.$$

The following theorem is very useful in the differentiation of composite functions. Composite function might be not familiar to you, briefly speaking, it is a function of a function. Suppose we have two functions $f(x) = x^2$ and $g(x) = \sin x$, then the composite function is $f(g(x)) = \sin^2 x$.

Theorem 3.5.5 (Derivative of a composite function): If $f$ is differentiable at $g(x)$ and $g$ is differentiable at $x$, then $f \circ g$ is differentiable at $x$ and

$$(f \circ g)'(x) = f'(g(x))g'(x).$$

Proof: By definition, we have

$$\begin{aligned} (f \circ g)'(x) &= \lim_{\Delta x \to 0} \frac{f(g(x + \Delta x)) - f(g(x))}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{f(g(x + \Delta x)) - f(g(x))}{g(x + \Delta x) - g(x)} \cdot \frac{g(x + \Delta x) - g(x)}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{f(g(x + \Delta x)) - f(g(x))}{g(x + \Delta x) - g(x)} \cdot \lim_{\Delta x \to 0} \frac{g(x + \Delta x) - g(x)}{\Delta x} \\ &= f'(g(x))g'(x). \end{aligned}$$

Remark: One may call the theorem as the chain rule for differentiation. In some books they might give you the notation of the chain rule as

$$\frac{d}{dx} f(g(x)) = f'(g(x))g'(x).$$

or

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.$$

For a new beginner, it might be a little bit confusing for what is $du$ and $dy$. But one may think of $dy$ as the derivative of $y$ with respect to $u$ and $du$ as the derivative of $u$ with respect to $x$. Then the chain rule is just the product of the two derivatives.

However, students should not mistake the form $\frac{dy}{dx}$ as a fraction. It is not a fraction, but a notation for the derivative of $y$ with respect to $x$. In your further study in mathematics (If you are really interested in), you will know that it is 1-form in differential geometry.

Example 3.5.3: Differentiate the function $f(x) = \sin(x^2)$.

Solution: We observe that $f(x) = \sin(x^2) = \sin(g(x))$ where $g(x) = x^2$. Then we have

$$f'(x) = \cos(x^2) \cdot 2x = 2x \cos(x^2).$$

Example 3.5.4: Differentiate the function $f(x) = e^{x^2}$.

Solution: We observe that $f(x) = e^{x^2} = e^{g(x)}$ where $g(x) = x^2$. Then we have

$$f'(x) = e^{x^2} \cdot 2x = 2x e^{x^2}.$$

Exercises for this section

1. Differentiate the following:
   1. $(x-a)(x-b)$
   2. $\frac{1}{x^2}$
   3. $\frac{1}{2}x^2\cos x$
   4. $\frac{1}{x^2+1}$
   5. $\sin(\sin(\sin x))$ (Hint: An extension of the chain rule)
2. Prove the idenity

   $$1+x+x^2+\cdots+x^n = \frac{x^{n+1}-1}{x-1}$$

   where $x \neq 1$. Then differentiate both sides with respect to $x$ to deduce a formula for the following sum:

   $$1+2x+3x^2+\cdots+nx^{n-1}.$$

3.6 Differentials. Further Notation

Suppose $f$ is differentiable at $x$. Then the differential of $f$ at $x$ is defined by

$$df(x) = f'(x)dx.$$

The differential $df(x)$ is a linear function of $dx$. It is a very useful concept in the study of differential equations. For numerical purposes, we may approximate the increment $\Delta y$ of $f$ by the differential $dy$ of $f$ as

$$\Delta y \approx dy = f'(x)dx.$$

Let us consider the following example.

Example 3.6.1: Find the increment $\Delta y$ and the differential $dy$ of the function $y = x^2$ for $x = 20$ and $\Delta x = 0.1$. What is the percentage error in the approximation of $\Delta y$ by $dy$?

Solution: We have

$$\begin{aligned} \Delta y &= (20.1)^2 - 20^2 = (20.1 + 20)(20.1 - 20) = 40.1 \cdot 0.1 = 4.01, \\ dy &= 2 \cdot 20 \cdot 0.1 = 4. \end{aligned}$$

The percentage error is

$$\frac{4.01 - 4}{4.01} \times 100\% = 0.25\%.$$

Remark: The differential $dy$ is a linear approximation to the increment $\Delta y$. The percentage error in the approximation is small, which shows that the differential is a good approximation to the increment.

Example 3.6.2: Estimate the value of $\sqrt{16.1}$.

Solution: First, we know that $f(x) = \sqrt{x}$ is differentiable at $x = 16$. Then we have

$$\begin{aligned} \Delta f &= \sqrt{x + \Delta x} - \sqrt{x}\\ df &= (\sqrt{x})' \Delta x = \frac{1}{2\sqrt{x}} \Delta x. \end{aligned}$$

Then we have

$$\sqrt{x+\Delta x} \approx \sqrt{x} + \frac{1}{2\sqrt{x}} \Delta x.$$

Substitute $x = 16$ and $\Delta x = 0.1$, we have

$$\sqrt{16.1} \approx 4 + \frac{1}{2 \cdot 4} \cdot 0.1 = 4.0125.$$

Remark: This section is important for a taste of the concept of numerical analysis. In the study of numerical analysis, one may encounter the concept of Taylor series which is a generalization of the concept of differentials.

3.7 Implicit Differentiation

Given an equation $F(x,y) = 0$, we may differentiate the equation with respect to $x$ to find the derivative of $y$ with respect to $x$. This is called implicit differentiation.

Example 3.7.1: Find $y'$ if $y^6-y-x^2 = 0$.

Solution: Differentiating the equation with respect to $x$, we have

$$6y^5 \frac{dy}{dx} - \frac{dy}{dx} - 2x = 0.$$

Then we have

$$\frac{dy}{dx} = \frac{2x}{6y^5 - 1}.$$

The following is a more practical or real-world example.

Example 3.7.2: A spherical balloon is being inflated so that its radius is increasing at a rate of $0.1$ cm/s. How fast is the volume of the balloon increasing when the radius is $10$ cm?

Solution: Let $r$ be the radius of the balloon and $V$ be the volume of the balloon. Then we have

$$V = \frac{4}{3}\pi r^3.$$

Differentiating the equation with respect to $t$, we have

$$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.$$

Substitute $r = 10$ and $\frac{dr}{dt} = 0.1$, we have

$$\frac{dV}{dt} = 4\pi \cdot 10^2 \cdot 0.1 = 40\pi.$$

Therefore the volume of the balloon is increasing at a rate of $40\pi$ cm$^3$/s.

Exercises for this section

1. Find $y'$ if
   1. $y^3-3y+2ax = 0$.
   2. $y^2-2xy+b^2=0$.
2. Water is slowly poured at the rate of 3 cm$^3$/s into a conical container of height 10 cm and base radius 5 cm. How fast is the water level rising when the water is 4 cm deep?
3. A ladder of length 10 m is leaning against a wall. If the foot of the ladder is being pulled away from the wall at the rate of 0.1 m/s, how fast is the top of the ladder sliding down the wall when the foot of the ladder is 6 m away from the wall?

3.8 Higher Derivatives

Let $f$ be a function defined in an interval $I$(closed, open or half open, finite or infinte), with endpoints $a$ and $b$ ($a<b$). Then $f$ is said to be differentiable on $I$ if the derivative $f'(x)$ exists at every point $x$ in $I$ which has been dicussed in the previous sections. If $f'$ is differentiable on $I$, then the derivative of $f'$ is called the second derivative of $f$ and is denoted by $f''$. If $f''$ is differentiable on $I$, then the derivative of $f''$ is called the third derivative of $f$ and is denoted by $f'''$. In general, the $n$th derivative of $f$ is denoted by $f^{(n)}$.

One may wonder why we have to introduce the concept of higher derivatives. The reason is that the higher derivatives of a function may give us some information about the function. For example, the second derivative of a function may give us the concavity of the function. The third derivative of a function may give us the rate of change of the concavity of the function. The fourth derivative of a function may give us the rate of change of the rate of change of the concavity of the function. And so on.

Example 3.8.1: Find the second derivative of the function $f(x) = x^3$.

Solution: We have $f'(x) = 3x^2$. Then we have $f''(x) = 6x$.

Example 3.8.2: Find the third derivative of the function $f(x) = \sin x$.

Solution: We have $f'(x) = \cos x$. Then we have $f''(x) = -\sin x$. Finally we have $f'''(x) = -\cos x$.

Example 3.8.3: Evaluate $(f(x)g(x))'''$

Solution: We have

$$\begin{aligned} (f(x)g(x))''' &= ((f(x)g(x))')'' \\ &= (f'(x)g(x) + f(x)g'(x))'' \\ &= (f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x))' \\ &= f'''(x)g(x) + 3f''(x)g'(x) + 3f'(x)g''(x) + f(x)g'''(x). \end{aligned}$$

when written out full. The binomial coefficients are used in the expansion of the product of two binomials.

Remark: The Example 3.8.3 is not a coincidence. It is a general rule for the differentiation of the product of two functions. The rule is called the Leibniz rule.

Theorem 3.8.1 (Leibniz rule): If $f$ and $g$ are $n$ times differentiable at $x$, then the $n$th derivative of the product $f(x)g(x)$ is

$$(f(x)g(x))^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x).$$

Proof: We may prove the theorem by induction. Obviously the theorem is true for $n = 1$. Suppose the theorem is true for $n = m$. Then we have

$$\begin{aligned} (f(x)g(x))^{(m+1)} &= \left[ \sum_{k=0}^m \binom{m}{k} f^{(m-k)}(x)g^{(k)}(x) \right]'\\ &= \sum_{k=0}^m \binom{m}{k} f^{(m+1-k)}(x)g^{(k)}(x) + \sum_{k=0}^m \binom{m}{k} f^{(m-k)}(x)g^{(k+1)}(x)\\ &= \sum_{k=0}^m \binom{m}{k} f^{(m+1-k)}(x)g^{(k)}(x) + \sum_{k=1}^{m+1} \binom{m}{k-1} f^{(m+1-k)}(x)g^{(k)}(x)\\ &= \binom{m}{0} f^{(m+1)}(x)g(x) + \binom{m}{m} f(x)g^{(m+1)}(x) + \sum_{k=1}^m \left[ \binom{m}{k} + \binom{m}{k-1} \right] f^{(m+1-k)}(x)g^{(k)}(x)\\ &= f^{(m+1)}(x)g(x) + f(x)g^{(m+1)}(x) + \sum_{k=1}^m \binom{m+1}{k} f^{(m+1-k)}(x)g^{(k)}(x)\\ &= \sum_{k=0}^{m+1} \binom{m+1}{k} f^{(m+1-k)}(x)g^{(k)}(x). \end{aligned}$$

It completes the proof.

Example 3.8.4: Find the tenth derivative of the function $f(x) = x^4\sin x$.

Solution: By Leibniz rule, we have

$$\begin{aligned} (f(x)g(x))^{(10)} &= \sum_{k=0}^{10} \binom{10}{k} f^{(10-k)}(x)g^{(k)}(x)\\ &= \binom{10}{0} f^{(10)}(x)g(x) + \binom{10}{1} f^{(9)}(x)g'(x) + \cdots + \binom{10}{10} f(x)g^{(10)}(x)\\ &= -x^4\sin x + 40x^3\cos x + 540x^2\sin x - 2280x\cos x -5040\sin x. \end{aligned}$$

Exercises for this section

1. Find $y''$ if
   1. $y=\sin^2 x$.
   2. $y=e^{6x}$.
   3. $y=\cos x$.
2. A moving particle has equation of motion $s = 10 + 20t – 5t^2$. Find the particle’s velocity and acceleration at time $t = 2$. Does the particle’s acceleration ever change?

3.9 Maxima and Minima

Differentiation can be used to help locate maxima and minima of functions. There are two different uses of the word "maximum" in mathematics. One is the absolute maximum or minimum of a function, which is the largest or smallest value of the function over its entire domain. The other is the local maximum or minimum of a function, which is the largest or smallest value of the function in some neighborhood of a point. The local maximum or minimum is also called the relative maximum or minimum.

Theorem 3.9.1 (First derivative test): If $f$ is differentiable at $x_0$ and $f'(x_0) = 0$, then $f$ has a local maximum or minimum at $x_0$.

Proof: If $f'(x_0) = 0$, then the tangent to the curve $y = f(x)$ at $x_0$ is horizontal. Therefore the curve has a horizontal tangent at $x_0$. This implies that the curve has a local maximum or minimum at $x_0$. This completes the proof.

Remark: The first derivative test is a very useful tool in the study of maxima and minima of functions. The first derivative test is also called the stationary point test. One may also wonder why the first derivative test can only give us the local maximum or minimum of a function. The reason is that the first derivative test can only give us the critical points of a function. The critical points of a function are the points where the derivative of the function is zero or undefined. The critical points of a function are the points where the function may have a local maximum or minimum.

Theorem 3.9.2 (Second derivative test): If $f$ is twice differentiable at $x_0$ and $f'(x_0) = 0$, then

1. If $f''(x_0) > 0$, then $f$ has a local minimum at $x_0$.
2. If $f''(x_0) < 0$, then $f$ has a local maximum at $x_0$.
3. If $f''(x_0) = 0$, then the test is inconclusive.
4. If $f''(x_0)$ does not exist, then the test is inconclusive.

Proof: If $f''(x_0) > 0$, then the curve $y = f(x)$ is concave up at $x_0$. Therefore the curve has a local minimum at $x_0$. If $f''(x_0) < 0$, then the curve $y = f(x)$ is concave down at $x_0$. Therefore the curve has a local maximum at $x_0$. If $f''(x_0) = 0$, then the test is inconclusive. If $f''(x_0)$ does not exist, then the test is inconclusive. This completes the proof.

Example 3.9.1: Determine the local maximum and minimum of the function $f(x) = x^3 - 3x^2 + 2$.

Solution: We have $f'(x) = 3x^2 - 6x = 3x(x-2)$. Therefore $f'(x) = 0$ at $x = 0$ and $x = 2$. Then we have $f''(x) = 6x - 6$. Therefore $f''(0) = -6$ and $f''(2) = 6$. Therefore $f$ has a local maximum at $x = 0$ and a local minimum at $x = 2$.

Example 3.9.2: Determine the local maximum and minimum of the function $f(x) = \sin x$.