FP3 Chapter 3: Differentiation

FP3 Lecture Notes: Differentiation

Prior Knowledge Check

Before we begin, let’s review some key concepts:

Exercise 1: Basic Trigonometric Derivatives

Without referring to any formula, prove that:

1. $\frac{d}{dx}(\sin^2 x) = \sin 2x$
2. $\frac{d}{dx}(\tan x) = \sec^2 x$
3. Given that $y = x^2 \cos 3x$, find $\frac{dy}{dx}$

Exercise 2: Basic Hyperbolic Functions

Write down the definitions of the following functions:

$\sinh x, \cosh x, \tanh x, \operatorname{sech} x, \operatorname{csch} x, \coth x$

1. Differentiation: Geometric and Algebraic Perspectives

1.1 The Geometric View

The derivative of a function at a point represents the gradient of the tangent line at that point. This can be understood as the limit of the gradient of secant lines.

Definition: Derivative as a Limit from First Principle

The derivative of a function $f$ at a point $x = a$ is defined as:

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

provided this limit exists.

1.2 The Algebraic View

From an algebraic perspective, differentiation can be viewed as finding the best linear approximation to a function near a point.

Theorem: Linear Approximation

Near $x = a$, a differentiable function $f(x)$ can be approximated by:

$f(x) \approx f(a) + f'(a)(x - a)$

This is called the linear approximation or tangent line approximation.

Example: Application to Error Analysis

Consider estimating $\sqrt{4.1}$ using the linear approximation to $f(x) = \sqrt{x}$ at $x = 4$.

Since $f'(x) = \frac{1}{2\sqrt{x}}$, we have:

$\sqrt{4.1} \approx \sqrt{4} + \frac{1}{2\sqrt{4}}(4.1 - 4) = 2 + \frac{0.1}{4} = 2.025$

2. Differentiation of Trigonometric Functions and Their Inverses

2.1 Deriving Inverse Trigonometric Derivatives

Example: Derivative of arcsin

Show that $\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}}$.

Implicit Differentiation:

1. Let $y = \arcsin x$, then $\sin y = x$
2. Differentiate both sides: $\cos y \frac{dy}{dx} = 1$
3. Therefore $\frac{dy}{dx} = \frac{1}{\cos y}$
4. Since $\sin^2 y + \cos^2 y = 1$, we have $\cos^2 y = 1 - \sin^2 y = 1 - x^2$
5. Thus $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$

Guided Exercise: Derivative of arccos

Show that $\frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1 - x^2}}$.

Method 1 (Using $\arccos x + \arcsin x = \frac{\pi}{2}$):

Method 2 (Implicit Differentiation): Try this yourself using $y = \arccos x \implies \cos y = x$.

Guided Exercise: Derivative of arctan

Show that $\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}$.

3. Differentiation of Hyperbolic Functions

3.1 Deriving Basic Hyperbolic Derivatives

Guided Exercise: Basic Hyperbolic Derivatives

Using the definitions:

$\sinh x = \frac{e^x - e^{-x}}{2}, \qquad \cosh x = \frac{e^x + e^{-x}}{2}$

show that:

$\frac{d}{dx}(\sinh x) = \cosh x, \quad \frac{d}{dx}(\cosh x) = \sinh x, \quad \frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x$

Guided Exercise: Reciprocal Hyperbolic Derivatives

Show that:

1. $\frac{d}{dx}(\operatorname{sech} x) = -\tanh x \operatorname{sech} x$
2. $\frac{d}{dx}(\operatorname{csch} x) = -\coth x \operatorname{csch} x$
3. $\frac{d}{dx}(\coth x) = -\operatorname{csch}^2 x$

Hint: Use the quotient rule and previous results.

3.2 Inverse Hyperbolic Functions

Guided Exercise: Derivative of arsinh

Show that $\frac{d}{dx}(\operatorname{arsinh} x) = \frac{1}{\sqrt{x^2 + 1}}$.

Method 1 (Using definition):

1. Use $\operatorname{arsinh} x = \ln(x + \sqrt{x^2 + 1})$
2. Apply chain rule to differentiate

Method 2 (Implicit Differentiation):

1. Let $y = \operatorname{arsinh} x$, then $\sinh y = x$
2. Differentiate both sides: $\cosh y \frac{dy}{dx} = 1$
3. Use $\cosh^2 y - \sinh^2 y = 1$ to eliminate $\cosh y$

Guided Exercise: Derivative of arcosh

Show that $\frac{d}{dx}(\operatorname{arcosh} x) = \frac{1}{\sqrt{x^2 - 1}}$, for $x > 1$.

Method 1 (Using definition $\operatorname{arcosh} x = \ln(x + \sqrt{x^2 - 1})$):

Method 2 (Implicit Differentiation): Try this yourself using $y = \operatorname{arcosh} x \implies \cosh y = x$.

Guided Exercise: Derivative of artanh

Show that $\frac{d}{dx}(\operatorname{artanh} x) = \frac{1}{1 - x^2}$, for $|x| < 1$.

Method 1 (Using definition $\operatorname{artanh} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$):

Method 2 (Implicit Differentiation): Try this yourself using $y = \operatorname{artanh} x \implies \tanh y = x$.

4. Applications and Past Paper Questions

Example: Q04 WFM03/01, June 2021

(a) Given that $f(x) = x \arccos x$ where $-1 \leq x \leq 1$, find the exact value of $f'(0.5)$.

(b) Given that $g(x) = \arctan(e^{2x})$, show that

$g''(x) = k \operatorname{sech}(2x) \tanh(2x)$

where $k$ is a constant to be found.

Example: Q03 WFM03/01, Jan 2022

(a) Given that $y = \operatorname{arsech}\left(\frac{x}{2}\right)$, where $0 < x \leq 2$, show that

$\frac{dy}{dx} = \frac{p}{x\sqrt{q - x^2}}$

where $p$ and $q$ are constants to be determined.

(b) Given that $f(x) = \operatorname{artanh}(x) + \operatorname{arsech}\left(\frac{x}{2}\right)$ where $0 < x \leq 1$, determine, in simplest form, the exact value of $x$ for which $f'(x) = 0$.

Example: Q04 WFM03/01, June 2022

Given that $y = \operatorname{artanh}\left(\frac{\cos x + a}{\cos x - a}\right)$ where $a$ is a non-zero constant, show that

$\frac{dy}{dx} = k \tan x$

where $k$ is a constant to be determined.

Summary

Key Points

• The derivative can be understood both geometrically (as slope) and algebraically (as rate of change)
• Inverse trigonometric derivatives involve the form $\frac{1}{\sqrt{1-x^2}}$
• Hyperbolic function derivatives mirror trigonometric ones, but with some sign differences
• Inverse hyperbolic derivatives can be derived using logarithmic definitions

Common Mistakes

• Forgetting the minus sign in $\frac{d}{dx}(\arccos x)$
• Confusing domains of inverse functions
• Not recognizing hyperbolic identities
• Incorrect use of chain rule with composite functions

Homework Exercises

Past Paper Question: 6669/s10/01/5

Given that $y = (\operatorname{arcosh} 3x)^2$, where $3x > 1$, show that:

(a) $(9x^2 - 1)\left(\frac{dy}{dx}\right)^2 = 36y$ (5 marks)

(b) $(9x^2 - 1)\frac{d^2y}{dx^2} + 9x\frac{dy}{dx} = 18$ (4 marks)

Past Paper Question: 6669/s14/01/5

Given that $y = \operatorname{artanh}\frac{x}{\sqrt{1+x^2}}$, show that

$\frac{dy}{dx} = \frac{1}{\sqrt{1+x^2}}$

(4 marks)

Challenge Problem: Proof of $(\sin x)' = \cos x$

Background Review

Key Concept 1: Unit Circle and Trigonometric Functions

For a point $P(\cos\theta, \sin\theta)$ on the unit circle:

• $\cos\theta$ is the $x$-coordinate
• $\sin\theta$ is the $y$-coordinate
• As $\theta$ increases, $P$ moves counterclockwise

Key Concept 2: Definition of Derivative

The derivative of a function $f(x)$ at point $x$ is defined as the limit of the difference quotient

$\frac{f(x+h) - f(x)}{h}$

as $h$ approaches $0$.

Key Concept 3: Arclength Definition of Arcsin

On the unit circle, $\arcsin x$ represents the arclength of the arc from the point $(1, 0)$ to the point $(\sqrt{1-x^2}, x)$.

Key Properties:

• The angle $\theta = \arcsin(x)$ is measured in radians
• For any point $(\sqrt{1-x^2}, x)$ on the unit circle, $x = \sin(\theta)$
• The arclength equals the radian measure $\theta$ (since radius = 1)
• Domain restriction $|x| \leq 1$ comes from the unit circle geometry

Task 1: Physical Method

Task 1: Physical Method

1. Consider a particle moving on the plane with parametric equations: $x(t) = \cos t, \quad y(t) = \sin t$ What does this mean physically?

2. The velocity vector $\vec{v}(t)$ is given by $\vec{v}(t) = (x'(t), y'(t))$.

   • What is the direction of the velocity vector?
   • Why must it be perpendicular to the position vector?
   • What is its magnitude?

3. How does this physical interpretation help us understand why $\frac{d}{dt}\sin t = \cos t$?

Task 2: Understanding the Derivative Through Limits

Task 2: Understanding the Derivative Through Limits

Let’s break down the derivative of sine function step by step:

1. At any point $a$, we want to find $(\sin x)'$ at $x = a$. This means we need to evaluate:

   $\frac{\sin(a+h) - \sin a}{h} \quad \text{as $h$ gets closer and closer to 0}$

2. We can use the trigonometric addition formula:

   $\sin(a+h) = \sin a \cos h + \cos a \sin h$

3. Q: Substitute this into our difference quotient and show that:

   $\frac{\sin(a+h) - \sin a}{h} = \sin a \cdot \frac{(\cos h - 1)}{h} + \cos a \cdot \frac{\sin h}{h}$

4. To show this approaches $\cos a$, we need to understand what happens to:

   • $\frac{\sin h}{h}$ as $h$ approaches $0$
   • $\frac{\cos h - 1}{h}$ as $h$ approaches $0$

5. Q: Using the identity $\cos h - 1 = -2\sin^2(h/2)$, show that:

   $\frac{\cos h - 1}{h} = -\sin(h/2) \cdot \frac{\sin(h/2)}{h/2}$

   Hence explain why $\frac{\cos h - 1}{h}$ approaches $0$ as long as $\frac{\sin h}{h}$ approaches $1$ as $h$ approaches $0$.

6. Therefore, if we can show $\frac{\sin h}{h}$ approaches $1$ as $h$ approaches $0$ (which we’ll do in Task 3), we’ll have proved that the derivative of $\sin x$ at any point $a$ is $\cos a$.

Task 3: Analytical Method

Task 3: Analytical Method

In this task, we will explore an alternative definition of the sine function through integration and use such definition to prove that $\frac{\sin x}{x}$ approaches $1$ as $x$ approaches $0$.

Part 1: Understanding Arclength and Arcsine

1. Recall from Task 1 that when a point moves along the unit circle with position vector $(\cos t, \sin t)$:

   • Its velocity vector is $(-\sin t, \cos t)$
   • The speed (magnitude of velocity) is constant = 1
   • Therefore, the parameter $t$ equals the arclength traveled

2. Consider the arclength definition of $\arcsin x$ (see Key Concept 3):

   • On the unit circle, $\arcsin x$ is the arclength from $(1, 0)$ to $(\sqrt{1-x^2}, x)$
   • For a curve given by $(f(t), g(t))$, its arclength is $\int \sqrt{(f'(t))^2 + (g'(t))^2}\,dt$
   • Remark: this is how “arc”sin is related to “arc”length and we will discuss the computation of arclength in detail in the next chapter.

3. Q: Using the arclength formula above:

   • (a) Express the arc representing $\arcsin x$ as $(x(t), y(t)) = (\sqrt{1-t^2}, t)$ for $0 \leq t \leq x$.
   • (b) Show that $\arcsin x = \int_0^x \frac{1}{\sqrt{1-t^2}}\,dt$.

Part 2: Properties of the Integral

1. For the function $f(x) = \int_0^x \frac{1}{\sqrt{1-t^2}}\,dt$, explain why:

   • The value of $f(0)$ must be $0$
   • As $x$ increases, $f(x)$ must increase (think about the area under a positive curve)
   • The function is symmetric about the origin (what happens if you replace $x$ with $-x$?)

2. For any $x$ between $0$ and $1$:

   • Find the smallest value of $\frac{1}{\sqrt{1-t^2}}$ on $[0, x]$
   • Find the largest value of $\frac{1}{\sqrt{1-t^2}}$ on $[0, x]$

   Use these facts to explain why:

   $x \leq f(x) \leq \frac{x}{\sqrt{1-x^2}}$

Part 3: The Fundamental Property

The Squeeze Theorem states: if $a(x) \leq b(x) \leq c(x)$ for all $x$ near some value $p$, and if $a(x)$ and $c(x)$ both approach the same value $L$ as $x$ approaches $p$, then $b(x)$ must also approach $L$.

1. Use the squeeze theorem to explain why $\frac{f(x)}{x}$ approaches $1$ as $x$ approaches $0$.

2. Let $y = f(x)$, then show that $\frac{\sin y}{y}$ also approaches $1$ as $y$ approaches $0$.