Table of Contents

Differentiation in Calculus

Calculus, specifically differentiation, is a fundamental concept in IB Mathematics Analysis and Approaches at the Standard Level (SL). It involves understanding how functions change, which is critical for modeling and solving problems in physics, engineering, economics, and many other fields.

What is Differentiation?

Differentiation is the process of finding the derivative of a function. The derivative measures how a function's output value changes as its input value changes. In practical terms, it tells us the rate at which one quantity changes with respect to another.
For example, if we consider a moving object, such as a vehicle, an airplane, or a bicycle, all of these move at some average speed over a certain period of time. When we divide the change in position by the change in time,

$a v e r a g e v e l o c i t y = \frac{c h a n g e i n p o s i t i o n}{c h a n g e i n t i m e} = \frac{x_{2} (t) - x_{1} (t)}{t_{2} - t_{1}}$

we get the velocity, which is the speed of the object with a direction. In calculus, we are able to find the instantaneous speed, i.e. speed at a specific moment in time, by taking the derivative of a given Position function, x(t). There are various rules of differentiation which one will learn in a Calculus course. Rules such as the Power Rule, the Product Rule, Quotient Rule, Chain Rule and so on, will be learned and applied over the duration of this course. Initially, however, all students will have to learn to find derivatives of given functions using the definition of the derivative.

Basic Concepts in Differentiation :

Definition of the Derivative:

The derivative of a function at a point – the exact change of the y value at a given x-value - gives the slope of the tangent line to the graph of the function at that point.

An example of what this indicates is this; assume we are given the position function of a car that starts from rest as y(t) = t².

If we create a table of values for this function for a few values of t, we get:

Table Example table { width: 100px; border-collapse: collapse; /* Ensures borders between cells are merged / } th, td { border: 1px solid black; / Adds border to table cells / padding: 8px; / Adds space between text and border in cells / text-align: center; / Centers the text in cells */ }

t y(t)

0 0

1 1

2 4

3 9

4 16

t	y(t)
0	0
1	1
2	4
3	9
4	16

The average velocity over a certain interval can be calculated using the formula:

$v e l o c i t y = \frac{c h a n g e i n p o s i t i o n}{c h a n g e i n t i m e} = \frac{x_{2} (t) - x_{1} (t)}{t_{2} - t_{1}}$

For example, the average velocity of this object over the time interval 0 ≤ t ≤ 4 seconds is:

$v e l o c i t y = \begin{matrix} \frac{y (4) - y (0)}{4 - 0} = \frac{16 - 0}{4 - 0} = \frac{16}{4} = 4 \end{matrix}$

This means that the car had an average speed of 4 m/s over the first 4 seconds.

Since the car starts at rest and starts speeding up, its velocity is increasing at each instant over the first 4 seconds. This means that its speed is always increasing over this time interval and is never the same at any point in time. This is where calculus comes in and helps us solve these problems.

The slope of the tangent line at t = 1 indicates the speed of the car after 1 second.

t = 1

The slope of the tangent line at t = 2 indicates the speed of the car after 2 seconds.

t = 2

The slope of the tangent line at t = 3 indicates the speed of the car after 3 seconds.

t = 3

The slope of the tangent line at t = 4 indicates the speed of the car after 4 seconds.

t = 4

As we can see from the graphs above, the speed of the car at each specific time increases as time increases, and the derivative is used to find the exact value of this speed at that point in time.

The graph below shows the tangent line to the graph on the interval 0 ≤ x ≤ 4. The slope of each tangent line at specific moments in time is known as the instantaneous rate of change and is the speed of the car at that moment in time.

To find the slope of each tangent line, we need to take the derivative of the function – also known as derive or differentiate the function – and substitute a value for t to calculate the speed at that time.

Before we learn the rules of derivation, we need to learn how to find the derivative using the Definition of the Derivative. This is done using limits.

There are three forms of the Definition of the Derivative:

It is defined as the limit of the difference quotient as the increment approaches zero:

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

This expression gives the derivative in the form of an equation with which you can find any slope to any tangent line to any equation that is differentiable at that point.

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

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

The two expressions to the lift find the slope of the tangent line at some value of x, which is x = a.

Where a function is differentiable.

In order for a function to be differentiable, it has to meet certain criteria. One of these criteria is that the function must be continuous.

Here are a few examples of functions that are continuous in their domain.

$y = x^{3} - 9 x$

$y = \sqrt{x} {x \geq 0}$

$y = \ln x {x > 0}$

$y = x^{4} - 3 x^{2} + 2 x + 1$

THEOREM:

If a function f(x) is differentiable at x = c, then f is continuous at x = c.

BUT…

If f(x) is continuous at x = c, it may or may not be differentiable at x = c.

A function is not differentiable when one of the following occurs on a graph;

(i) There is a sudden change in the slope of a tangent line at a point (may be a sharp corner or a cusp)

(ii) There is a hole in the graph

(iii) There is a vertical asymptote x = k, in which case the function is not defined at that x value.

(iv) There is a jump in the graph. This usually happens when the function is a piecewise function.

Differentiability:

A function is differentiable at a point if its derivative exists at that point. For the derivative to exist, the function must be continuous at that point, and the slope of the tangent line has to be defined.
In order to better understand where a function is differentiable, we can take a look at examples where a function is not differentiable.

Rational Function Example

y = x³ - 3x² - 2x + 6 x - 3

    y = 
        x - 3(x2 - 2)
        x - 3
    

Hole at x = 3

This graph is not continuous at x = 3 due to a hole at this
x-value. A hole occurs when we cross out the same polynomial I the numerator and denominator.

y = \frac{x - 2}{x - 3}

Vertical Asymptote Example

vertical asymptote x = 3

This graph is not continuous at x = 3 due to a vertical asymptote at this x-value. Therefore the function is not differentiable at x = 3.

\begin{aligned} y = x^{\frac{2}{3}} \\ Cusp x = 0 \end{aligned}

This graph is continuous at x = 0, however, it is not differentiable due to a cusp at this x-value. There is a sudden and abrupt change in the slope at x = 0.

Sharp Corner Example

y = |x + 2|

Sharp corner x = -2

This graph is continuous at x = -2 however, it is not differentiable due to a sharp corner at this
x-value. There is a sudden and abrupt change in the slope at x = -2.

y = x^{\frac{1}{3}} or y = \sqrt[3]{x}

Vertical Tangent Example

Vertical tangent x = 0

This graph is continuous at x = 0 , however, it is not differentiable at this x-value because the tangent line at this point is vertical, and all vertical lines have slopes that are undefined.

y = \{\begin{matrix} x + 4 & x < 2 \\ 3 & x \geq 2 \end{matrix}

Vertical Tangent Example

Jump x = 2

This graph is not continuous at x = 2 , however, it is not differentiable at this x-value because there is a jump at this
x-value. As we can see from the graph, the slopes of the lines to the left of at x = 2 and to the right of at x = 2 are not equal, so we conclude the slope at this point does not exist and we say the function is not differentiable at x = 2.

y = \sin (\frac{1}{x})

Vertical Tangent Example

Oscillating x = 0

This graph is continuous throughout its domain except at
x = 0. The function is not defined at x = 0. Plugging x = 0 into the function $y = \sin (\frac{1}{x})$ gives $y = \sin (\frac{1}{0})$ and $\frac{1}{0}$ is undefined, so the entire function is undefined at x = 0.

y = [[x]]

Vertical Tangent Example

Greatest integer functions x = 0

This graph is known as the Greatest Integer Function, or Step Function. This function is not continuous at an infinite number of x-values due to ‘jumps’ in the graph.

y = \tan x

x = \frac{π}{2} \pm n π (n \in Z)

This graph represents the parent tan function, $y = \tan x$ . As we can see from the graph of $y = \tan x$ , the graph has an infinite number of vertical asymptotes, in which case it is not continuous, and therefore not differentiable. The asymptotes can be represented by the equations $x = \frac{π}{2} \pm n π (n \in Z)$ Where n is an integer.

Higher-Order Derivatives:

These are derivatives of derivatives, often used to understand the curvature and concavity of functions (e.g., the second derivative).

Rules of Differentiation:

Here are some basic rules that are essential for differentiation:

1. Power Rule:
$If f (x) = x^{n}, t h e n f^{'} (x) = n x^{n - 1} .$

2. Constant Rule:
The derivative of a constant is zero.

3. Sum Rule:
The derivative of a sum is the sum of the derivatives.

4. Product Rule:
$If f (x) = u (x) v (x), then f^{'} (x) = u^{'} (x) v (x) + u (x) v^{'} (x) .$

5. Quotient Rule:
$If f (x) = \frac{u (x)}{v (x)},then f^{'} (x) = \frac{u^{'} (x) v (x) - u (x) v^{'} (x)}{v (x)^{2}} .$

6. Chain Rule:
Used to differentiate composite functions.
$If f (x) = g (h (x)), then f^{'} (x) = g^{'} (h (x)) \cdot h^{'} (x) .$

Problem 1:

Differentiate the function $f (x) = (3 x^{2} - 4 x + 5) (2 x^{3} - x) .$

Solution:

Identify the Components

$\begin{aligned} L e t u (x) = 3 x^{2} - 4 x + 5 \\ L e t v (x) = 2 x^{3} - x \end{aligned}$

Use the Product Rule

$f^{'} (x) = u^{'} (x) v (x) + u (x) v^{'} (x)$

Differentiate Each Component Differentiation Example body { font-family: Arial, sans-serif; font-size: 16px; color: #333; } .math { font-style: italic; margin-left: 20px; /* Adjust margin to align equations neatly / } ul { list-style: none; / Removes default bullet points / padding: 0; / Removes default padding / } li { margin-bottom: 10px; / Adds space between list items / } .equation { font-family: 'Times New Roman', serif; color: #0F9ED5; / Black color for emphasis on equations */ }
Differentiate u(x):
u'(x) = 6x - 4

Differentiate v(x):
v'(x) = 6x² - 1

Apply the Product Rule

• Substitute the derivatives back into the product rule formula:

$f^{'} (x) = (6 x - 4) (2 x^{3} - x) + (3 x^{2} - 4 x + 5) (6 x^{2} - 1)$

Simplify

• Expand and simplify the expression:

$f^{'} (x) = (12 x^{4} - 6 x^{2} - 8 x^{3} + 4 x) + (18 x^{4} - 3 x^{2} - 24 x^{3} + 4 x + 30 x f^{'} (x)$

$= 30 x^{4} - 32 x^{3} + 27 x^{2} - 1$

Conclusion:

The derivative of the function

$\begin{array}{r} f (x) = (3 x^{2} - 4 x + 5) (2 x^{3} - x) \\ is f^{'} (x) = 30 x^{4} - 32 x^{3} + 27 x^{2} - 1 \end{array}$
This process highlights the use of the product rule and basic algebraic manipulation in calculus differentiation.

Here is a clearer and more detailed graph of the function $f (x) = (3 x^{2} - 4 x + 5) (2 x^{3} - x)$ and its derivative $f^{'} (x) = 30 x^{4} - 32 x^{3} + 27 x^{2} - 1$

• The blue line represents the function f(x), showing the behavior and changes in value more prominently.

• The red line represents the derivative f'(x), which helps visualize the rate of change of the function and the points where it reaches local maxima or minima.

See More Blogs

wikkipedia link

Applications of Differentiation in Calculus

Differentiation is a fundamental concept in calculus with numerous applications across various fields, including science, engineering, economics, and more. Here are some key applications: Differentiation in Calculus

1. Physics

Motion Analysis: Differentiation is used to analyze motion. The derivative of the position function with respect to time gives the velocity, and the derivative of the velocity function gives the acceleration.Differentiation in Calculus
Optics: Differentiation helps in understanding how light changes direction, such as calculating the rate of change of the angle of incidence or refraction using Snell’s law.Differentiation in Calculus

2. Engineering

Design and Manufacturing: Engineers use differentiation to model and optimize designs. For example, in the design of mechanical parts, differentiation helps in finding the maximum stress points or optimizing shapes for minimal material use.Differentiation in Calculus
Electrical Engineering: In circuit design, differentiation is used to analyze how current and voltage change over time, which is essential for designing stable and efficient circuits.Differentiation in Calculus

3. Economics

Marginal Analysis: Differentiation is used to determine marginal cost, marginal revenue, and marginal profit. These concepts are crucial for making decisions about production levels and pricing strategies.Differentiation in Calculus
Elasticity of Demand: The price elasticity of demand, which measures how the quantity demanded of a good responds to changes in price, is calculated using differentiation.Differentiation in Calculus

4. Biology

Population Growth: Differentiation is used to model the growth rates of populations. The derivative of the population function with respect to time gives the growth rate.Differentiation in Calculus

5. Medicine

Pharmacokinetics: Differentiation helps in modeling how drugs are absorbed, distributed, metabolized, and excreted in the body. The rate at which drug concentration changes in the bloodstream is analyzed using differential equations.Differentiation in Calculus

6. Computer Science

Algorithm Optimization: Differentiation is used in machine learning and artificial intelligence to optimize algorithms. For example, gradient descent is an optimization technique that uses derivatives to minimize a function.Differentiation in Calculus

7. Environmental Science

Modeling Climate Change: Differentiation is used to model and predict changes in environmental factors, such as temperature and pollution levels, over time.
- Example: The rate of change of atmospheric CO2 levels can be studied using differential equations.Differentiation in Calculus

Differentiation in Calculus

Calculus, specifically differentiation, is a fundamental concept in IB Mathematics Analysis and Approaches at the Standard Level (SL). It involves understanding how functions change, which is critical for modeling and solving problems in physics, engineering, economics, and many other fields.

a v e r a g e v e l o c i t y = c h a n g e i n p o s i t i o n c h a n g e i n t i m e = x 2 ( t ) − x 1 ( t ) t 2 − t 1

Basic Concepts in Differentiation : Definition of the Derivative: The derivative of a function at a point – the exact change of the y value at a given x-value - gives the slope of the tangent line to the graph of the function at that point.

An example of what this indicates is this; assume we are given the position function of a car that starts from rest as y(t) = t2.

If we create a table of values for this function for a few values of t, we get:

The average velocity over a certain interval can be calculated using the formula:

v e l o c i t y = c h a n g e i n p o s i t i o n c h a n g e i n t i m e = x 2 ( t ) − x 1 ( t ) t 2 − t 1

For example, the average velocity of this object over the time interval 0 ≤ t ≤ 4 seconds is:

The slope of the tangent line at t = 1 indicates the speed of the car after 1 second.

t = 1

The slope of the tangent line at t = 2 indicates the speed of the car after 2 seconds.

t = 2

The slope of the tangent line at t = 3 indicates the speed of the car after 3 seconds.

t = 3

The slope of the tangent line at t = 4 indicates the speed of the car after 4 seconds.

t = 4

As we can see from the graphs above, the speed of the car at each specific time increases as time increases, and the derivative is used to find the exact value of this speed at that point in time.

The graph below shows the tangent line to the graph on the interval 0 ≤ x ≤ 4. The slope of each tangent line at specific moments in time is known as the instantaneous rate of change and is the speed of the car at that moment in time.

To find the slope of each tangent line, we need to take the derivative of the function – also known as derive or differentiate the function – and substitute a value for t to calculate the speed at that time.

Before we learn the rules of derivation, we need to learn how to find the derivative using the Definition of the Derivative. This is done using limits. There are three forms of the Definition of the Derivative:

It is defined as the limit of the difference quotient as the increment approaches zero:

f ′ ( x ) = lim h → 0 ⁡ f ( x + h ) − f ( x ) h

This expression gives the derivative in the form of an equation with which you can find any slope to any tangent line to any equation that is differentiable at that point.

f ′ ( x ) = lim h → 0 f ( a + h ) − f ( a ) h

f ′ ( x ) = lim h → 0 f ( x ) − f ( a ) x − a

The two expressions to the lift find the slope of the tangent line at some value of x, which is x = a.

Where a function is differentiable. In order for a function to be differentiable, it has to meet certain criteria. One of these criteria is that the function must be continuous. Here are a few examples of functions that are continuous in their domain.

y = x 3 − 9 x

y = x { x ≥ 0 }

y = ln ⁡ x { x > 0 }

y = x 4 − 3 x 2 + 2 x + 1

THEOREM: If a function f(x) is differentiable at x = c, then f is continuous at x = c.

BUT…

If f(x) is continuous at x = c, it may or may not be differentiable at x = c. A function is not differentiable when one of the following occurs on a graph;

This graph is not continuous at x = 3 due to a hole at this x-value. A hole occurs when we cross out the same polynomial I the numerator and denominator.

This graph is not continuous at x = 3 due to a vertical asymptote at this x-value. Therefore the function is not differentiable at x = 3.

This graph is continuous at x = 0, however, it is not differentiable due to a cusp at this x-value. There is a sudden and abrupt change in the slope at x = 0.

This graph is continuous at x = -2 however, it is not differentiable due to a sharp corner at this x-value. There is a sudden and abrupt change in the slope at x = -2.

This graph is continuous at x = 0 , however, it is not differentiable at this x-value because the tangent line at this point is vertical, and all vertical lines have slopes that are undefined.

This graph is continuous throughout its domain except at x = 0. The function is not defined at x = 0. Plugging x = 0 into the function y = sin ⁡ ( 1 x ) gives y = sin ⁡ ( 1 0 ) and ⁡ 1 0 is undefined, so the entire function is undefined at x = 0.

This graph is known as the Greatest Integer Function, or Step Function. This function is not continuous at an infinite number of x-values due to ‘jumps’ in the graph.

Higher-Order Derivatives: These are derivatives of derivatives, often used to understand the curvature and concavity of functions (e.g., the second derivative).

Problem 1: Differentiate the function f ( x ) = ( 3 x 2 − 4 x + 5 ) ( 2 x 3 − x ) .

Solution: Identify the Components L e t u ( x ) = 3 x 2 − 4 x + 5 L e t v ( x ) = 2 x 3 − x

Use the Product Rule f ′ ( x ) = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x )

Apply the Product Rule • Substitute the derivatives back into the product rule formula: f ′ ( x ) = ( 6 x − 4 ) ( 2 x 3 − x ) + ( 3 x 2 − 4 x + 5 ) ( 6 x 2 − 1 )

Simplify • Expand and simplify the expression: f ′ ( x ) = ( 12 x 4 − 6 x 2 − 8 x 3 + 4 x ) + ( 18 x 4 − 3 x 2 − 24 x 3 + 4 x + 30 x f ′ ( x ) = 30 x 4 − 32 x 3 + 27 x 2 − 1

Conclusion: The derivative of the function f ( x ) = ( 3 x 2 − 4 x + 5 ) ( 2 x 3 − x ) is f ′ ( x ) = 30 x 4 − 32 x 3 + 27 x 2 − 1 This process highlights the use of the product rule and basic algebraic manipulation in calculus differentiation.

Applications of Differentiation in Calculus

1. Physics

2. Engineering

3. Economics

4. Biology

5. Medicine

6. Computer Science

7. Environmental Science

Leave a comment Cancel reply

Recent Posts

Recent Comments

AP Subjects | NUM8ERS

College Board AP Spanish Language & Culture: Past Exams

College Board AP Research: Past Exams

College Board AP Latin: Past Exams

Word to JPEG – Convert Word files to JPEG Images Online

WebP to JPEG – Convert WebP to JPEG

HEIC to JPG – Convert HEIC Images to JPEG Format

PNG to JPG – Convert PNG images to JPEG

$a v e r a g e v e l o c i t y = \frac{c h a n g e i n p o s i t i o n}{c h a n g e i n t i m e} = \frac{x_{2} (t) - x_{1} (t)}{t_{2} - t_{1}}$

Basic Concepts in Differentiation :

Definition of the Derivative:

The derivative of a function at a point – the exact change of the y value at a given x-value - gives the slope of the tangent line to the graph of the function at that point.

An example of what this indicates is this; assume we are given the position function of a car that starts from rest as y(t) = t².

$v e l o c i t y = \frac{c h a n g e i n p o s i t i o n}{c h a n g e i n t i m e} = \frac{x_{2} (t) - x_{1} (t)}{t_{2} - t_{1}}$

Before we learn the rules of derivation, we need to learn how to find the derivative using the Definition of the Derivative. This is done using limits.

There are three forms of the Definition of the Derivative:

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

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

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

Where a function is differentiable.

In order for a function to be differentiable, it has to meet certain criteria. One of these criteria is that the function must be continuous.

Here are a few examples of functions that are continuous in their domain.

$y = x^{3} - 9 x$

$y = \sqrt{x} {x \geq 0}$

$y = \ln x {x > 0}$

$y = x^{4} - 3 x^{2} + 2 x + 1$

THEOREM:

If a function f(x) is differentiable at x = c, then f is continuous at x = c.

If f(x) is continuous at x = c, it may or may not be differentiable at x = c.

A function is not differentiable when one of the following occurs on a graph;

This graph is not continuous at x = 3 due to a hole at this
x-value. A hole occurs when we cross out the same polynomial I the numerator and denominator.

This graph is continuous at x = -2 however, it is not differentiable due to a sharp corner at this
x-value. There is a sudden and abrupt change in the slope at x = -2.

This graph is continuous throughout its domain except at
x = 0. The function is not defined at x = 0. Plugging x = 0 into the function $y = \sin (\frac{1}{x})$ gives $y = \sin (\frac{1}{0})$ and $\frac{1}{0}$ is undefined, so the entire function is undefined at x = 0.

Higher-Order Derivatives:

These are derivatives of derivatives, often used to understand the curvature and concavity of functions (e.g., the second derivative).

Problem 1:

Differentiate the function $f (x) = (3 x^{2} - 4 x + 5) (2 x^{3} - x) .$

Solution:

Identify the Components

$\begin{aligned} L e t u (x) = 3 x^{2} - 4 x + 5 \\ L e t v (x) = 2 x^{3} - x \end{aligned}$

Use the Product Rule

$f^{'} (x) = u^{'} (x) v (x) + u (x) v^{'} (x)$

Apply the Product Rule

• Substitute the derivatives back into the product rule formula:

$f^{'} (x) = (6 x - 4) (2 x^{3} - x) + (3 x^{2} - 4 x + 5) (6 x^{2} - 1)$

Simplify

• Expand and simplify the expression:

$f^{'} (x) = (12 x^{4} - 6 x^{2} - 8 x^{3} + 4 x) + (18 x^{4} - 3 x^{2} - 24 x^{3} + 4 x + 30 x f^{'} (x)$

$= 30 x^{4} - 32 x^{3} + 27 x^{2} - 1$

Conclusion:

The derivative of the function

$\begin{array}{r} f (x) = (3 x^{2} - 4 x + 5) (2 x^{3} - x) \\ is f^{'} (x) = 30 x^{4} - 32 x^{3} + 27 x^{2} - 1 \end{array}$
This process highlights the use of the product rule and basic algebraic manipulation in calculus differentiation.