AOS4 Topic 1: Differentiation

Differentiation


Definition

Derivatives, the building blocks of differentiation, represent the rate of change or slope of a function at a specific point. They offer precise information about how functions behave locally near a given point. Symbolically, derivatives are denoted using symbols like \( f'(x) \) or \( \frac{df}{dx} \), where \( f(x) \) is the original function. 


Rules of Differentiation

Rule Derivative Formula
Constant Rule \[ \frac{d}{dx}[c]=0 \]
Power Rule \[ \frac{d}{dx}[x^n] = n \cdot x^{n-1} \]
Sum Rule \[ \frac{d}{dx}[f(x) + g(x)] = \frac{df}{dx} + \frac{dg}{dx} \]
Product Rule \[ \frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x) \]
Quotient Rule \[ \frac{d}{dx}\left[\frac{g(x)}{f(x)}\right] = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{(f(x))^2} \]
Chain Rule \(\frac{d}{dx} g(h(x)) = g'(h(x)) \cdot h'(x) \)
Exponential Functions \[\frac{d}{dx}(e^x) = e^x \]
Logarithmic Functions
  • \[\frac{d}{dx}(\log_a(x)) = \frac{1}{x \ln(a)} \]
  • \[\frac{d}{dx}(\ln(x)) = \frac{1}{x} \]
Trigonometric Functions
  • \( \frac{d}{dx}(\sin(x)) = \cos(x) \)
  • \( \frac{d}{dx}(\cos(x)) = -\sin(x) \)
  • \( \frac{d}{dx}(\tan(x)) = \sec^2(x) \)
  • \( \frac{d}{dx}(\csc(x)) = -\csc(x) \cot(x) \)
  • \( \frac{d}{dx}(\sec(x)) = \sec(x) \tan(x) \)
  • \( \frac{d}{dx}(\cot(x)) = -\csc^2(x) \)



Relationship between Limits and Differentiation

The relationship between limits and differentiation is fundamental in calculus. In fact, differentiation is defined in terms of limits.

When we differentiate a function \( f(x) \) with respect to \( x \), we are essentially finding the limit of the rate of change of the function as the change in \( x \) approaches zero. Mathematically, the derivative of \( f(x) \) at a point \( x = a \) is defined as:

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

Here, the limit as \( h \) approaches zero represents the slope of the tangent line to the graph of \( f(x) \) at \( x = a \).

In essence, differentiation involves computing the limit of the average rate of change of the function over smaller and smaller intervals around a point. As the interval approaches zero, we get the instantaneous rate of change, which is the derivative.

So, limits are crucial in the process of differentiation, providing the foundation for understanding how functions change and behave at specific points.



Explanation of Derivatives by Diagrams

Derivatives can be explained visually using diagrams, particularly graphs of functions. Let's consider a simple example to illustrate this concept.

  1. Graph of the Function: Start by plotting the graph of the function \( f(x) = x^2 \) on a coordinate system.

  4. Tangent Line: Choose a specific point \( x = 1 \) on the graph. Draw a tangent line to the curve at that point.
  5. The slope of this tangent line represents the derivative of \( f(x) \) at \( x = 1 \).


  1. Slope of the Tangent Line: The slope of the tangent line at \( x = 1 \) is \( 2 \). This indicates that the derivative of \( f(x) \) at \( x = 1 \) is \( 2 \), meaning that the function is increasing at that point.

  4. Change in Slope: Move along the curve from left to right or right to left, observing how the slope of the tangent line changes. This change in slope represents how the rate of change of the function varies as \( x \) changes.

  7. Connecting Derivative to Function Behavior: Relate the slope of the tangent line to the behavior of the function. For example, if the derivative is positive, the function is increasing. If the derivative is negative, the function is decreasing. If the derivative is zero, the function may have a maximum, minimum, or point of inflection.



Examples of Derivatives

  1. Linear Functions: Consider the function \( f(x) = mx + b \), where \( m \) and \( b \) are constants. The derivative of this function with respect to \( x \) is simply the slope \( m \) of the line, as the function represents a straight line.
    \[ f'(x) = m \]

  4. Polynomials: For polynomial functions like \( f(x) = ax^n + bx^{n-1} + \ldots + c \), where \( a \), \( b \), \( c \), etc., are constants and \( n \) is a positive integer, the derivative can be found using the power rule. For each term, you multiply the coefficient by the exponent, then subtract one from the exponent.
    For example, if \( f(x) = 3x^2 - 2x + 5 \), then:
    \[ f'(x) = 2 \cdot 3x^{2-1} - 1 \cdot 2x^{1-1} + 0 = 6x - 2 \]

  7. Trigonometric Functions: The derivatives of trigonometric functions can be found using trigonometric identities.
  8. For example, the derivative of \( \sin(x) \) is \( \cos(x) \), and the derivative of \( \cos(x) \) is \( -\sin(x) \).

  11. Exponential Functions: For functions of the form \( f(x) = e^x \), where \( e \) is Euler's number, the derivative is the function itself:
    \[ \frac{d}{dx} e^x = e^x \]

  14. Logarithmic Functions: The derivative of \( \ln(x) \), the natural logarithm function, is \( \frac{1}{x} \).



Derivatives of Inverse Circular Functions

Definition: The derivatives of inverse circular functions are the rates of change of the inverse trigonometric functions with respect to their arguments. These derivatives are essential in calculus for analyzing functions involving angles and trigonometric ratios.

;
Function Derivative
\[f(x) = \sin^{-1}\left(\frac{x}{a}\right)\] \[ f'(x) = \frac{1}{\sqrt{a^2 - x^2}} \]
\[f(x) = \cos^{-1}\left(\frac{x}{a}\right)\] \[ f'(x) = -\frac{1}{\sqrt{a^2 - x^2}} \]
\[f(x) = \tan^{-1}\left(\frac{x}{a}\right)\] \[ f'(x) = \frac{a}{a^2 + x^2} \]



Second derivatives

For the function \( f \) with rule \( f(x) \), the derivative is denoted by \( f' \) and has rule \( f'(x) \). This notation is extended to taking the derivative of the derivative: the new function is denoted by \( f'' \) and has rule \( f''(x) \). This new function is known as the second derivative.

Consider the function \( g \) with rule \( g(x) = 2x^3 - 4x^2 \). The derivative has rule \( g'(x) = 6x^2 - 8x \), and the second derivative has rule \( g''(x) = 12x - 8 \).

Note: The second derivative might not exist at a point even if the first derivative does.

For example, let \( f(x) = x^\frac{4}{3} \). Then \( f'(x) = \frac{4}{3}x^\frac{1}{3} \) and \( f''(x) = \frac{4}{9}x^{-\frac{2}{3}} \).

We see that \( f'(0) = 0 \), but the second derivative \( f''(x) \) is not defined at \( x = 0 \).

In Leibniz notation, the second derivative of \( y \) with respect to \( x \) is denoted by \( \frac{d^2y}{dx^2} \).



Points of Inflection

Points of inflection are locations on the graph of a function where the concavity changes. More formally, a point of inflection occurs at a specific \( x \)-value \( c \) if the second derivative of the function changes sign at that point, i.e., if the concavity changes from concave up to concave down or vice versa.

Mathematically, for a function \( f(x) \), a point of inflection occurs at \( x = c \) if:

  1. \( f''(c) = 0 \) (the second derivative exists at \( c \)).
  2. The sign of \( f''(x) \) changes from negative to positive or from positive to negative as \( x \) crosses \( c \).

Graphically, a point of inflection appears as a location where the curve changes from being bowed upward (concave up) to being bowed downward (concave down), or vice versa.

Graph of y = 4x^3 - x^4 nd 2nd derivative



Concavity and Points of Inflection

Concavity refers to the direction in which a curve bends, either upward (concave upward) or downward (concave downward).

Points of inflection are locations on a curve where the concavity changes.

At a point of inflection, the curve transitions from being concave upward to concave downward, or vice versa.



Concave Up and Concave Down

For a curve \( y = f(x) \):

  • If \( f''(x) > 0 \) for all \( x \) in the interval \( (a, b) \), then the gradient of the curve is strictly increasing over the interval \( (a, b) \). The curve is said to be concave up.
  • If \( f''(x) < 0 \) for all \( x \) in the interval \( (a, b) \), then the gradient of the curve is strictly decreasing over the interval \( (a, b) \). The curve is said to be concave down.

For example, consider the function \( f(x) = x^3 \). This function exhibits concavity and has a point of inflection at the origin (0, 0). At \( x = 0 \), the concavity changes from being concave downward for \( x < 0 \) to concave upward for \( x > 0 \). Therefore, the origin is a point of inflection for this function.

This graph shows the different examples where we can understand concavity of different function whether it is concave up or concave down


Points of inflection are significant in analyzing the behavior of functions, as they indicate changes in the curvature of the curve. They are often associated with changes in the direction of a function's concavity, providing insights into its overall shape and behavior.


Use the simulation below to see these effects

Created with GeoGebra®, by Ravinder Kumar, Link

Example 1
Differentiate the function

     \[ g(x) = 4x^3 - 2x^2 + 7x - 9 \]

Solution:

\[ g'(x) = 12x^2 - 4x + 7 \]

Example 2

Given \( h(x) = (x^2 + 3x)(2x - 1) \), find \( h'(x) \).

Solution: Apply the product rule:

\( h'(x) = (x^2 + 3x)'(2x - 1) + (x^2 + 3x)(2x - 1)' \).

\( h'(x) = (2x + 3)(2x - 1) + (x^2 + 3x)(2) \).

\( h'(x) = 4x^2 + 6x - 2x - 3 + 2x^2 + 6x \).

Example 3

Find the derivative of \( y(x) = \frac{2x - 1}{x^2 + 3x + 2} \).

Solution: Apply the quotient rule:

\( y'(x) = \frac{(x^2 + 3x + 2)'}{(2x - 1)} - \frac{(x^2 + 3x + 2)(2x - 1)'}{(2x - 1)^2} \).

\( y'(x) = \frac{(2x + 3)(2x - 1) - (x^2 + 3x + 2)(2)}{(2x - 1)^2} \).

\( y'(x) = \frac{4x^2 + 6x - 2x - 3 - 2x^2 - 6x - 4}{(2x - 1)^2} \).

\( y'(x) = \frac{2x^2 - 7}{(2x - 1)^2} \).

Example 4

If \( u(x) = 2x - 1 \) and \( f(u) = u^3 \), find \( \frac{d}{dx}f(u) \).

Solution: Apply the chain rule:

\( \frac{d}{dx}f(u) = \frac{d}{du}f(u) \cdot \frac{dx}{du} \).

\( \frac{d}{dx}f(u) = 3u^2 \cdot 2 \).

\( \frac{d}{dx}f(u) = 6(2x - 1)^2 \).

Example 5
{Linear Functions}:
Consider the function \( f(x) = mx + b \), where \( m \) and \( b \) are constants. The derivative of this function with respect to \( x \) is simply the slope \( m \) of the line, as the function represents a straight line.

     \[ f'(x) = m \]

Example 6

Find the gradient of the curve \( x = y^2 - 4y \) at \( x = 5 \).

\( x = y^2 - 4y \)

\( \frac{dx}{dy} = 2y - 4 \)

\( \frac{dy}{dx} = \frac{1}{2y - 4} \)

Therefore, \( y = 2 \)

Substituting \( x = 5 \) into \( x = y^2 - 4y \) yields

\( y^2 - 4y = 5 \)

\( y^2 - 4y - 5 = 0 \)

\( (y - 5)(y + 1) = 0 \)

Therefore, \( y = 5 \) or \( y = -1 \)

Substituting these two \( y \)-values into the derivative gives

\( \frac{dy}{dx} = \frac{1}{6} \) or \( \frac{dy}{dx} = -\frac{1}{6} \)

Example 7

Find the derivative of \( \tan^{-1}\left(\frac{2x}{3}\right) \) with respect to \( x \).

To find the derivative, we use the chain rule:

Let \( u = \frac{2x}{3} \), so \( \frac{du}{dx} = \frac{2}{3} \).

The derivative of \( \tan^{-1}(u) \) with respect to \( u \) is \( \frac{1}{1 + u^2} \).

Using the chain rule, we have:

\[ \frac{d}{dx} \left( \tan^{-1}\left(\frac{2x}{3}\right) \right) = \frac{1}{1 + \left(\frac{2x}{3}\right)^2} \cdot \frac{du}{dx} \]

Substituting \( \frac{2}{3} \) for \( \frac{du}{dx} \), we get:

\[ \frac{d}{dx} \left( \tan^{-1}\left(\frac{2x}{3}\right) \right) = \frac{1}{1 + \left(\frac{2x}{3}\right)^2} \cdot \frac{2}{3} \]

So, the derivative of \( \tan^{-1}\left(\frac{2x}{3}\right) \) with respect to \( x \) is:

\[ \frac{2}{3(1 + \frac{4x^2}{9})} \]

This simplifies to:

\[ \frac{2}{3 + \frac{4x^2}{9}} \]

Further simplification yields:

\[ \frac{2}{\frac{9 + 4x^2}{9}} = \frac{18}{9 + 4x^2} = \frac{18}{4x^2 + 9} \]

So, the correct derivative is \( \frac{18}{4x^2 + 9} \).

Example 8

Find the second derivative of \( 2\tan^{-1}\left(\frac{2x}{3}\right) \).

To find the second derivative of \( 2\tan^{-1}\left(\frac{2x}{3}\right) \) with respect to \( x \), we first need to find the first derivative and then differentiate it again.

Given function: \( f(x) = 2\tan^{-1}\left(\frac{2x}{3}\right) \)

First derivative:

\[ f'(x) = \frac{4}{3\left(1+\frac{4x^2}{9}\right)} \]

Now, to find the second derivative, we differentiate \( f'(x) \) with respect to \( x \):

\[ f''(x) = \frac{32x}{27\left(1+\frac{4x^2}{9}\right)^2} \]

So, the second derivative of \( 2\tan^{-1}\left(\frac{2x}{3}\right) \) with respect to \( x \) is \( \frac{32x}{27\left(1+\frac{4x^2}{9}\right)^2} \).

Example 9

A car is traveling along a straight road. Its velocity function is given by \( v(t) = 3t^2 - 6t + 5 \), where \( v \) is measured in meters per second and \( t \) is measured in seconds. Find the time when the car changes direction.

To find the time when the car changes direction using the derivative, we look for the critical points of the velocity function \( v(t) \). A critical point occurs where the derivative \( v'(t) \) equals zero or is undefined.

Given the velocity function \( v(t) = 3t^2 - 6t + 5 \), we find its derivative \( v'(t) \):

\[ v'(t) = \frac{d}{dt} (3t^2 - 6t + 5) \]

\[ v'(t) = 6t - 6 \]

To find critical points, we set \( v'(t) \) equal to zero and solve for \( t \):

\[ 6t - 6 = 0 \]

\[ 6t = 6 \]

\[ t = 1 \]

Now, we check the second derivative \( v''(t) \) to determine whether the critical point \( t = 1 \) corresponds to a maximum, minimum, or an inflection point:

\[ v''(t) = \frac{d}{dt} (6t - 6) = 6 \]

Since \( v''(t) \) is positive for all \( t \), the function \( v(t) \) is concave up everywhere, and the critical point \( t = 1 \) corresponds to a minimum value of \( v(t) \). Therefore, the car changes direction at \( t = 1 \) second.

Example 10
The position function of a particle moving along a straight line is given by \( s(t) = t^3 - 6t^2 + 9t \), where \( s \) is measured in meters and \( t \) is measured in seconds. Determine the intervals on which the particle's path is concave up and concave down, and find any points of inflection.


To find the velocity function (the first derivative of the position function), differentiate \( s(t) \) with respect to \( t \):

\( s'(t) = \frac{d}{dt} (t^3 - 6t^2 + 9t) \)

\( s'(t) = 3t^2 - 12t + 9 \)

  • Second Derivative \( s''(t) \):

    To determine concavity, we find the acceleration function (the second derivative of the position function), which is the derivative of the velocity function:

    \( s''(t) = \frac{d}{dt} (3t^2 - 12t + 9) \)

    \( s''(t) = 6t - 12 \)


  • Analysis of \( s''(t) \):

    Now, we analyze the sign of \( s''(t) \) to identify intervals of concavity:

    • For \( s''(t) > 0 \), the particle's path is concave up.
    • For \( s''(t) < 0 \), the particle's path is concave down.
    • Points where \( s''(t) = 0 \) or is undefined are potential points of inflection.
  • Interval Analysis:

    To find intervals of concavity and points of inflection, we solve \( s''(t) = 0 \) to find the critical points:

    \( 6t - 12 = 0 \)

    \( t = 2 \)

    Now, we test the intervals around \( t = 2 \) using the second derivative test:

    • For \( t < 2 \), \( s''(t) < 0 \), so the curve is concave down.
    • For \( t > 2 \), \( s''(t) > 0 \), so the curve is concave up.
    • At \( t = 2 \), the concavity changes, indicating a point of inflection.

    • So, the particle's path is concave down for \( t \in (-\infty, 2) \) and concave up for \( t \in (2, +\infty) \). The point of inflection is at \( t = 2 \).

    Exercise &&1&& (&&1&& Question)

    Find the Derivative

    Find the derivative of \( f(x) = 3\ln(x) \).

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    Exercise &&2&& (&&1&& Question)

    Compute the derivative of \( g(x) = \sqrt{2x + 1} \).

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    Exercise &&3&& (&&1&& Question)

    If \( x = t^2 \) and \( y = t^3 \), then \( \frac{dx}{dy} \) =

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    Exercise &&4&& (&&1&& Question)

    Find the derivative of \( \sin^{-1}\left({0.2x}\right) \) with respect to \( x \).

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    Exercise &&5&& (&&1&& Question)

    Find the second derivative of \( y = e^x \sin(x) \).

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    Exercise &&6&& (&&1&& Question)

    Consider the function \( f(x) = x^3 - 3x^2 + 2 \). Identify any points of inflection for this function.

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    Exercise &&7&& (&&1&& Question)

    Consider the function \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. Find the derivative of this function with respect to \( x \).

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    Exercise &&8&& (&&1&& Question)

    The gradient of the line that is perpendicular to the curve \(4y^2 - 6xy - 2x^2 = 2\) at the point (1, 2) is:

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