Understanding Differentiation
Why Study Differentiation?
Differentiation is the mathematical tool used to determine the rate at which quantities change. It is essential across various fields such as physics, engineering, and economics for solving problems involving rates of change and optimization.
Applications of Differentiation
- Science and Engineering: Used to model and predict behaviors in systems, such as the stress on materials or fluid dynamics.
- Economics: Helps in finding cost minimization and profit maximization points.
- Medicine: Used in models of growth rates of tumors or the spread of diseases.
Basic Principles of Differentiation
The derivative of a function describes how the function's output changes in relation to changes in the input. It is fundamentally the slope of the tangent to the function at any point.
Example
Consider the function f(x) = x2. The derivative, f'(x) = 2x, helps determine how fast the function's value changes as x changes. At x = 3, the rate of change is 6, indicating that a small change in x around 3 causes about six times that change in the output.
Real World Example: Aerospace Engineering
In designing aircraft wings, differentiation helps in calculating the optimal angles and materials to use that balance lift and drag, crucial for efficient and safe aircraft design.
Learning Differentiation
To effectively learn and apply differentiation:
- Start with understanding the derivative of basic functions like polynomials.
- Practice solving problems using rules of differentiation such as the product rule, quotient rule, and chain rule.
- Apply differentiation to practical situations to see its real-world utility.
- Use computational tools to visualize derivatives and their impacts on functions.
Differentiation Concepts Explained
Notation
There are three different notations commonly used to describe the derivative of a function:
- If \( y = x^2 \), then \( \frac{dy}{dx} = 2x \).
- If \( f(x) = x^2 \), then \( f'(x) = 2x \).
- \( \frac{d}{dx}(x^2) = 2x \).
The derivative \( \frac{dy}{dx} \) is called the derivative of \( y \) with respect to \( x \). \( f'(x) \) is the derivative of \( f(x) \), sometimes referred to as the gradient function of this curve, indicating the slope of the tangent to the curve at any point.
Differentiation of Power Functions
The power rule for differentiation is given by:
\[ \frac{d}{dx}(x^n) = nx^{n-1} \]
This rule applies to any real power \( n \), not only for positive integers.
Examples
- \( \frac{d}{dx}(x^4) = 4x^3 \)
- \( \frac{d}{dx}(x^5) = 5x^4 \)
- \( \frac{d}{dx}(x^6) = 6x^5 \)
Key Point 7.1
The general rule for differentiating power functions is:
\[ \frac{d}{dx}(x^n) = nx^{n-1} \]
Find the Derivative of Each of the Following
- \( x^7 \): \( \frac{d}{dx}(x^7) = 7x^6 \)
- \( \frac{1}{x^2} \): Rewrite as \( x^{-2} \), then \( \frac{d}{dx}(x^{-2}) = -2x^{-3} = -\frac{2}{x^3} \)
- \( \sqrt{x} \): Rewrite as \( x^{1/2} \), then \( \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \)
- \( y = 2 \): The derivative of a constant is zero.
Detailed Solutions for Differentiation
Example 1: Differentiating a Polynomial and Rational Function
Function: \( f(x) = 3x^4 - \frac{1}{2x^2} + \frac{4}{\sqrt{x}} + 5 \)
Differentiation:
\[ f'(x) = \frac{d}{dx}(3x^4) - \frac{d}{dx}\left(\frac{1}{2x^2}\right) + \frac{d}{dx}\left(\frac{4}{x^{1/2}}\right) + \frac{d}{dx}(5) \]
\[ = 12x^3 + \frac{1}{x^3} - \frac{2}{\sqrt{x^3}} \]
Example 2: Gradient of the Tangent to the Curve
Function: \( y = x(2x-1)(x+3) \)
Expanded: \( y = 2x^3 + 5x^2 - 3x \)
Differentiate: \( \frac{dy}{dx} = 6x^2 + 10x - 3 \)
At \( x = 1 \): \( \frac{dy}{dx} = 6(1)^2 + 10(1) - 3 = 13 \)
Gradient at \( (1, 4) \) is 13.
Example 3: Coefficients of a Polynomial
Function: \( y = ax^4 + bx^2 + x \)
Differentiate: \( \frac{dy}{dx} = 4ax^3 + 2bx + 1 \)
Given gradients:
\( 4a(1)^3 + 2b(1) + 1 = 3 \)
\( 4a(-2)^3 + 2b(-2) + 1 = -51 \)
Solving these equations:
\( 4a + 2b = 2 \)
\( -32a - 4b = -52 \)
\( a = 2, b = -3 \)
Detailed Solutions for DifferentiationGradient Calculations
Example a: y = x^4 at (1, 1) results in dy/dx = 4.
Example b: y = x^2 - 2x + 3 at (0, 3) results in dy/dx = -2.
Differentiation of Functions
Example a: \( y = x^5 \) results in \( dy/dx = 5x^4 \).
Example b: \( y = x^9 \) results in \( dy/dx = 9x^8 \).
Example c: \( y = \frac{1}{x} \) results in \( dy/dx = -\frac{1}{x^2} \).
Derivatives for Given Functions
Example a: \( f(x) = 2x^4 \) results in \( f'(x) = 8x^3 \).
Example b: \( f(x) = 3x^5 \) results in \( f'(x) = 15x^4 \).
Derivatives at Specific Points
Example a: \( y = x^2 + x - 4 \) at (1, -2) results in \( dy/dx = 3 \).
Example b: \( y = 5 - \frac{2}{x} \) at (2, 4) results in \( dy/dx = \frac{1}{2} \).
Detailed Solutions for Differentiation
Finding Gradients and Differentiating Functions
Example a: \( y = x^4 \) at \( (1, 1) \), gradient: \( \frac{dy}{dx} = 4x^3 \rightarrow 4 \).
Example b: \( y = x^2 - 2x + 3 \) at \( (0, 3) \), gradient: \( \frac{dy}{dx} = 2x - 2 \rightarrow -2 \).
Basic Differentiation
Example a: \( x^5 \), derivative: \( \frac{dx^5}{dx} = 5x^4 \).
Example b: \( x^9 \), derivative: \( \frac{dx^9}{dx} = 9x^8 \).
Example c: \( x^{-4} \), derivative: \( \frac{dx^{-4}}{dx} = -4x^{-5} = -\frac{4}{x^5} \).
Example d: \( \frac{1}{x} \), derivative: \( \frac{d}{dx}(\frac{1}{x}) = -\frac{1}{x^2} \).
Complex Function Differentiation
Example e: \( \frac{5}{3x^2} \), derivative: \( \frac{d}{dx}(\frac{5}{3x^2}) = -\frac{10}{3x^3} \).
Example f: \( -2 \), derivative: \( \frac{d}{dx}(-2) = 0 \).
Example g: \( \frac{4x}{\sqrt{x}} \), derivative: \( \frac{d}{dx}(4\sqrt{x}) = 2\sqrt{x} \).
Example h: \( \frac{2x}{3x^3} \), derivative: \( \frac{d}{dx}(\frac{2}{3x^2}) = -\frac{4}{3x^3} \).
Differentiating Polynomial and Rational Functions
Example g: \( y = 7x^2 - \frac{3}{x} + \frac{2}{x^2} \), derivative: \( \frac{dy}{dx} = 14x + \frac{3}{x^2} - \frac{4}{x^3} \).
Example h: \( y = 3x + \frac{5}{x} - \frac{1}{\sqrt{x}} \), derivative: \( \frac{dy}{dx} = 3 - \frac{5}{x^2} + \frac{1}{2x^{3/2}} \).
Example i: \( y = 4x^2 + 3x - \frac{2}{\sqrt{x}} \), derivative: \( \frac{dy}{dx} = 8x + 3 + \frac{1}{x^{3/2}} \).
Further Problems with solutions
Problem 1: Coordinates of Points on the Curve
\( y = x^3 - 3x - 8 \) with gradient 9. Solutions: Points \((2, -9)\) and \((-2, -2)\).
Problem 2: Gradient at x-axis Crossing
Gradient at \( x = 2 \) for \( y = \frac{5x - 10}{x^2} \) is 1.25.
Problem 3: Intersection Points and Gradients
Points A and B: \((3, -8)\) and \((-2, 7)\).
Problem 4: Gradient for Quadratic Equation
\( y = ax^2 + bx \) at \( (3, -3) \), gradient is 5, solving for \( a \) and \( b \).
Problem 5: Derivative and Coefficients
\( y = x^3 + ax^2 + bx + 7 \) at \( (1, 5) \), gradient -5, solve for \( a \) and \( b \).
Chain Rule
Solution for the Curve Problem
The curve \( y = \sqrt{ax + b} \) passes through the point (12, 4) and has gradient \( \frac{1}{4} \) at this point. Find the value of \( a \) and the value of \( b \).
Given: \( y = \sqrt{ax + b} \), passes through \( (12, 4) \), gradient \( \frac{1}{4} \) at this point.
Start by substituting \( x = 12 \) and \( y = 4 \):
\[ 4 = \sqrt{12a + b} \quad \text{(1)} \]
Square both sides to eliminate the square root:
\[ 16 = 12a + b \quad \text{(rearranged from 1)} \]
Differentiate \( y = \sqrt{ax + b} \) with respect to \( x \) using the chain rule:
\[ \frac{dy}{dx} = \frac{a}{2\sqrt{ax + b}} \]
Set the derivative equal to the known gradient at \( x = 12 \):
\[ \frac{1}{4} = \frac{a}{2\sqrt{12a + b}} \quad \text{(2)} \]
Substitute \( b = 16 - 12a \) from (1) into (2) and solve for \( a \) and \( b \):
\[ \frac{1}{4} = \frac{a}{8} \Rightarrow a = 2, \quad b = -8 \text{ (from solving the system)} \]
Conclusion: \( a = 2 \), \( b = -8 \).
TextBook solutions
h1>Detailed Calculus SolutionsDifferentiation Problems
1a. \( \frac{d}{dx}\left(x + 4\right)^6 = 6\left(x + 4\right)^5 \)
1b. \( \frac{d}{dx}\left(2x + 3\right)^8 = 16\left(2x + 3\right)^7 \)
1c. \( \frac{d}{dx}\left(3 - 4x\right)^5 = -20\left(3 - 4x\right)^4 \)
1d. \( \frac{d}{dx}\left(\frac{x}{2} + 1\right)^9 = \frac{9}{2}\left(\frac{x}{2} + 1\right)^8 \)
Tangents and Normals
Equations of Tangent and Normal Lines
Find the equation of the tangent and the normal to the curve \( y = 2x^2 + \frac{8}{x^2} - 9 \) at the point where \( x = 2 \).
Answer:
Given the function:
\[ y = 2x^2 + \frac{8}{x^2} - 9 \]
First, differentiate with respect to \( x \) to find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = 4x - \frac{16}{x^3} \]
At \( x = 2 \), calculate \( y \) and \( \frac{dy}{dx} \):
\[ y = 2(2)^2 + \frac{8}{(2)^2} - 9 = 1 \]
\[ \frac{dy}{dx} = 4(2) - \frac{16}{(2)^3} = 6 \]
The equation of the tangent line at \( (2, 1) \) with a gradient of 6:
\[ y - 1 = 6(x - 2) \]
\[ y = 6x - 11 \]
The equation of the normal line at \( (2, 1) \) with a gradient of \( -\frac{1}{6} \):
\[ y - 1 = -\frac{1}{6}(x - 2) \]
\[ x + 6y = 8 \]
Question
Worksheet 1: Basic Differentiation
Differentiate the following functions:
- \( f(x) = 6x^5 \)
- \( g(x) = 9x^3 - 4x \)
- \( h(x) = x^2 + 3x + 7 \)
- \( p(x) = 5 \)
- \( q(x) = 8x^{-2} \)
- \( r(x) = \sqrt{x} \)
- \( s(x) = \frac{1}{x} \)
- \( t(x) = x^{3/2} \)
- \( u(x) = 3x^4 - 5x^2 + 6 \)
- \( v(x) = 4x \)
- \( w(x) = 7 \)
- \( y(x) = 3x^3 - 2x + 4x^{-1} \)
- \( z(x) = \frac{1}{\sqrt{x}} \)
- \( a(x) = x^7 - 7x^3 + 5x \)
- \( b(x) = 9x^{2/3} - 3x^{1/3} \)
- \( c(x) = 5x^{1/4} \)
- \( d(x) = 2x^2 + 5x^3 - x^{-3} \)
- \( e(x) = \frac{2}{x^2} - 3x^4 \)
- \( f(x) = x(x^2 + 1) \)
- \( g(x) = (2x + 3)(x - 5) \)
Worksheet 2: Chain Rule
Differentiate the following functions using the chain rule:
- \( f(x) = (3x+4)^5 \)
- \( g(x) = \sqrt{5x-1} \)
- \( h(x) = \left(\frac{x}{x+1}\right)^2 \)
- \( p(x) = \sin(2x + 3) \)
- \( q(x) = e^{2x-3} \)
- \( r(x) = \ln(x^2 + 1) \)
- \( s(x) = \cos^2(x) \)
- \( t(x) = e^{x^2 + 2x} \)
- \( u(x) = \tan(x + \frac{\pi}{4}) \)
- \( v(x) = \sqrt{x^4 + 3x} \)
- \( w(x) = (x^3 - x)^4 \)
- \( y(x) = \ln(\sin x) \)
- \( z(x) = e^{\sqrt{x}} \)
- \( a(x) = (2x^2 - 3x + 4)^3 \)
- \( b(x) = \sin^2(3x) \)
- \( c(x) = \cos(x^2 + x) \)
- \( d(x) = \tan^2(x + 2) \)
- \( e(x) = \sqrt{\cos x} \)
- \( f(x) = (5x + 3)^{1/2} \)
- \( g(x) = \ln(x^3 + 2x) \)
Worksheet 3: Tangents and Normals
Determine the equations for the tangents and normals to the following functions at the given points:
- \( y = x^3 - 4x \) at \( x = 2 \)
- \( y = \sqrt{x} \) at \( x = 4 \)
- \( y = \sin x \) at \( x = \frac{\pi}{4} \)
- \( y = \ln(x) \) at \( x = e \)
- \( y = e^{x} \) at \( x = 1 \)
- \( y = \tan x \) at \( x = \frac{\pi}{4} \)
- \( y = \frac{1}{x} \) at \( x = 1 \)
- \( y = x^2 \) at \( x = -1 \)
- \( y = 3x^3 - 5x + 1 \) at \( x = 0 \)
- \( y = 4\sqrt{x} \) at \( x = 9 \)
- \( y = \cos x \) at \( x = \frac{\pi}{3} \)
- \( y = e^{-x} \) at \( x = 2 \)
- \( y = \frac{x}{x+1} \) at \( x = 1 \)
- \( y = \frac{1}{x^2} \) at \( x = 2 \)
- \( y = x^4 - 2x^2 + x \) at \( x = 1 \)
- \( y = \sqrt{x+1} \) at \( x = 0 \)
- \( y = \frac{1}{\sqrt{x}} \) at \( x = 4 \)
- \( y = x^3 \) at \( x = -2 \)
- \( y = \ln(x^2 + 1) \) at \( x = 0 \)
- \( y = \sin^2(x) \) at \( x = \frac{\pi}{6} \)