O Level (Sec 3 & 4)Additional Mathematics
Techniques and Applications of Differentiation
Master differentiation techniques and applications with this comprehensive study guide. Learn chain rule, product rule, quotient rule, finding tangents and normals, and solving real-world rate of change problems. Includes 14 worked examples with step-by-step video solutions.
By Timothy Gan14 September 2021Singapore O-Level syllabus
Free revision pack
Use the notes online, then download the worksheet to practise without distractions.
- 17 concept sections
- 14 worked questions
- 29 video lessons
17
Concept sections
14
Practice questions
15
Concept videos
14
Solution videos
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Understanding Differentiation and Its Applications
Hi students, let's explore the powerful concept of differentiation today! Differentiation is one of the most fundamental practices in mathematics. It directly relates to the knowledge we acquired during Elementary Mathematics, particularly the concept of gradients and rates of change.
Differentiation allows us to find the instantaneous rate of change of a function at any point, which has countless real-world applications - from finding the velocity of a moving object to optimizing business profits to determining the maximum height of a projectile.
In this comprehensive guide, I will explain the key differentiation techniques (chain rule, product rule, quotient rule), show you how to find tangents and normals to curves, solve connected rate of change problems, and analyze stationary points.
Differentiation is a very common topic that can be found in many mathematics syllabi such as the IGCSE Additional Mathematics (0606) and Singapore SEAB Additional Mathematics.
1
Concept
Derivative as a Gradient Function
The derivative of a function $y = f(x)$ represents the gradient (or slope) of the curve at any point. We write this as $\frac{dy}{dx}$ or $f'(x)$.
Geometric Interpretation:
- If $\frac{dy}{dx} > 0$, the function is increasing (gradient is positive)
- If $\frac{dy}{dx} < 0$, the function is decreasing (gradient is negative)
- If $\frac{dy}{dx} = 0$, there is a stationary point (horizontal tangent)
Understanding the derivative as a gradient function is crucial for applications like finding tangent lines and analyzing the behavior of functions.
Video lesson
Derivative As Gradient Function
2
Concept
Derivative as Power Functions
The Power Rule is the fundamental rule for differentiating powers of $x$.
Power Rule: If $y = x^n$, then $\frac{dy}{dx} = nx^{n-1}$
This rule works for all values of $n$ - positive, negative, fractions, and even irrational numbers.
Examples:
- $y = x^3 \Rightarrow \frac{dy}{dx} = 3x^2$
- $y = x^{-2} \Rightarrow \frac{dy}{dx} = -2x^{-3}$
- $y = \sqrt{x} = x^{1/2} \Rightarrow \frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$
- $y = \frac{1}{x^3} = x^{-3} \Rightarrow \frac{dy}{dx} = -3x^{-4} = -\frac{3}{x^4}$
Constant Rule: If $y = c$ (constant), then $\frac{dy}{dx} = 0$
This makes sense because a constant doesn't change, so its rate of change is zero.
Video lesson
Derivative As Power Functions
3
Concept
Scalar Multiple Rule
The Scalar Multiple Rule allows us to "pull out" constants when differentiating.
Scalar Multiple Rule: If $y = cf(x)$ where $c$ is a constant, then:
$\frac{dy}{dx} = c \cdot f'(x)$
In other words, we can differentiate the function and multiply by the constant.
Examples:
- $y = 5x^3 \Rightarrow \frac{dy}{dx} = 5 \cdot 3x^2 = 15x^2$
- $y = -2x^4 \Rightarrow \frac{dy}{dx} = -2 \cdot 4x^3 = -8x^3$
- $y = \frac{3}{x^2} = 3x^{-2} \Rightarrow \frac{dy}{dx} = 3 \cdot (-2)x^{-3} = -6x^{-3} = -\frac{6}{x^3}$
This rule makes differentiation much simpler by allowing us to handle constants separately.
Video lesson
Scalar Multiple
4
Concept
Addition and Subtraction Rule
The Addition and Subtraction Rule allows us to differentiate each term separately.
Sum/Difference Rule: If $y = f(x) \pm g(x)$, then:
$\frac{dy}{dx} = f'(x) \pm g'(x)$
This means we can differentiate each term individually and then add or subtract the results.
Examples:
$y = x^3 + x^2$
$\frac{dy}{dx} = 3x^2 + 2x$
$\frac{dy}{dx} = 3x^2 + 2x$
$y = 5x^4 - 3x^2 + 7x$
$\frac{dy}{dx} = 20x^3 - 6x + 7$
$\frac{dy}{dx} = 20x^3 - 6x + 7$
$y = 3x^4 - 5x^2 + 7x - 2$
$\frac{dy}{dx} = 12x^3 - 10x + 7$
$\frac{dy}{dx} = 12x^3 - 10x + 7$
Combining the Power Rule, Scalar Multiple Rule, and Addition/Subtraction Rule allows us to differentiate any polynomial function quickly and efficiently.
Video lesson
Addition and Subtraction Rule
5
Concept
Chain Rule
The Chain Rule is used to differentiate composite functions (a function within a function).
Formula: If $y = f(u)$ and $u = g(x)$, then:
$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$
Or more directly: If $y = [f(x)]^n$, then $\frac{dy}{dx} = n[f(x)]^{n-1} \cdot f'(x)$
How to Apply the Chain Rule:
- Identify the outer function and inner function
- Differentiate the outer function (keeping the inner function unchanged)
- Multiply by the derivative of the inner function
Example: $y = (3x^2 + 5)^4$
Outer function: $u^4$ where $u = 3x^2 + 5$
Inner function: $u = 3x^2 + 5$, so $\frac{du}{dx} = 6x$
Inner function: $u = 3x^2 + 5$, so $\frac{du}{dx} = 6x$
$\frac{dy}{dx} = 4(3x^2 + 5)^3 \times 6x = 24x(3x^2 + 5)^3$
Example with Negative Power: $y = \frac{1}{\sqrt{2x+1}} = (2x+1)^{-1/2}$
$\frac{dy}{dx} = -\frac{1}{2}(2x+1)^{-3/2} \times 2 = -(2x+1)^{-3/2} = -\frac{1}{(2x+1)^{3/2}}$
The chain rule is one of the most important differentiation techniques you'll use!
Video lesson
Chain Rule
6
Concept
Product Rule
The Product Rule is used to differentiate the product of two functions.
Formula: If $y = u \cdot v$ where both $u$ and $v$ are functions of $x$, then:
$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$
Or in shorthand: $\frac{dy}{dx} = uv' + vu'$
Mnemonic: "First times derivative of second, plus second times derivative of first"
When to Use Product Rule:
Use the product rule when you have two separate functions multiplied together that can't be easily combined.
Example: $y = x^3(2x+1)^5$
Let $u = x^3$ and $v = (2x+1)^5$
$\frac{du}{dx} = 3x^2$
$\frac{dv}{dx} = 5(2x+1)^4 \times 2 = 10(2x+1)^4$ (using chain rule)
$\frac{dy}{dx} = x^3 \cdot 10(2x+1)^4 + (2x+1)^5 \cdot 3x^2$
$= 10x^3(2x+1)^4 + 3x^2(2x+1)^5$
$= x^2(2x+1)^4[10x + 3(2x+1)]$ (factoring out common terms)
$= x^2(2x+1)^4(16x + 3)$
Video lesson
Product Rule
7
Concept
Quotient Rule
The Quotient Rule is used to differentiate a fraction where both numerator and denominator are functions of $x$.
Formula: If $y = \frac{u}{v}$ where both $u$ and $v$ are functions of $x$, then:
$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
Or in shorthand: $\frac{dy}{dx} = \frac{vu' - uv'}{v^2}$
Mnemonic: "Bottom times derivative of top, minus top times derivative of bottom, all over bottom squared" or "vdu minus udv over v squared"
When to Use Quotient Rule:
Use the quotient rule when you have a fraction with both numerator and denominator containing variables.
Alternative Approach: You can also rewrite as $y = u \cdot v^{-1}$ and use the product rule, but the quotient rule is often more direct.
Example: $y = \frac{x+1}{\sqrt{3x^2-1}}$
Let $u = x+1$ and $v = \sqrt{3x^2-1} = (3x^2-1)^{1/2}$
$\frac{du}{dx} = 1$
$\frac{dv}{dx} = \frac{1}{2}(3x^2-1)^{-1/2} \times 6x = \frac{3x}{\sqrt{3x^2-1}}$ (using chain rule)
$\frac{dy}{dx} = \frac{\sqrt{3x^2-1} \cdot 1 - (x+1) \cdot \frac{3x}{\sqrt{3x^2-1}}}{(3x^2-1)}$
$= \frac{\sqrt{3x^2-1} - \frac{3x(x+1)}{\sqrt{3x^2-1}}}{3x^2-1}$
$= \frac{(3x^2-1) - 3x(x+1)}{(3x^2-1)^{3/2}}$
$= \frac{3x^2-1-3x^2-3x}{(3x^2-1)^{3/2}} = \frac{-3x-1}{(3x^2-1)^{3/2}}$
Video lesson
Quotient Rule
8
Concept
Equations of Tangents and Normals
One of the most important applications of differentiation is finding equations of tangent and normal lines to curves.
Tangent Line: A line that touches the curve at exactly one point and has the same gradient as the curve at that point.
Normal Line: A line perpendicular to the tangent at the point of contact.
Key Steps to Find Tangent Equation:
- Find the $y$-coordinate at the given $x$-value
- Find $\frac{dy}{dx}$ (the gradient function)
- Evaluate $\frac{dy}{dx}$ at the given $x$-value to get the gradient $m$
- Use point-slope form: $y - y_1 = m(x - x_1)$
Key Steps to Find Normal Equation:
- Follow steps 1-3 above to find the tangent gradient $m$
- The normal gradient is $m_{normal} = -\frac{1}{m}$ (negative reciprocal)
- Use point-slope form with the normal gradient
Example: Find the tangent to $y = x^3 + 2x^2 - 5x + 1$ at $x = 1$
Step 1: $y = 1^3 + 2(1)^2 - 5(1) + 1 = -1$, so point is $(1, -1)$
Step 2: $\frac{dy}{dx} = 3x^2 + 4x - 5$
Step 3: At $x = 1$: $\frac{dy}{dx} = 3(1)^2 + 4(1) - 5 = 2$
Step 4: Tangent: $y - (-1) = 2(x - 1)$
$y + 1 = 2x - 2$
$y = 2x - 3$
$y + 1 = 2x - 2$
$y = 2x - 3$
Normal: $m_{normal} = -\frac{1}{2}$
$y + 1 = -\frac{1}{2}(x - 1)$
$y = -\frac{1}{2}x - \frac{1}{2}$
$y + 1 = -\frac{1}{2}(x - 1)$
$y = -\frac{1}{2}x - \frac{1}{2}$
9
Concept
Connected Rates of Change
Connected rates of change problems involve finding how one quantity changes with respect to time when it's related to another quantity that's also changing with time.
The Key Relationship: Using the chain rule:
$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
This allows us to connect the rate of change of $y$ with respect to time to the rate of change of $x$ with respect to time.
Typical Problem Structure:
- You're given a relationship between two variables (e.g., $V = \frac{4}{3}\pi r^3$ for a sphere)
- You're told how one variable changes with time (e.g., radius increasing at 0.5 cm/s)
- You need to find how another variable changes with time (e.g., rate of change of volume)
General Steps:
- Write down the relationship between the variables
- Differentiate both sides with respect to time $t$ (using chain rule)
- Substitute known values
- Solve for the unknown rate
Example: A spherical balloon is being inflated. Its radius $r$ is increasing at 2 cm/s. Find the rate of increase of volume when $r = 5$ cm.
$V = \frac{4}{3}\pi r^3$
Differentiate with respect to $t$:
$\frac{dV}{dt} = \frac{4}{3}\pi \times 3r^2 \times \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$
Given: $\frac{dr}{dt} = 2$ cm/s and $r = 5$ cm
$\frac{dV}{dt} = 4\pi(5)^2(2) = 200\pi \approx 628.3$ cm³/s
10
Concept
Increasing and Decreasing Functions
A function is increasing when its gradient is positive, and decreasing when its gradient is negative. Understanding when functions increase or decrease is crucial for optimization problems and curve sketching.
Key Concepts:
- If $\frac{dy}{dx} > 0$ for all $x$ in an interval, then $y$ is increasing in that interval
- If $\frac{dy}{dx} < 0$ for all $x$ in an interval, then $y$ is decreasing in that interval
- If $\frac{dy}{dx} = 0$ at a point, that point might be a stationary point
Finding Intervals of Increase/Decrease:
- Find $\frac{dy}{dx}$
- Solve $\frac{dy}{dx} = 0$ to find critical points
- Test the sign of $\frac{dy}{dx}$ in each interval between critical points
- Determine where the function is increasing (positive gradient) or decreasing (negative gradient)
Always Increasing/Decreasing:
For a function to be always increasing, we need $\frac{dy}{dx} \geq 0$ for all values of $x$.
For a quadratic expression $ax^2 + bx + c$ to be always positive (when $a > 0$), the discriminant must satisfy $b^2 - 4ac \leq 0$.
Example: Determine if $y = 2x^3 + 3x^2 - 12x + 5$ is always increasing.
$\frac{dy}{dx} = 6x^2 + 6x - 12 = 6(x^2 + x - 2) = 6(x+2)(x-1)$
Critical points: $x = -2$ and $x = 1$
Test intervals:
- $x < -2$: $\frac{dy}{dx} = 6(-)(-) = (+)$ → increasing
- $-2 < x < 1$: $\frac{dy}{dx} = 6(+)(-) = (-)$ → decreasing
- $x > 1$: $\frac{dy}{dx} = 6(+)(+) = (+)$ → increasing
The function is NOT always increasing - it decreases between $x = -2$ and $x = 1$.
Video lesson
Increasing and Decreasing Functions
11
Concept
Stationary Points
A stationary point occurs where the gradient of the curve is zero, i.e., where $\frac{dy}{dx} = 0$.
Types of Stationary Points:
- Maximum Point: The curve changes from increasing to decreasing
- Minimum Point: The curve changes from decreasing to increasing
- Point of Inflection: The gradient is zero but doesn't change sign
Finding Stationary Points:
- Find $\frac{dy}{dx}$
- Set $\frac{dy}{dx} = 0$ and solve for $x$
- Substitute $x$-values back into original equation to find $y$-coordinates
- Determine the nature using one of the nature tests (first derivative test or second derivative test)
Example: Find the stationary points of $y = x^3 - 3x + 2$
$\frac{dy}{dx} = 3x^2 - 3$
Set $\frac{dy}{dx} = 0$: $3x^2 - 3 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$
At $x = 1$: $y = 1 - 3 + 2 = 0$ → Point $(1, 0)$
At $x = -1$: $y = -1 + 3 + 2 = 4$ → Point $(-1, 4)$
At $x = -1$: $y = -1 + 3 + 2 = 4$ → Point $(-1, 4)$
Once we've found the stationary points, we need to determine their nature using either the first derivative test or second derivative test.
Video lesson
Stationary Points
12
Concept
First Derivative Test
The First Derivative Test determines the nature of a stationary point by checking how the gradient changes on either side of that point.
Method:
Test the sign of $\frac{dy}{dx}$ on either side of the stationary point:
- If $\frac{dy}{dx}$ changes from $+$ to $-$: Maximum point
- If $\frac{dy}{dx}$ changes from $-$ to $+$: Minimum point
- If $\frac{dy}{dx}$ doesn't change sign: Point of inflection
How to Apply:
- Find the stationary points by solving $\frac{dy}{dx} = 0$
- Choose test points slightly to the left and right of each stationary point
- Evaluate $\frac{dy}{dx}$ at these test points
- Observe the sign change to determine the nature
Example: For $y = x^3 - 3x + 2$, we found stationary points at $x = -1$ and $x = 1$.
$\frac{dy}{dx} = 3x^2 - 3$
First derivative test:
For $x = -1$:
- Test $x = -2$: $\frac{dy}{dx} = 3(-2)^2 - 3 = 9 > 0$ (positive, increasing)
- Test $x = 0$: $\frac{dy}{dx} = 3(0)^2 - 3 = -3 < 0$ (negative, decreasing)
- Gradient changes from $+$ to $-$ → Maximum point at $(-1, 4)$
For $x = 1$:
- Test $x = 0$: $\frac{dy}{dx} = 3(0)^2 - 3 = -3 < 0$ (negative, decreasing)
- Test $x = 2$: $\frac{dy}{dx} = 3(2)^2 - 3 = 9 > 0$ (positive, increasing)
- Gradient changes from $-$ to $+$ → Minimum point at $(1, 0)$
The first derivative test is particularly useful when the second derivative is difficult to compute or when $\frac{d^2y}{dx^2} = 0$ (which makes the second derivative test inconclusive).
Video lesson
First Derivative Test
13
Concept
Second Derivative Test
The Second Derivative Test uses the second derivative $\frac{d^2y}{dx^2}$ to determine the nature of a stationary point. This test is often quicker than the first derivative test when the second derivative is easy to calculate.
Method:
Find $\frac{d^2y}{dx^2}$ and evaluate at the stationary point:
- If $\frac{d^2y}{dx^2} < 0$: Maximum point (curve is concave down, like ∩)
- If $\frac{d^2y}{dx^2} > 0$: Minimum point (curve is concave up, like ∪)
- If $\frac{d^2y}{dx^2} = 0$: Test is inconclusive (use first derivative test instead)
How to Apply:
- Find the stationary points by solving $\frac{dy}{dx} = 0$
- Find the second derivative $\frac{d^2y}{dx^2}$
- Evaluate $\frac{d^2y}{dx^2}$ at each stationary point
- Interpret the sign to determine the nature
Example: For $y = x^3 - 3x + 2$, we found stationary points at $x = -1$ and $x = 1$.
$\frac{dy}{dx} = 3x^2 - 3$
$\frac{d^2y}{dx^2} = 6x$
Second derivative test:
At $x = 1$:
$\frac{d^2y}{dx^2} = 6(1) = 6 > 0$ → Minimum point at $(1, 0)$
$\frac{d^2y}{dx^2} = 6(1) = 6 > 0$ → Minimum point at $(1, 0)$
At $x = -1$:
$\frac{d^2y}{dx^2} = 6(-1) = -6 < 0$ → Maximum point at $(-1, 4)$
$\frac{d^2y}{dx^2} = 6(-1) = -6 < 0$ → Maximum point at $(-1, 4)$
Why It Works:
The second derivative measures the rate of change of the gradient:
- When $\frac{d^2y}{dx^2} > 0$, the gradient is increasing, which means the curve is bending upward (minimum)
- When $\frac{d^2y}{dx^2} < 0$, the gradient is decreasing, which means the curve is bending downward (maximum)
The second derivative test is generally faster than the first derivative test, but it can't be used when $\frac{d^2y}{dx^2} = 0$ at the stationary point.
Video lesson
Second Derivative Test
14
Concept
Maximum and Minimum Values
Beyond finding stationary points on curves, we often need to find the absolute maximum or minimum value of a function over a given domain or interval. This is crucial for optimization problems in real-world applications.
Key Concepts:
- Local Maximum/Minimum: The highest/lowest value in a small region around a point (what we find at stationary points)
- Absolute Maximum/Minimum: The highest/lowest value over the entire domain
- The absolute extrema can occur at:
- Stationary points (where $\frac{dy}{dx} = 0$)
- Endpoints of the domain
- Points where the derivative doesn't exist
Finding Absolute Maximum/Minimum on an Interval $a \leq x \leq b$:
- Find all stationary points by solving $\frac{dy}{dx} = 0$
- Evaluate $y$ at:
- Both endpoints $x = a$ and $x = b$
- The largest value is the absolute maximum
- The smallest value is the absolute minimum
Example: Find the maximum and minimum values of $y = x^3 - 3x$ on the interval $-2 \leq x \leq 3$.
Step 1: Find stationary points:
$\frac{dy}{dx} = 3x^2 - 3 = 0$
$x^2 = 1$
$x = \pm 1$ (both are in the interval $-2 \leq x \leq 3$)
$\frac{dy}{dx} = 3x^2 - 3 = 0$
$x^2 = 1$
$x = \pm 1$ (both are in the interval $-2 \leq x \leq 3$)
Step 2: Evaluate at stationary points and endpoints:
- At $x = -2$: $y = (-2)^3 - 3(-2) = -8 + 6 = -2$
- At $x = -1$: $y = (-1)^3 - 3(-1) = -1 + 3 = 2$
- At $x = 1$: $y = 1^3 - 3(1) = 1 - 3 = -2$
- At $x = 3$: $y = 3^3 - 3(3) = 27 - 9 = 18$
Step 3: Compare values:
- Maximum value: $y = 18$ at $x = 3$
- Minimum value: $y = -2$ at $x = -2$ and $x = 1$
Optimization Problems:
Many real-world problems involve finding maximum profit, minimum cost, maximum area, etc. The general approach is:
- Define the variable and express the quantity to be optimized as a function
- Find the domain of the problem
- Find critical points using $\frac{dy}{dx} = 0$
- Determine which critical point gives the maximum or minimum using nature tests
- Check endpoints if the domain is a closed interval
Video lesson
Maximum and Minimum Value
15
Concept
Basics of Trigonometric Derivatives
Trigonometric functions have specific differentiation rules that are essential for solving problems involving periodic motion, waves, and oscillations.
Basic Trigonometric Derivatives:
- $\frac{d}{dx}(\sin x) = \cos x$
- $\frac{d}{dx}(\cos x) = -\sin x$
- $\frac{d}{dx}(\tan x) = \sec^2 x$
These are fundamental results that you should memorize.
Important Notes:
- When differentiating trigonometric functions, the angle must be in radians, not degrees
- For composite functions like $\sin(3x)$ or $\cos(x^2)$, use the chain rule
- Always look for opportunities to simplify using trigonometric identities
Chain Rule with Trigonometric Functions:
For $y = \sin(f(x))$: $\frac{dy}{dx} = \cos(f(x)) \cdot f'(x)$
For $y = \cos(f(x))$: $\frac{dy}{dx} = -\sin(f(x)) \cdot f'(x)$
For $y = \tan(f(x))$: $\frac{dy}{dx} = \sec^2(f(x)) \cdot f'(x)$
Examples:
Example 1: $y = \sin(3x)$
Using chain rule:
$\frac{dy}{dx} = \cos(3x) \times 3 = 3\cos(3x)$
$\frac{dy}{dx} = \cos(3x) \times 3 = 3\cos(3x)$
Example 2: $y = \cos(x^2 + 1)$
Using chain rule:
$\frac{dy}{dx} = -\sin(x^2 + 1) \times 2x = -2x\sin(x^2 + 1)$
$\frac{dy}{dx} = -\sin(x^2 + 1) \times 2x = -2x\sin(x^2 + 1)$
Example 3: $y = \tan(2x - 5)$
Using chain rule:
$\frac{dy}{dx} = \sec^2(2x - 5) \times 2 = 2\sec^2(2x - 5)$
$\frac{dy}{dx} = \sec^2(2x - 5) \times 2 = 2\sec^2(2x - 5)$
Example 4: $y = \sin^3 x = (\sin x)^3$
Using chain rule (outer function is power, inner is sin):
$\frac{dy}{dx} = 3(\sin x)^2 \times \cos x = 3\sin^2 x \cos x$
$\frac{dy}{dx} = 3(\sin x)^2 \times \cos x = 3\sin^2 x \cos x$
Product and Quotient Rules with Trig Functions:
For $y = x \sin x$:
$\frac{dy}{dx} = x \cdot \cos x + \sin x \cdot 1 = x\cos x + \sin x$
$\frac{dy}{dx} = x \cdot \cos x + \sin x \cdot 1 = x\cos x + \sin x$
For $y = \frac{\sin x}{x}$:
$\frac{dy}{dx} = \frac{x \cdot \cos x - \sin x \cdot 1}{x^2} = \frac{x\cos x - \sin x}{x^2}$
$\frac{dy}{dx} = \frac{x \cdot \cos x - \sin x \cdot 1}{x^2} = \frac{x\cos x - \sin x}{x^2}$
Video lesson
Basics of Trigonometric Derivative
16
Concept
Basics of Logarithmic Derivatives
Logarithmic functions have unique differentiation properties that are particularly useful in calculus and applications involving exponential growth and decay.
Basic Logarithmic Derivative:
- $\frac{d}{dx}(\ln x) = \frac{1}{x}$ for $x > 0$
This is one of the most important differentiation rules in calculus.
Important Notes:
- The derivative formula only applies to natural logarithm ($\ln x$, base $e$)
- For logarithms with other bases, convert to natural logarithm first: $\log_a x = \frac{\ln x}{\ln a}$
- The domain restriction $x > 0$ is crucial - you can't take the logarithm of zero or negative numbers
Chain Rule with Logarithmic Functions:
For $y = \ln(f(x))$ where $f(x) > 0$:
$\frac{dy}{dx} = \frac{1}{f(x)} \times f'(x) = \frac{f'(x)}{f(x)}$
This is an extremely useful formula!
Examples:
Example 1: $y = \ln(3x)$
$\frac{dy}{dx} = \frac{1}{3x} \times 3 = \frac{1}{x}$
Interesting note: $\frac{d}{dx}(\ln(3x)) = \frac{d}{dx}(\ln x)$ because $\ln(3x) = \ln 3 + \ln x$ and $\ln 3$ is a constant.
Example 2: $y = \ln(x^2 + 1)$
$\frac{dy}{dx} = \frac{1}{x^2 + 1} \times 2x = \frac{2x}{x^2 + 1}$
Example 3: $y = \ln(\sin x)$
$\frac{dy}{dx} = \frac{1}{\sin x} \times \cos x = \frac{\cos x}{\sin x} = \cot x$
Example 4: $y = (\ln x)^3$
Using chain rule (power of a logarithm):
$\frac{dy}{dx} = 3(\ln x)^2 \times \frac{1}{x} = \frac{3(\ln x)^2}{x}$
$\frac{dy}{dx} = 3(\ln x)^2 \times \frac{1}{x} = \frac{3(\ln x)^2}{x}$
Logarithmic Differentiation:
For complicated products or quotients, taking the natural logarithm first can simplify differentiation:
Example: $y = \frac{x^3(x+1)^2}{\sqrt{x-1}}$
Take natural log of both sides:
$\ln y = 3\ln x + 2\ln(x+1) - \frac{1}{2}\ln(x-1)$
$\ln y = 3\ln x + 2\ln(x+1) - \frac{1}{2}\ln(x-1)$
Differentiate both sides:
$\frac{1}{y}\frac{dy}{dx} = \frac{3}{x} + \frac{2}{x+1} - \frac{1}{2(x-1)}$
$\frac{1}{y}\frac{dy}{dx} = \frac{3}{x} + \frac{2}{x+1} - \frac{1}{2(x-1)}$
$\frac{dy}{dx} = y\left[\frac{3}{x} + \frac{2}{x+1} - \frac{1}{2(x-1)}\right]$
Substitute back $y = \frac{x^3(x+1)^2}{\sqrt{x-1}}$ if needed.
Video lesson
Basics of Logarithmic Derivative
17
Concept
Basics of Exponential Derivatives
Exponential functions, particularly those with base $e$ (Euler's number), have remarkable differentiation properties that make them uniquely important in mathematics.
Basic Exponential Derivative:
- $\frac{d}{dx}(e^x) = e^x$
This is the most remarkable derivative in calculus - the function $e^x$ is its own derivative! This property makes $e^x$ the natural choice for modeling growth and decay.
For Other Bases:
- $\frac{d}{dx}(a^x) = a^x \ln a$ for $a > 0, a \neq 1$
However, in most advanced mathematics, we work with base $e$ because of its simpler derivative.
Chain Rule with Exponential Functions:
For $y = e^{f(x)}$:
$\frac{dy}{dx} = e^{f(x)} \times f'(x)$
The exponential function stays the same; we just multiply by the derivative of the exponent.
Examples:
Example 1: $y = e^{3x}$
$\frac{dy}{dx} = e^{3x} \times 3 = 3e^{3x}$
Example 2: $y = e^{x^2}$
$\frac{dy}{dx} = e^{x^2} \times 2x = 2xe^{x^2}$
Example 3: $y = e^{\sin x}$
$\frac{dy}{dx} = e^{\sin x} \times \cos x = e^{\sin x}\cos x$
Example 4: $y = e^{-x}$
$\frac{dy}{dx} = e^{-x} \times (-1) = -e^{-x}$
This represents exponential decay.
Example 5: $y = x^2 e^x$ (Product Rule)
Using product rule:
$\frac{dy}{dx} = x^2 \cdot e^x + e^x \cdot 2x = e^x(x^2 + 2x) = xe^x(x + 2)$
$\frac{dy}{dx} = x^2 \cdot e^x + e^x \cdot 2x = e^x(x^2 + 2x) = xe^x(x + 2)$
Example 6: $y = \frac{e^x}{x}$ (Quotient Rule)
$\frac{dy}{dx} = \frac{x \cdot e^x - e^x \cdot 1}{x^2} = \frac{e^x(x-1)}{x^2}$
Exponential and Logarithmic Connection:
Since $e^x$ and $\ln x$ are inverse functions:
- $e^{\ln x} = x$ for $x > 0$
- $\ln(e^x) = x$ for all $x$
This relationship is useful for solving equations and simplifying expressions.
Applications:
Exponential functions with derivatives appear in:
- Population growth models: $P(t) = P_0 e^{rt}$
- Radioactive decay: $N(t) = N_0 e^{-\lambda t}$
- Compound interest: $A = Pe^{rt}$
- Natural phenomena involving continuous growth or decay
Video lesson
Basics of Exponential Derivative
Guided practice
Practice Questions with Video Solutions
Attempt each question before opening the solution. Then compare your method with the worked steps and video explanation.
Question 1Video solution
Chain Rule with Polynomial and Fraction
Question
Find the derivative of each of the following:
(a) $y = (7x^3 + x)^4$
(b) $f(x) = \frac{1}{\sqrt{3x^2 - 1}}$
(c) $y = \frac{2}{(3 - \sqrt{x})^3}$
(a) $y = (7x^3 + x)^4$
(b) $f(x) = \frac{1}{\sqrt{3x^2 - 1}}$
(c) $y = \frac{2}{(3 - \sqrt{x})^3}$
Video lesson
Study Guide - Differentiation Question 1
Step-by-step solution
- 1(a) Using chain rule: $y = (7x^3 + x)^4$
- 2Let $u = 7x^3 + x$, then $y = u^4$
- 3$\frac{dy}{du} = 4u^3$ and $\frac{du}{dx} = 21x^2 + 1$
- 4$\frac{dy}{dx} = 4(7x^3 + x)^3(21x^2 + 1)$
- 5(b) Rewrite as $f(x) = (3x^2 - 1)^{-1/2}$
- 6Using chain rule: $f'(x) = -\frac{1}{2}(3x^2 - 1)^{-3/2} \times 6x$
- 7$f'(x) = -3x(3x^2 - 1)^{-3/2} = -\frac{3x}{(3x^2-1)^{3/2}}$
- 8(c) Rewrite as $y = 2(3 - x^{1/2})^{-3}$
- 9Using chain rule: $\frac{dy}{dx} = 2 \times (-3)(3 - x^{1/2})^{-4} \times (-\frac{1}{2}x^{-1/2})$
- 10$= -6(3 - \sqrt{x})^{-4} \times (-\frac{1}{2\sqrt{x}}) = \frac{3}{\sqrt{x}(3-\sqrt{x})^4}$
Final answer
(a) $4(7x^3 + x)^3(21x^2 + 1)$, (b) $-\frac{3x}{(3x^2-1)^{3/2}}$, (c) $\frac{3}{\sqrt{x}(3-\sqrt{x})^4}$
Question 2Video solution
Product Rule Application
Question
Differentiate $\sqrt{x}(x^2 + 2)^5$ with respect to $x$.
Video lesson
Study Guide - Differentiation Question 2
Step-by-step solution
- 1Let $u = \sqrt{x} = x^{1/2}$ and $v = (x^2 + 2)^5$
- 2Using product rule: $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$
- 3$\frac{du}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$
- 4Using chain rule for $v$: $\frac{dv}{dx} = 5(x^2 + 2)^4 \times 2x = 10x(x^2 + 2)^4$
- 5$\frac{dy}{dx} = \sqrt{x} \cdot 10x(x^2 + 2)^4 + (x^2 + 2)^5 \cdot \frac{1}{2\sqrt{x}}$
- 6$= 10x^{3/2}(x^2 + 2)^4 + \frac{(x^2 + 2)^5}{2\sqrt{x}}$
- 7Factor out common terms: $= \frac{(x^2+2)^4}{2\sqrt{x}}[20x^2 + (x^2+2)]$
- 8$= \frac{(x^2+2)^4}{2\sqrt{x}}(21x^2 + 2)$
Final answer
$\frac{(x^2+2)^4(21x^2 + 2)}{2\sqrt{x}}$ or $10x^{3/2}(x^2 + 2)^4 + \frac{(x^2 + 2)^5}{2\sqrt{x}}$
Question 3Video solution
Quotient Rule Application
Question
Find the derivative of $y = \frac{x+1}{\sqrt{3x^2-1}}$.
Video lesson
Study Guide - Differentiation Question 3
Step-by-step solution
- 1Let $u = x + 1$ and $v = \sqrt{3x^2 - 1} = (3x^2 - 1)^{1/2}$
- 2Using quotient rule: $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
- 3$\frac{du}{dx} = 1$
- 4Using chain rule for $v$: $\frac{dv}{dx} = \frac{1}{2}(3x^2-1)^{-1/2} \times 6x = \frac{3x}{\sqrt{3x^2-1}}$
- 5$\frac{dy}{dx} = \frac{\sqrt{3x^2-1} \cdot 1 - (x+1) \cdot \frac{3x}{\sqrt{3x^2-1}}}{3x^2-1}$
- 6Multiply numerator by $\sqrt{3x^2-1}$:
- 7$= \frac{(3x^2-1) - 3x(x+1)}{(3x^2-1)^{3/2}}$
- 8$= \frac{3x^2-1-3x^2-3x}{(3x^2-1)^{3/2}} = \frac{-3x-1}{(3x^2-1)^{3/2}}$
Final answer
$\frac{-3x-1}{(3x^2-1)^{3/2}}$
Question 4Video solution
Finding Gradient at a Point
Question
Find the gradient of the curve $y = (x-1)(2x+3)$ at $x = 2$.
Video lesson
Study Guide - Differentiation Question 4
Step-by-step solution
- 1First, expand the expression: $y = (x-1)(2x+3) = 2x^2 + 3x - 2x - 3 = 2x^2 + x - 3$
- 2Differentiate: $\frac{dy}{dx} = 4x + 1$
- 3Evaluate at $x = 2$:
- 4$\frac{dy}{dx}\Big|_{x=2} = 4(2) + 1 = 9$
Final answer
The gradient at $x = 2$ is $9$
Question 5Video solution
Finding Unknown Constants
Question
The curve $y = ax^2 + bx$ has a gradient of $5$ at the point $(1, 3)$. Find the values of $a$ and $b$.
Video lesson
Study Guide - Differentiation Question 5
Step-by-step solution
- 1Since $(1, 3)$ lies on the curve: $3 = a(1)^2 + b(1) = a + b$ ... (1)
- 2Differentiate: $\frac{dy}{dx} = 2ax + b$
- 3At $x = 1$, gradient is $5$: $5 = 2a(1) + b = 2a + b$ ... (2)
- 4Subtract equation (1) from equation (2):
- 5$(2a + b) - (a + b) = 5 - 3$
- 6$a = 2$
- 7Substitute $a = 2$ into equation (1):
- 8$2 + b = 3$
- 9$b = 1$
Final answer
$a = 2$, $b = 1$
Question 6Video solution
Equations of Tangent and Normal
Question
Find the equation of the tangent and the normal to the curve $y = x^3 + x^2 - 4x - 3$ at the point where $x = -2$.
Video lesson
Study Guide - Differentiation Question 6
Step-by-step solution
- 1Find the $y$-coordinate at $x = -2$:
- 2$y = (-2)^3 + (-2)^2 - 4(-2) - 3 = -8 + 4 + 8 - 3 = 1$
- 3Point is $(-2, 1)$
- 4Find the gradient: $\frac{dy}{dx} = 3x^2 + 2x - 4$
- 5At $x = -2$: $\frac{dy}{dx} = 3(-2)^2 + 2(-2) - 4 = 12 - 4 - 4 = 4$
- 6Tangent equation using $y - y_1 = m(x - x_1)$:
- 7$y - 1 = 4(x - (-2))$
- 8$y - 1 = 4x + 8$
- 9$y = 4x + 9$
- 10Normal gradient = $-\frac{1}{4}$ (negative reciprocal)
- 11Normal equation:
- 12$y - 1 = -\frac{1}{4}(x + 2)$
- 13$y - 1 = -\frac{1}{4}x - \frac{1}{2}$
- 14$y = -\frac{1}{4}x + \frac{1}{2}$
Final answer
Tangent: $y = 4x + 9$; Normal: $y = -\frac{1}{4}x + \frac{1}{2}$
Question 7Video solution
Normal Intersecting Curve Again
Question
Find the equation of the normal to the curve $y = 3x^2 + 5x - 9$ at the point where $x = -2$. Hence, find the $y$-coordinate of the point where this normal meets the curve again.
Video lesson
Study Guide - Differentiation Question 7
Step-by-step solution
- 1At $x = -2$: $y = 3(-2)^2 + 5(-2) - 9 = 12 - 10 - 9 = -7$
- 2Point is $(-2, -7)$
- 3$\frac{dy}{dx} = 6x + 5$
- 4At $x = -2$: $\frac{dy}{dx} = 6(-2) + 5 = -7$
- 5Normal gradient = $\frac{1}{7}$ (negative reciprocal of $-7$)
- 6Normal equation: $y - (-7) = \frac{1}{7}(x - (-2))$
- 7$y + 7 = \frac{1}{7}(x + 2)$
- 8$7y + 49 = x + 2$
- 9$x = 7y + 47$ or $y = \frac{x-47}{7}$
- 10To find intersection with curve, substitute into curve equation:
- 11$y = 3x^2 + 5x - 9$ and $y = \frac{1}{7}x - \frac{45}{7}$
- 12$3x^2 + 5x - 9 = \frac{1}{7}x - \frac{45}{7}$
- 13$21x^2 + 35x - 63 = x - 45$
- 14$21x^2 + 34x - 18 = 0$
- 15Using quadratic formula or factoring: $x = -2$ (known point) or $x = \frac{3}{7}$
- 16At $x = \frac{3}{7}$: $y = 3(\frac{3}{7})^2 + 5(\frac{3}{7}) - 9 = \frac{27}{49} + \frac{15}{7} - 9 = -\frac{390}{49} = -\frac{390}{49}$
Final answer
Normal: $y = \frac{1}{7}x - \frac{45}{7}$; Second intersection: $y = -\frac{390}{49}$ or $y \approx -7.96$
Question 8Video solution
Normal Parallel to a Line
Question
Find the coordinates of point $P$ on the curve $y = 3x^2 - 2x + 1$ for which the normal at $P$ is parallel to the line $y = 2x - 3$.
Video lesson
Study Guide - Differentiation Question 8
Step-by-step solution
- 1The line $y = 2x - 3$ has gradient $2$
- 2For the normal to be parallel, normal gradient = $2$
- 3If normal gradient = $2$, then tangent gradient = $-\frac{1}{2}$ (negative reciprocal)
- 4Find $\frac{dy}{dx}$: $\frac{dy}{dx} = 6x - 2$
- 5Set $\frac{dy}{dx} = -\frac{1}{2}$:
- 6$6x - 2 = -\frac{1}{2}$
- 7$6x = 2 - \frac{1}{2} = \frac{3}{2}$
- 8$x = \frac{1}{4}$
- 9Find $y$-coordinate at $x = \frac{1}{4}$:
- 10$y = 3(\frac{1}{4})^2 - 2(\frac{1}{4}) + 1 = \frac{3}{16} - \frac{1}{2} + 1 = \frac{3}{16} - \frac{8}{16} + \frac{16}{16} = \frac{11}{16}$
Final answer
Point $P$ is at $(\frac{1}{4}, \frac{11}{16})$
Question 9Video solution
Connected Rate of Change - Volume of Hemisphere
Question
The radius, $r$ cm, of a hemisphere is increasing at a constant rate of $0.5$ cm/s. Find the rate of increase of the volume of the hemisphere when $r = 3$ cm.
Video lesson
Study Guide - Differentiation Question 9
Step-by-step solution
- 1Volume of hemisphere: $V = \frac{2}{3}\pi r^3$
- 2Differentiate with respect to time $t$ using chain rule:
- 3$\frac{dV}{dt} = \frac{2}{3}\pi \times 3r^2 \times \frac{dr}{dt} = 2\pi r^2 \frac{dr}{dt}$
- 4Given: $\frac{dr}{dt} = 0.5$ cm/s and $r = 3$ cm
- 5Substitute:
- 6$\frac{dV}{dt} = 2\pi(3)^2(0.5) = 2\pi(9)(0.5) = 9\pi$
- 7$\frac{dV}{dt} = 9\pi \approx 28.3$ cm³/s
Final answer
$9\pi$ cm³/s or approximately $28.3$ cm³/s
Question 10Video solution
Connected Rate of Change with Implicit Equation
Question
The values of $x$ and $y$ are related by the equation $xy = 23x - 8$. If $x$ increases at the rate of $0.03$ unit/s, find the rate of change of $y$ when $y = 21$.
Video lesson
Study Guide - Differentiation Question 10
Step-by-step solution
- 1Given: $xy = 23x - 8$
- 2When $y = 21$: $21x = 23x - 8 \Rightarrow 2x = 8 \Rightarrow x = 4$
- 3Differentiate the equation with respect to $t$ using product rule:
- 4$\frac{d}{dt}(xy) = \frac{d}{dt}(23x - 8)$
- 5$x\frac{dy}{dt} + y\frac{dx}{dt} = 23\frac{dx}{dt}$
- 6Given: $\frac{dx}{dt} = 0.03$ unit/s, $x = 4$, $y = 21$
- 7Substitute:
- 8$4\frac{dy}{dt} + 21(0.03) = 23(0.03)$
- 9$4\frac{dy}{dt} + 0.63 = 0.69$
- 10$4\frac{dy}{dt} = 0.06$
- 11$\frac{dy}{dt} = 0.015$ unit/s
Final answer
$\frac{dy}{dt} = 0.015$ unit/s
Question 11Video solution
Connected Rate of Change with Area
Question
The figure shows part of the curve $y = 2x^2 + 3$. Point $B(x, y)$ moves along the curve for $0 < x < 6$. $C$ is on the $x$-axis such that $BC$ is parallel to the $y$-axis, and $A(6, 0)$ lies on the $x$-axis. Express the area of triangle $ABC$, $T$ units², in terms of $x$, and find $\frac{dT}{dx}$. Given that when $x = 2$, $T$ is increasing at $0.8$ units²/s, find the rate of change of $x$ at this instant.
Video lesson
Study Guide - Differentiation Question 11
Step-by-step solution
- 1Point $B$ is at $(x, 2x^2 + 3)$, point $C$ is at $(x, 0)$, point $A$ is at $(6, 0)$
- 2Base of triangle $AC = 6 - x$
- 3Height of triangle $BC = y = 2x^2 + 3$
- 4Area: $T = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(6-x)(2x^2+3)$
- 5$T = \frac{1}{2}(12x^2 + 18 - 2x^3 - 3x)$
- 6$T = -x^3 + 6x^2 - \frac{3}{2}x + 9$
- 7Find $\frac{dT}{dx}$:
- 8$\frac{dT}{dx} = -3x^2 + 12x - \frac{3}{2}$
- 9Using chain rule: $\frac{dT}{dt} = \frac{dT}{dx} \times \frac{dx}{dt}$
- 10At $x = 2$: $\frac{dT}{dx} = -3(2)^2 + 12(2) - \frac{3}{2} = -12 + 24 - 1.5 = 10.5$
- 11Given $\frac{dT}{dt} = 0.8$ units²/s:
- 12$0.8 = 10.5 \times \frac{dx}{dt}$
- 13$\frac{dx}{dt} = \frac{0.8}{10.5} = \frac{8}{105}$ units/s
Final answer
$T = -x^3 + 6x^2 - \frac{3}{2}x + 9$; $\frac{dT}{dx} = -3x^2 + 12x - \frac{3}{2}$; $\frac{dx}{dt} = \frac{8}{105}$ units/s
Question 12Video solution
Increasing and Decreasing Functions
Question
The equation of a curve is $y = x^3 + 4x^2 + kx + 3$, where $k$ is a constant. Find the set of values of $k$ for which the curve is always an increasing function.
Video lesson
Study Guide - Differentiation Question 12
Step-by-step solution
- 1For the curve to be always increasing, we need $\frac{dy}{dx} \geq 0$ for all $x$
- 2$\frac{dy}{dx} = 3x^2 + 8x + k$
- 3For $3x^2 + 8x + k \geq 0$ for all $x$, the discriminant must be $\leq 0$
- 4(A quadratic $ax^2 + bx + c \geq 0$ for all $x$ when $a > 0$ and $b^2 - 4ac \leq 0$)
- 5Here $a = 3$, $b = 8$, $c = k$
- 6Discriminant: $b^2 - 4ac = 8^2 - 4(3)(k) = 64 - 12k$
- 7For always increasing: $64 - 12k \leq 0$
- 8$64 \leq 12k$
- 9$k \geq \frac{64}{12} = \frac{16}{3}$
Final answer
$k \geq \frac{16}{3}$ or $k \geq 5\frac{1}{3}$
Question 13Video solution
Stationary Points - First Derivative Test
Question
Find the coordinates of the stationary points on the curve $y = x^3 - 3x + 2$. Determine their nature by using the first derivative test. Hence, sketch the curve.
Video lesson
Study Guide - Differentiation Question 13
Step-by-step solution
- 1$\frac{dy}{dx} = 3x^2 - 3$
- 2For stationary points, set $\frac{dy}{dx} = 0$:
- 3$3x^2 - 3 = 0$
- 4$x^2 = 1$
- 5$x = \pm 1$
- 6At $x = 1$: $y = 1^3 - 3(1) + 2 = 0$ → Point $(1, 0)$
- 7At $x = -1$: $y = (-1)^3 - 3(-1) + 2 = 4$ → Point $(-1, 4)$
- 8First derivative test:
- 9Test $x = -2$: $\frac{dy}{dx} = 3(-2)^2 - 3 = 9 > 0$ (increasing)
- 10Test $x = 0$: $\frac{dy}{dx} = 3(0)^2 - 3 = -3 < 0$ (decreasing)
- 11Test $x = 2$: $\frac{dy}{dx} = 3(2)^2 - 3 = 9 > 0$ (increasing)
- 12At $x = -1$: gradient changes from $+$ to $-$ → Maximum point at $(-1, 4)$
- 13At $x = 1$: gradient changes from $-$ to $+$ → Minimum point at $(1, 0)$
Final answer
Maximum at $(-1, 4)$; Minimum at $(1, 0)$
Question 14Video solution
Stationary Points - Second Derivative Test
Question
Find the coordinates of the stationary points on the curve $y = 2x^3 + 3x^2 - 120x + 4$. Determine their nature by using the second derivative test. Hence, sketch the curve.
Video lesson
Study Guide - Differentiation Question 14
Step-by-step solution
- 1$\frac{dy}{dx} = 6x^2 + 6x - 120$
- 2For stationary points, set $\frac{dy}{dx} = 0$:
- 3$6x^2 + 6x - 120 = 0$
- 4$x^2 + x - 20 = 0$
- 5$(x+5)(x-4) = 0$
- 6$x = -5$ or $x = 4$
- 7At $x = -5$: $y = 2(-5)^3 + 3(-5)^2 - 120(-5) + 4 = -250 + 75 + 600 + 4 = 429$
- 8Point: $(-5, 429)$
- 9At $x = 4$: $y = 2(4)^3 + 3(4)^2 - 120(4) + 4 = 128 + 48 - 480 + 4 = -300$
- 10Point: $(4, -300)$
- 11Second derivative test:
- 12$\frac{d^2y}{dx^2} = 12x + 6$
- 13At $x = -5$: $\frac{d^2y}{dx^2} = 12(-5) + 6 = -54 < 0$ → Maximum point at $(-5, 429)$
- 14At $x = 4$: $\frac{d^2y}{dx^2} = 12(4) + 6 = 54 > 0$ → Minimum point at $(4, -300)$
Final answer
Maximum at $(-5, 429)$; Minimum at $(4, -300)$
Revision summary
Key Formulas to Remember
Power Rule
$\frac{d}{dx}(x^n) = nx^{n-1}$
Fundamental rule for differentiating powers of $x$
Chain Rule
$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$
Used for composite functions (function within a function)
Product Rule
$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$
Used when differentiating a product of two functions
Quotient Rule
$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
Used when differentiating a quotient of two functions
Connected Rates
$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
Relates rates of change using the chain rule
Identities and Results to Memorise
- $\frac{d}{dx}(c) = 0$ (constant)
- $\frac{d}{dx}(x) = 1$
- Tangent gradient at point: $m = \frac{dy}{dx}$
- Normal gradient: $m_{normal} = -\frac{1}{m_{tangent}}$
- Stationary point: $\frac{dy}{dx} = 0$
- Maximum: $\frac{d^2y}{dx^2} < 0$
- Minimum: $\frac{d^2y}{dx^2} > 0$
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