Tim Gan Math
O Level (Sec 3 & 4)Additional Mathematics

Differentiation Techniques & Applications for O Level A Math

Master differentiation for O Level & IGCSE Additional Mathematics (4047/0606). Learn chain rule, product rule, quotient rule, tangent/normal equations, and stationary points. Includes 14 practice questions with step-by-step video solutions by Timothy Gan.

Timothy Gan
September 14, 2021
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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.
Table of Contents
Jump to any section to learn at your own pace

Special Functions

Practice Questions with Video Solutions (14 questions)

📐Basic Differentiation Rules

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.

Derivative As Gradient Function

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.

Derivative As Power Functions

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.

Scalar Multiple

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:
$y = x^3 + x^2$ $\frac{dy}{dx} = 3x^2 + 2x$
$y = 5x^4 - 3x^2 + 7x$ $\frac{dy}{dx} = 20x^3 - 6x + 7$
$y = 3x^4 - 5x^2 + 7x - 2$ $\frac{dy}{dx} = 12x^3 - 10x + 7$
Combining the Power Rule, Scalar Multiple Rule, and Addition/Subtraction Rule allows us to differentiate any polynomial function quickly and efficiently.

Addition and Subtraction Rule

🎯Advanced Techniques

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:
1. Identify the outer function and inner function 2. Differentiate the outer function (keeping the inner function unchanged) 3. 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 $\frac{du}{dx} = 6x$
$\frac{dy}{dx} = 4(3x^2 + 5)^3 \times 6x = 24x(3x^2 + 5)^3$
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!

Chain Rule

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)
$= x^2(2x+1)^4(16x + 3)$

Product Rule

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)
$\frac{dy}{dx} = \frac{\sqrt{3x^2-1} \cdot 1 - (x+1) \cdot \frac{3x}{\sqrt{3x^2-1}}}{(3x^2-1)}$
$= \frac{\sqrt{3x^2-1} - \frac{3x(x+1)}{\sqrt{3x^2-1}}}{3x^2-1}$
$= \frac{(3x^2-1) - 3x(x+1)}{(3x^2-1)^{3/2}}$
$= \frac{3x^2-1-3x^2-3x}{(3x^2-1)^{3/2}} = \frac{-3x-1}{(3x^2-1)^{3/2}}$

Quotient Rule

⚙️Applications

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:
1. Find the $y$-coordinate at the given $x$-value 2. Find $\frac{dy}{dx}$ (the gradient function) 3. Evaluate $\frac{dy}{dx}$ at the given $x$-value to get the gradient $m$ 4. Use point-slope form: $y - y_1 = m(x - x_1)$
Key Steps to Find Normal Equation:
1. Follow steps 1-3 above to find the tangent gradient $m$ 2. The normal gradient is $m_{normal} = -\frac{1}{m}$ (negative reciprocal) 3. 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: $y - (-1) = 2(x - 1)$ $y + 1 = 2x - 2$ $y = 2x - 3$
Normal: $m_{normal} = -\frac{1}{2}$ $y + 1 = -\frac{1}{2}(x - 1)$ $y = -\frac{1}{2}x - \frac{1}{2}$

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:
1. You're given a relationship between two variables (e.g., $V = \frac{4}{3}\pi r^3$ for a sphere) 2. You're told how one variable changes with time (e.g., radius increasing at 0.5 cm/s) 3. You need to find how another variable changes with time (e.g., rate of change of volume)
General Steps:
1. Write down the relationship between the variables 2. Differentiate both sides with respect to time $t$ (using chain rule) 3. Substitute known values 4. 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

📊Stationary Points & Optimization

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:
1. Find $\frac{dy}{dx}$ 2. Solve $\frac{dy}{dx} = 0$ to find critical points 3. Test the sign of $\frac{dy}{dx}$ in each interval between critical points 4. 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 $x = 1$.

Increasing and Decreasing Functions

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:
1. Maximum Point: The curve changes from increasing to decreasing 2. Minimum Point: The curve changes from decreasing to increasing 3. Point of Inflection: The gradient is zero but doesn't change sign
Finding Stationary Points:
1. Find $\frac{dy}{dx}$ 2. Set $\frac{dy}{dx} = 0$ and solve for $x$ 3. Substitute $x$-values back into original equation to find $y$-coordinates 4. 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 $(-1, 4)$
Once we've found the stationary points, we need to determine their nature using either the first derivative test or second derivative test.

Stationary Points

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:
1. Find the stationary points by solving $\frac{dy}{dx} = 0$ 2. Choose test points slightly to the left and right of each stationary point 3. Evaluate $\frac{dy}{dx}$ at these test points 4. 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).

First Derivative Test

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:
1. Find the stationary points by solving $\frac{dy}{dx} = 0$ 2. Find the second derivative $\frac{d^2y}{dx^2}$ 3. Evaluate $\frac{d^2y}{dx^2}$ at each stationary point 4. 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 $(1, 0)$
At $x = -1$: $\frac{d^2y}{dx^2} = 6(-1) = -6 < 0$ → 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.

Second Derivative Test

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:
1. Stationary points (where $\frac{dy}{dx} = 0$) 2. Endpoints of the domain 3. Points where the derivative doesn't exist
Finding Absolute Maximum/Minimum on an Interval $a \leq x \leq b$:
1. Find all stationary points by solving $\frac{dy}{dx} = 0$ 2. Evaluate $y$ at: - All stationary points within the interval $a \leq x \leq b$ - Both endpoints $x = a$ and $x = b$ 3. The largest value is the absolute maximum 4. 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 $-2 \leq x \leq 3$)
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:
1. Define the variable and express the quantity to be optimized as a function 2. Find the domain of the problem 3. Find critical points using $\frac{dy}{dx} = 0$ 4. Determine which critical point gives the maximum or minimum using nature tests 5. Check endpoints if the domain is a closed interval

Maximum and Minimum Value

✨Special Functions

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:
1. When differentiating trigonometric functions, the angle must be in radians, not degrees 2. For composite functions like $\sin(3x)$ or $\cos(x^2)$, use the chain rule 3. 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: $\frac{dy}{dx} = \cos(3x) \times 3 = 3\cos(3x)$
Example 2: $y = \cos(x^2 + 1)$
Using chain rule: $\frac{dy}{dx} = -\sin(x^2 + 1) \times 2x = -2x\sin(x^2 + 1)$
Example 3: $y = \tan(2x - 5)$
Using chain rule: $\frac{dy}{dx} = \sec^2(2x - 5) \times 2 = 2\sec^2(2x - 5)$
Example 4: $y = \sin^3 x = (\sin x)^3$
Using chain rule (outer function is power, inner is sin): $\frac{dy}{dx} = 3(\sin x)^2 \times \cos x = 3\sin^2 x \cos x$
Product and Quotient Rules with Trig Functions:
For $y = x \sin x$: $\frac{dy}{dx} = x \cdot \cos x + \sin x \cdot 1 = x\cos x + \sin x$
For $y = \frac{\sin x}{x}$: $\frac{dy}{dx} = \frac{x \cdot \cos x - \sin x \cdot 1}{x^2} = \frac{x\cos x - \sin x}{x^2}$

Basics of Trigonometric Derivative

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:
1. The derivative formula only applies to natural logarithm ($\ln x$, base $e$) 2. For logarithms with other bases, convert to natural logarithm first: $\log_a x = \frac{\ln x}{\ln a}$ 3. 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): $\frac{dy}{dx} = 3(\ln x)^2 \times \frac{1}{x} = \frac{3(\ln x)^2}{x}$
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: $\ln y = 3\ln x + 2\ln(x+1) - \frac{1}{2}\ln(x-1)$
Differentiate both sides: $\frac{1}{y}\frac{dy}{dx} = \frac{3}{x} + \frac{2}{x+1} - \frac{1}{2(x-1)}$
$\frac{dy}{dx} = y\left[\frac{3}{x} + \frac{2}{x+1} - \frac{1}{2(x-1)}\right]$
Substitute back $y = \frac{x^3(x+1)^2}{\sqrt{x-1}}$ if needed.

Basics of Logarithmic Derivative

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: $\frac{dy}{dx} = x^2 \cdot e^x + e^x \cdot 2x = e^x(x^2 + 2x) = xe^x(x + 2)$
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

Basics of Exponential Derivative

Practice Questions with Video Solutions

Watch step-by-step video explanations for each question to master the concepts

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) $y = \frac{2}{(3 - \sqrt{x})^3}$

Video Solution

Step-by-Step Solution

  1. 1
    (a) Using chain rule: $y = (7x^3 + x)^4$
  2. 2
    Let $u = 7x^3 + x$, then $y = u^4$
  3. 3
    $\frac{dy}{du} = 4u^3$ and $\frac{du}{dx} = 21x^2 + 1$
  4. 4
    $\frac{dy}{dx} = 4(7x^3 + x)^3(21x^2 + 1)$
  5. 5
    (b) Rewrite as $f(x) = (3x^2 - 1)^{-1/2}$
  6. 6
    Using chain rule: $f'(x) = -\frac{1}{2}(3x^2 - 1)^{-3/2} \times 6x$
  7. 7
    $f'(x) = -3x(3x^2 - 1)^{-3/2} = -\frac{3x}{(3x^2-1)^{3/2}}$
  8. 8
    (c) Rewrite as $y = 2(3 - x^{1/2})^{-3}$
  9. 9
    Using chain rule: $\frac{dy}{dx} = 2 \times (-3)(3 - x^{1/2})^{-4} \times (-\frac{1}{2}x^{-1/2})$
  10. 10
    $= -6(3 - \sqrt{x})^{-4} \times (-\frac{1}{2\sqrt{x}}) = \frac{3}{\sqrt{x}(3-\sqrt{x})^4}$
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 Solution

Step-by-Step Solution

  1. 1
    Let $u = \sqrt{x} = x^{1/2}$ and $v = (x^2 + 2)^5$
  2. 2
    Using product rule: $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$
  3. 3
    $\frac{du}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$
  4. 4
    Using chain rule for $v$: $\frac{dv}{dx} = 5(x^2 + 2)^4 \times 2x = 10x(x^2 + 2)^4$
  5. 5
    $\frac{dy}{dx} = \sqrt{x} \cdot 10x(x^2 + 2)^4 + (x^2 + 2)^5 \cdot \frac{1}{2\sqrt{x}}$
  6. 6
    $= 10x^{3/2}(x^2 + 2)^4 + \frac{(x^2 + 2)^5}{2\sqrt{x}}$
  7. 7
    Factor out common terms: $= \frac{(x^2+2)^4}{2\sqrt{x}}[20x^2 + (x^2+2)]$
  8. 8
    $= \frac{(x^2+2)^4}{2\sqrt{x}}(21x^2 + 2)$
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 Solution

Step-by-Step Solution

  1. 1
    Let $u = x + 1$ and $v = \sqrt{3x^2 - 1} = (3x^2 - 1)^{1/2}$
  2. 2
    Using quotient rule: $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
  3. 3
    $\frac{du}{dx} = 1$
  4. 4
    Using chain rule for $v$: $\frac{dv}{dx} = \frac{1}{2}(3x^2-1)^{-1/2} \times 6x = \frac{3x}{\sqrt{3x^2-1}}$
  5. 5
    $\frac{dy}{dx} = \frac{\sqrt{3x^2-1} \cdot 1 - (x+1) \cdot \frac{3x}{\sqrt{3x^2-1}}}{3x^2-1}$
  6. 6
    Multiply numerator by $\sqrt{3x^2-1}$:
  7. 7
    $= \frac{(3x^2-1) - 3x(x+1)}{(3x^2-1)^{3/2}}$
  8. 8
    $= \frac{3x^2-1-3x^2-3x}{(3x^2-1)^{3/2}} = \frac{-3x-1}{(3x^2-1)^{3/2}}$
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 Solution

Step-by-Step Solution

  1. 1
    First, expand the expression: $y = (x-1)(2x+3) = 2x^2 + 3x - 2x - 3 = 2x^2 + x - 3$
  2. 2
    Differentiate: $\frac{dy}{dx} = 4x + 1$
  3. 3
    Evaluate at $x = 2$:
  4. 4
    $\frac{dy}{dx}\Big|_{x=2} = 4(2) + 1 = 9$
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 Solution

Step-by-Step Solution

  1. 1
    Since $(1, 3)$ lies on the curve: $3 = a(1)^2 + b(1) = a + b$ ... (1)
  2. 2
    Differentiate: $\frac{dy}{dx} = 2ax + b$
  3. 3
    At $x = 1$, gradient is $5$: $5 = 2a(1) + b = 2a + b$ ... (2)
  4. 4
    Subtract equation (1) from equation (2):
  5. 5
    $(2a + b) - (a + b) = 5 - 3$
  6. 6
    $a = 2$
  7. 7
    Substitute $a = 2$ into equation (1):
  8. 8
    $2 + b = 3$
  9. 9
    $b = 1$
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 Solution

Step-by-Step Solution

  1. 1
    Find the $y$-coordinate at $x = -2$:
  2. 2
    $y = (-2)^3 + (-2)^2 - 4(-2) - 3 = -8 + 4 + 8 - 3 = 1$
  3. 3
    Point is $(-2, 1)$
  4. 4
    Find the gradient: $\frac{dy}{dx} = 3x^2 + 2x - 4$
  5. 5
    At $x = -2$: $\frac{dy}{dx} = 3(-2)^2 + 2(-2) - 4 = 12 - 4 - 4 = 4$
  6. 6
    Tangent equation using $y - y_1 = m(x - x_1)$:
  7. 7
    $y - 1 = 4(x - (-2))$
  8. 8
    $y - 1 = 4x + 8$
  9. 9
    $y = 4x + 9$
  10. 10
    Normal gradient = $-\frac{1}{4}$ (negative reciprocal)
  11. 11
    Normal equation:
  12. 12
    $y - 1 = -\frac{1}{4}(x + 2)$
  13. 13
    $y - 1 = -\frac{1}{4}x - \frac{1}{2}$
  14. 14
    $y = -\frac{1}{4}x + \frac{1}{2}$
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 Solution

Step-by-Step Solution

  1. 1
    At $x = -2$: $y = 3(-2)^2 + 5(-2) - 9 = 12 - 10 - 9 = -7$
  2. 2
    Point is $(-2, -7)$
  3. 3
    $\frac{dy}{dx} = 6x + 5$
  4. 4
    At $x = -2$: $\frac{dy}{dx} = 6(-2) + 5 = -7$
  5. 5
    Normal gradient = $\frac{1}{7}$ (negative reciprocal of $-7$)
  6. 6
    Normal equation: $y - (-7) = \frac{1}{7}(x - (-2))$
  7. 7
    $y + 7 = \frac{1}{7}(x + 2)$
  8. 8
    $7y + 49 = x + 2$
  9. 9
    $x = 7y + 47$ or $y = \frac{x-47}{7}$
  10. 10
    To find intersection with curve, substitute into curve equation:
  11. 11
    $y = 3x^2 + 5x - 9$ and $y = \frac{1}{7}x - \frac{45}{7}$
  12. 12
    $3x^2 + 5x - 9 = \frac{1}{7}x - \frac{45}{7}$
  13. 13
    $21x^2 + 35x - 63 = x - 45$
  14. 14
    $21x^2 + 34x - 18 = 0$
  15. 15
    Using quadratic formula or factoring: $x = -2$ (known point) or $x = \frac{3}{7}$
  16. 16
    At $x = \frac{3}{7}$: $y = 3(\frac{3}{7})^2 + 5(\frac{3}{7}) - 9 = \frac{27}{49} + \frac{15}{7} - 9 = -\frac{390}{49} = -\frac{390}{49}$
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 Solution

Step-by-Step Solution

  1. 1
    The line $y = 2x - 3$ has gradient $2$
  2. 2
    For the normal to be parallel, normal gradient = $2$
  3. 3
    If normal gradient = $2$, then tangent gradient = $-\frac{1}{2}$ (negative reciprocal)
  4. 4
    Find $\frac{dy}{dx}$: $\frac{dy}{dx} = 6x - 2$
  5. 5
    Set $\frac{dy}{dx} = -\frac{1}{2}$:
  6. 6
    $6x - 2 = -\frac{1}{2}$
  7. 7
    $6x = 2 - \frac{1}{2} = \frac{3}{2}$
  8. 8
    $x = \frac{1}{4}$
  9. 9
    Find $y$-coordinate at $x = \frac{1}{4}$:
  10. 10
    $y = 3(\frac{1}{4})^2 - 2(\frac{1}{4}) + 1 = \frac{3}{16} - \frac{1}{2} + 1 = \frac{3}{16} - \frac{8}{16} + \frac{16}{16} = \frac{11}{16}$
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 Solution

Step-by-Step Solution

  1. 1
    Volume of hemisphere: $V = \frac{2}{3}\pi r^3$
  2. 2
    Differentiate with respect to time $t$ using chain rule:
  3. 3
    $\frac{dV}{dt} = \frac{2}{3}\pi \times 3r^2 \times \frac{dr}{dt} = 2\pi r^2 \frac{dr}{dt}$
  4. 4
    Given: $\frac{dr}{dt} = 0.5$ cm/s and $r = 3$ cm
  5. 5
    Substitute:
  6. 6
    $\frac{dV}{dt} = 2\pi(3)^2(0.5) = 2\pi(9)(0.5) = 9\pi$
  7. 7
    $\frac{dV}{dt} = 9\pi \approx 28.3$ cm³/s
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 Solution

Step-by-Step Solution

  1. 1
    Given: $xy = 23x - 8$
  2. 2
    When $y = 21$: $21x = 23x - 8 \Rightarrow 2x = 8 \Rightarrow x = 4$
  3. 3
    Differentiate the equation with respect to $t$ using product rule:
  4. 4
    $\frac{d}{dt}(xy) = \frac{d}{dt}(23x - 8)$
  5. 5
    $x\frac{dy}{dt} + y\frac{dx}{dt} = 23\frac{dx}{dt}$
  6. 6
    Given: $\frac{dx}{dt} = 0.03$ unit/s, $x = 4$, $y = 21$
  7. 7
    Substitute:
  8. 8
    $4\frac{dy}{dt} + 21(0.03) = 23(0.03)$
  9. 9
    $4\frac{dy}{dt} + 0.63 = 0.69$
  10. 10
    $4\frac{dy}{dt} = 0.06$
  11. 11
    $\frac{dy}{dt} = 0.015$ unit/s
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 Solution

Step-by-Step Solution

  1. 1
    Point $B$ is at $(x, 2x^2 + 3)$, point $C$ is at $(x, 0)$, point $A$ is at $(6, 0)$
  2. 2
    Base of triangle $AC = 6 - x$
  3. 3
    Height of triangle $BC = y = 2x^2 + 3$
  4. 4
    Area: $T = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(6-x)(2x^2+3)$
  5. 5
    $T = \frac{1}{2}(12x^2 + 18 - 2x^3 - 3x)$
  6. 6
    $T = -x^3 + 6x^2 - \frac{3}{2}x + 9$
  7. 7
    Find $\frac{dT}{dx}$:
  8. 8
    $\frac{dT}{dx} = -3x^2 + 12x - \frac{3}{2}$
  9. 9
    Using chain rule: $\frac{dT}{dt} = \frac{dT}{dx} \times \frac{dx}{dt}$
  10. 10
    At $x = 2$: $\frac{dT}{dx} = -3(2)^2 + 12(2) - \frac{3}{2} = -12 + 24 - 1.5 = 10.5$
  11. 11
    Given $\frac{dT}{dt} = 0.8$ units²/s:
  12. 12
    $0.8 = 10.5 \times \frac{dx}{dt}$
  13. 13
    $\frac{dx}{dt} = \frac{0.8}{10.5} = \frac{8}{105}$ units/s
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 Solution

Step-by-Step Solution

  1. 1
    For the curve to be always increasing, we need $\frac{dy}{dx} \geq 0$ for all $x$
  2. 2
    $\frac{dy}{dx} = 3x^2 + 8x + k$
  3. 3
    For $3x^2 + 8x + k \geq 0$ for all $x$, the discriminant must be $\leq 0$
  4. 4
    (A quadratic $ax^2 + bx + c \geq 0$ for all $x$ when $a > 0$ and $b^2 - 4ac \leq 0$)
  5. 5
    Here $a = 3$, $b = 8$, $c = k$
  6. 6
    Discriminant: $b^2 - 4ac = 8^2 - 4(3)(k) = 64 - 12k$
  7. 7
    For always increasing: $64 - 12k \leq 0$
  8. 8
    $64 \leq 12k$
  9. 9
    $k \geq \frac{64}{12} = \frac{16}{3}$
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 Solution

Step-by-Step Solution

  1. 1
    $\frac{dy}{dx} = 3x^2 - 3$
  2. 2
    For stationary points, set $\frac{dy}{dx} = 0$:
  3. 3
    $3x^2 - 3 = 0$
  4. 4
    $x^2 = 1$
  5. 5
    $x = \pm 1$
  6. 6
    At $x = 1$: $y = 1^3 - 3(1) + 2 = 0$ → Point $(1, 0)$
  7. 7
    At $x = -1$: $y = (-1)^3 - 3(-1) + 2 = 4$ → Point $(-1, 4)$
  8. 8
    First derivative test:
  9. 9
    Test $x = -2$: $\frac{dy}{dx} = 3(-2)^2 - 3 = 9 > 0$ (increasing)
  10. 10
    Test $x = 0$: $\frac{dy}{dx} = 3(0)^2 - 3 = -3 < 0$ (decreasing)
  11. 11
    Test $x = 2$: $\frac{dy}{dx} = 3(2)^2 - 3 = 9 > 0$ (increasing)
  12. 12
    At $x = -1$: gradient changes from $+$ to $-$ → Maximum point at $(-1, 4)$
  13. 13
    At $x = 1$: gradient changes from $-$ to $+$ → Minimum point at $(1, 0)$
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 Solution

Step-by-Step Solution

  1. 1
    $\frac{dy}{dx} = 6x^2 + 6x - 120$
  2. 2
    For stationary points, set $\frac{dy}{dx} = 0$:
  3. 3
    $6x^2 + 6x - 120 = 0$
  4. 4
    $x^2 + x - 20 = 0$
  5. 5
    $(x+5)(x-4) = 0$
  6. 6
    $x = -5$ or $x = 4$
  7. 7
    At $x = -5$: $y = 2(-5)^3 + 3(-5)^2 - 120(-5) + 4 = -250 + 75 + 600 + 4 = 429$
  8. 8
    Point: $(-5, 429)$
  9. 9
    At $x = 4$: $y = 2(4)^3 + 3(4)^2 - 120(4) + 4 = 128 + 48 - 480 + 4 = -300$
  10. 10
    Point: $(4, -300)$
  11. 11
    Second derivative test:
  12. 12
    $\frac{d^2y}{dx^2} = 12x + 6$
  13. 13
    At $x = -5$: $\frac{d^2y}{dx^2} = 12(-5) + 6 = -54 < 0$ → Maximum point at $(-5, 429)$
  14. 14
    At $x = 4$: $\frac{d^2y}{dx^2} = 12(4) + 6 = 54 > 0$ → Minimum point at $(4, -300)$
Answer:
Maximum at $(-5, 429)$; Minimum at $(4, -300)$

Key Formulas to Remember

Power Rule
$\frac{d}{dx}(x^n) = nx^{n-1}$

Note: Fundamental rule for differentiating powers of $x$

Chain Rule
$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

Note: Used for composite functions (function within a function)

Product Rule
$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$

Note: Used when differentiating a product of two functions

Quotient Rule
$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Note: Used when differentiating a quotient of two functions

Connected Rates
$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$

Note: Relates rates of change using the chain rule

Special Identities to Memorize
  • $\frac{d}{dx}(c) = 0$ (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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