Derivative of a function - Mathematics M1 (Calculus and Statistics) - HKDSE

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M1 Calculus: Study Notes on the Derivative of a Function

Hey everyone! Welcome to your study notes for one of the most important topics in M1 Calculus: The Derivative. Don't worry if the name sounds scary. We're going to break it down into simple, easy-to-understand pieces.

So, what will you learn? We'll start with the idea of a 'limit', which is the foundation. Then, we'll discover what a derivative actually is (spoiler: it's all about the rate of change!) and learn the powerful rules that let us find derivatives easily. Think of this as learning the secret to finding the exact speed of a car at one instant, or the precise slope of a rollercoaster at any point on the track. It's a super useful tool!

Part 1: The Foundation - Understanding Limits

Before we can run, we need to learn to walk. In calculus, 'limits' are our first steps. The entire concept of the derivative is built on them.

So, what is a Limit?

Imagine you're walking towards a wall. You take steps that are always half the remaining distance. You get closer and closer, but you never actually touch the wall. The wall is your limit. It's the value a function "approaches" as the input gets closer and closer to a certain number.

We write the limit of a function f(x) as x approaches a number c like this:

$$ \lim_{x \to c} f(x) = L $$

This just means "As x gets really, really close to c, the value of f(x) gets really, really close to L".

How to Find Limits (The Easy Way)

For most functions you'll see (like polynomials), the easiest way to find a limit is by direct substitution. Just plug the number x is approaching into the function!

Example: Find $$ \lim_{x \to 2} (x^2 + 3x) $$

Step 1: Just substitute x = 2 into the expression.
Step 2: Calculate the result.
$$ (2)^2 + 3(2) = 4 + 6 = 10 $$

So, $$ \lim_{x \to 2} (x^2 + 3x) = 10 $$. Simple, right?

Watch Out! The 0/0 Problem

Sometimes, direct substitution gives you $$ \frac{0}{0} $$, which is called an indeterminate form. This is a red flag! It doesn't mean the limit doesn't exist; it just means we need to do some algebra first.

Example: Find $$ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} $$

If we substitute x = 3, we get $$ \frac{3^2 - 9}{3 - 3} = \frac{0}{0} $$. Time for algebra!

Step 1: Factor the expression. Notice that the numerator is a difference of squares.
$$ \frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} $$ Step 2: Cancel the common factor.
$$ \frac{\cancel{(x-3)}(x+3)}{\cancel{x-3}} = x+3 $$ Step 3: Now, try direct substitution again.
$$ \lim_{x \to 3} (x+3) = 3 + 3 = 6 $$

So, the limit is 6!

Key Takeaway: Limits

A limit is the value a function approaches. For most problems, you can find it with direct substitution. If you get $$ \frac{0}{0} $$, use your algebra skills (like factoring) to simplify the expression first.

Part 2: Introducing the Derivative

Now for the main event! What is this 'derivative' thing all about?

The Derivative is the Slope!

Imagine a winding, hilly road. Its steepness changes everywhere. The derivative is like a magic formula that tells you the exact steepness (slope) of the road at any single point you choose.

In math terms, the derivative of a function f(x) at a point gives the slope of the tangent line to the graph at that point. A tangent line is a straight line that just "touches" the curve at one point.

The derivative measures the instantaneous rate of change.

If f(x) is the position of a car, its derivative f'(x) is its instantaneous velocity.
If C(x) is the cost to produce items, its derivative C'(x) is the marginal cost (the cost to produce one more item).

The Formal Idea: Derivative from First Principles

The official definition of a derivative uses limits. It looks a bit complicated, but the idea is simple. We find the slope between two points that are super close together, and we see what happens as the distance between them approaches zero.

The definition is:

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

Don't Panic! The HKDSE syllabus says you only need to recognise this concept. You will NOT be asked to calculate derivatives using this formula. We have much easier rules for that, which we'll learn next!

Notation, Notation, Notation!

There are a few different ways to write "the derivative of a function". They all mean the same thing, so get familiar with them!

If our function is y = f(x), its derivative can be written as:

f'(x) (read as "f prime of x")
y' (read as "y prime")
$$ \frac{dy}{dx} $$ (read as "d y d x")

To show the derivative at a specific point, like x = 2, we write:

f'(2)
$$ \left. \frac{dy}{dx} \right|_{x=2} $$

Key Takeaway: The Derivative

The derivative gives you the slope of the tangent line at any point on a curve. It represents the instantaneous rate of change. There are several notations for it, like f'(x) and $$ \frac{dy}{dx} $$.

Part 3: Your Differentiation Toolkit (The Rules!)

This is where the real fun begins! These rules are your shortcuts to finding derivatives without using the complicated limit definition.

The Basic Rules (The Essentials)

1. The Constant Rule

The derivative of any constant number is zero.

$$ \frac{d}{dx}(C) = 0 $$

Analogy: Think of the graph of y = 5. It's a flat horizontal line. What's its slope? Zero!
Example: If f(x) = 10, then f'(x) = 0.

2. The Power Rule

This is your most-used rule! To find the derivative of x raised to a power, you bring the power down to the front and then subtract one from the original power.

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

Memory Aid: "Bring the power down, then take one away."

Example 1: Find the derivative of $$ y = x^4 $$
$$ \frac{dy}{dx} = 4x^{4-1} = 4x^3 $$

Example 2: Find the derivative of $$ f(x) = \sqrt{x} $$
First, rewrite the square root as a power: $$ f(x) = x^{1/2} $$
Now, use the Power Rule: $$ f'(x) = \frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2}x^{-1/2} $$

3. The Constant Multiple and Sum/Difference Rules

These rules are very intuitive.

Constant Multiple: You can pull constants out to the front. $$ \frac{d}{dx}(k \cdot f(x)) = k \cdot f'(x) $$
Sum/Difference: You can differentiate a function term by term. $$ \frac{d}{dx}(f(x) \pm g(x)) = f'(x) \pm g'(x) $$

Example: Find the derivative of $$ y = 5x^3 - 2x^2 + 7 $$
Differentiate each term separately:
$$ \frac{dy}{dx} = 5 \cdot (3x^2) - 2 \cdot (2x^1) + 0 $$ $$ \frac{dy}{dx} = 15x^2 - 4x $$

The 'Big Three' Advanced Rules

These rules are for more complex functions that are multiplied, divided, or composed together.

1. The Product Rule

Used when you have two functions multiplied together, like $$ f(x) = u(x) \cdot v(x) $$.

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

Memory Aid: "First times derivative of the second, PLUS second times derivative of the first."

Example: Find the derivative of $$ y = x^2 \ln(x) $$
Here, u = x² and v = ln(x).
So, $$ \frac{du}{dx} = 2x $$ and $$ \frac{dv}{dx} = \frac{1}{x} $$
Plug into the formula:
$$ \frac{dy}{dx} = (x^2) \cdot \left(\frac{1}{x}\right) + (\ln x) \cdot (2x) $$ $$ \frac{dy}{dx} = x + 2x \ln x $$

Common Mistake Alert! You CANNOT just multiply the derivatives. $$ (uv)' \neq u'v' $$

2. The Quotient Rule

Used when you have one function divided by another, like $$ f(x) = \frac{u(x)}{v(x)} $$.

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

Memory Aid (sing it!): "Low d(High) minus High d(Low), over the square of what's below!"

Example: Find the derivative of $$ y = \frac{e^x}{x^3} $$
Here, u = eˣ (High) and v = x³ (Low).
So, $$ \frac{du}{dx} = e^x $$ and $$ \frac{dv}{dx} = 3x^2 $$.
Plug into the formula:
$$ \frac{dy}{dx} = \frac{(x^3)(e^x) - (e^x)(3x^2)}{(x^3)^2} $$ $$ \frac{dy}{dx} = \frac{x^3e^x - 3x^2e^x}{x^6} = \frac{x^2e^x(x - 3)}{x^6} = \frac{e^x(x - 3)}{x^4} $$

3. The Chain Rule

Used for a "function inside a function" (a composite function). Think of it like a Russian nesting doll.

If $$ y = f(g(x)) $$, then its derivative is:

$$ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) $$

The process: "Differentiate the outer layer (leaving the inside alone), then multiply by the derivative of the inner layer."

Example: Find the derivative of $$ y = (x^2 + 5)^4 $$
Outer layer: $$ (\text{something})^4 $$
Inner layer: $$ x^2 + 5 $$

Step 1: Differentiate the outer layer. Use the Power Rule. Bring the 4 down, subtract 1, but keep the inside the same.
$$ 4(x^2 + 5)^3 $$ Step 2: Differentiate the inner layer. The derivative of $$ x^2 + 5 $$ is $$ 2x $$.
Step 3: Multiply them together.
$$ \frac{dy}{dx} = 4(x^2+5)^3 \cdot (2x) = 8x(x^2+5)^3 $$

Derivatives of Special Functions

These are formulas you just need to memorise. They often appear with the Chain Rule!

Natural Exponential: $$ \frac{d}{dx}(e^x) = e^x $$ (It's its own derivative!)
Natural Logarithm: $$ \frac{d}{dx}(\ln x) = \frac{1}{x} $$
General Exponential: $$ \frac{d}{dx}(a^x) = a^x \ln a $$
General Logarithm: $$ \frac{d}{dx}(\log_a x) = \frac{1}{x \ln a} $$

Did you know? The number 'e' (approx 2.718) is special in calculus precisely because the function eˣ has a slope equal to its value at every point!

Key Takeaway: Differentiation Rules

These rules are your essential toolkit. Power Rule for $$ x^n $$. Product Rule for multiplication. Quotient Rule for division. Chain Rule for functions inside functions. Memorise these and practice, practice, practice!

Part 4: Going One Step Further - The Second Derivative

What if we take the derivative... of the derivative? That's exactly what the second derivative is!

What is the Second Derivative?

The second derivative measures the rate of change of the first derivative. If the first derivative tells you the slope, the second derivative tells you how the slope is changing.

Real-world Analogy:

Function f(x): Your position.
First Derivative f'(x): Your velocity (how fast your position is changing).
Second Derivative f''(x): Your acceleration (how fast your velocity is changing).

The second derivative also helps us understand the concavity of a graph (whether it's shaped like a cup holding water or spilling water), which is very useful for finding maximum and minimum points.

Notation and Calculation

The notation for the second derivative simply adds another 'prime' or a '2'.

f''(x) (read as "f double-prime of x")
y'' (read as "y double-prime")
$$ \frac{d^2y}{dx^2} $$ (read as "d two y d x squared")

To find it, you just differentiate your function twice!

Example: Find the second derivative of $$ y = 2x^3 - 4x^2 + 10x $$

Step 1: Find the first derivative ($$ \frac{dy}{dx} $$).
$$ \frac{dy}{dx} = 2(3x^2) - 4(2x) + 10 = 6x^2 - 8x + 10 $$

Step 2: Differentiate the first derivative to get the second derivative ($$ \frac{d^2y}{dx^2} $$).
$$ \frac{d^2y}{dx^2} = 6(2x) - 8 + 0 = 12x - 8 $$

And that's it! You've found the second derivative.

Good news: For the M1 syllabus, you do not need to worry about third, fourth, or any higher derivatives!

Key Takeaway: Second Derivative

The second derivative is the "derivative of the derivative". It tells you how the slope is changing (think: acceleration). To find it, just apply your differentiation rules twice.