Master the foundations of calculus and learn how to find instantaneous rates of change using differentiation from first principles.
Calculus is one of the most powerful branches of mathematics, with applications in physics, engineering, economics, biology, and many other fields. At its core, calculus deals with two fundamental concepts:
Finding rates of change and gradients of curves at specific points. This is what we'll focus on in this lesson.
Finding areas under curves and accumulation of quantities over intervals (covered in separate lessons).
Real-world applications you encounter every day:
The key word in all these examples is "instantaneous". In Grade 10 and 11, you learned how to calculate the average gradient between two points on a curve. But what if we want to know the gradient at exactly one point? That's where calculus comes in!
Let's build our understanding by comparing what you already know with what we're about to learn.
The average gradient between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ on a curve is calculated using:
This gives you the gradient of the secant line (the straight line joining two points on the curve).
Example:
For $f(x) = x^2$, the average gradient between $x = 1$ and $x = 3$ is:
The instantaneous gradient is the gradient at exactly one point on the curve. It's the gradient of the tangent line to the curve at that point.
The Big Question:
How do we find the gradient at a single point when the formula $\frac{\Delta y}{\Delta x}$ requires two different points?
The Answer: We use limits!
We start with two points, but we make them get closer and closer together until they're infinitely close. The average gradient approaches the instantaneous gradient as the distance between the points approaches zero.
Imagine a curve $y = f(x)$ and a point $A$ on it at $x = a$. Now pick another point $B$ at $x = a + h$ where $h$ is a small distance.
Step 1: Average gradient from A to B
Step 2: Make h smaller and smaller
As $h \to 0$, point B gets closer to point A
Step 3: The limit gives the instantaneous gradient
For $f(x) = x^2$ at $x = 3$, we know $f(3) = 9$. As $h$ gets smaller, watch how the average gradient approaches $2x = 6$:
| h | x + h | f(x+h) = (x+h)² | f(x+h) - f(x) | Average Gradient | Getting closer to... |
|---|---|---|---|---|---|
| 2 | 5 | 25 | 16 | 8 | 6 |
| 1 | 4 | 16 | 7 | 7 | 6 |
| 0.5 | 3.5 | 12.25 | 3.25 | 6.5 | 6 |
| 0.1 | 3.1 | 9.61 | 0.61 | 6.1 | 6 |
| 0.01 | 3.01 | 9.0601 | 0.0601 | 6.01 | 6 |
| 0 (limit) | 3 | 9 | 0 | 6 | Instantaneous! |
Notice how as $h$ gets smaller, the average gradient gets closer and closer to 6. The limit tells us that when $h = 0$ (the two points become one), the gradient is exactly 6. This is the instantaneous gradient at $x = 3$!
What you're seeing:
The secant line connects points A (at x = 3) and B (at x = 3 + 2.0 = 5.0).
Gradient Calculation:
Gradient = rise ÷ run = Δy ÷ h = 0.400
Now we're ready for the formal definition of the derivative, which is the foundation of all differentiation.
The derivative of a function $f(x)$ at a point $x$ is defined as:
This is also written as: $\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
What does this mean in words?
To find the derivative (instantaneous rate of change) at any point $x$, we:
Many students confuse $f(x+h)$ with $f(x) + h$. These are completely different!
❌ WRONG: $f(x+h) = f(x) + h$
Example: If $f(x) = x^2$ and $x = 3$, $h = 2$:
$f(3+2) = f(3) + 2 = 9 + 2 = 11$ ❌
But $f(5) = 25$, not 11!
✅ CORRECT: $f(x+h) = f(x+h)$
Example: If $f(x) = x^2$ and $x = 3$, $h = 2$:
$f(3+2) = f(5) = 5^2 = 25$ ✅
Or: $f(3+2) = (3+2)^2 = 25$ ✅
💡 Remember:
$f(x+h)$ means: "substitute $(x+h)$ everywhere you see $x$ in the function"
For $f(x) = x^2 + 3x$, we get: $f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h$
Replace the function notation with the actual expressions
Expand, combine like terms, and factor if necessary
Factor out $h$ from the numerator so you can cancel it (this removes the $\frac{0}{0}$ problem)
Substitute $h = 0$ in the simplified expression
True derivative: f'(2) = 4
Error: 1.000
Observe:
The derivative can be written in several different ways. They all mean the same thing!
Read as "dee y dee x" or "the derivative of y with respect to x"
Read as "f prime of x" or "y prime"
Read as "D sub x of f"
Used when differentiating with respect to time
CAPS Examinations typically use:
$f(x) = 3x + 5$
$f(x+h) = 3(x+h) + 5 = 3x + 3h + 5$
Answer: $f'(x) = 3$
This makes sense! The gradient of the line $y = 3x + 5$ is 3 everywhere (it's a straight line).
$f(x) = x^2$
$f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$
$$f'(x) = \lim_{h \to 0} \frac{h(2x + h)}{h}$$
$$f'(x) = \lim_{h \to 0} (2x + h)$$
Answer: $f'(x) = 2x$
The gradient of $y = x^2$ at any point $x$ is $2x$. For example, at $x = 3$, the gradient is $2(3) = 6$.
$f(x) = 2x^2 + 3x - 1$
$f(x+h) = 2(x+h)^2 + 3(x+h) - 1$
$= 2(x^2 + 2xh + h^2) + 3x + 3h - 1$
$= 2x^2 + 4xh + 2h^2 + 3x + 3h - 1$
$$= \lim_{h \to 0} \frac{2x^2 + 4xh + 2h^2 + 3x + 3h - 1 - 2x^2 - 3x + 1}{h}$$
$$= \lim_{h \to 0} \frac{4xh + 2h^2 + 3h}{h}$$
$$= \lim_{h \to 0} \frac{h(4x + 2h + 3)}{h}$$
$$= \lim_{h \to 0} (4x + 2h + 3)$$
Answer: $f'(x) = 4x + 3$
Notice a pattern emerging? The derivative of $ax^2 + bx + c$ is $2ax + b$. You'll learn shortcuts for this later!
See the relationship between a function and its derivative. The derivative tells us the gradient (slope) of the original function at each point.
Key Observations:
Forgetting to expand $(x+h)^2$ or $(x+h)^3$ correctly
Always use binomial expansion: $(x+h)^2 = x^2 + 2xh + h^2$, not $x^2 + h^2$
Not factoring out $h$ from the numerator
You must factor out $h$ to cancel with the denominator, or you'll get $\frac{0}{0}$ when evaluating the limit
Sign errors when subtracting $f(x)$
Remember: $-(3x + 5) = -3x - 5$, not $-3x + 5$. Distribute the negative sign carefully!
Evaluating the limit too early
Only substitute $h = 0$ AFTER you've successfully canceled $h$ from numerator and denominator
Confusing $f(x+h)$ with $f(x) + h$
$f(x+h)$ means substitute $(x+h)$ everywhere you see $x$ in the function, not just add $h$ at the end
Work through these problems to master differentiation from first principles. Show all your steps and check your answers.
Test your understanding with these multiple-choice questions. Select your answer and check to see if you're correct.
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