The Logic of the “Moment”#
In Grade 10 and 11, you calculated the Average Gradient between two separate points ($m = \frac{y_2-y_1}{x_2-x_1}$).
In Calculus, we want to know the gradient at one single point.
The “Zooming In” Analogy#
Imagine you are looking at a curve on a screen.
- If you pick two points, you can see the slope between them.
- Now, imagine “zooming in” until those two points are so close they look like one single point.
- The distance between them ($h$) becomes almost zero. This is a Limit.
1. The Power of Limits#
A limit describes what a function is “approaching” as $x$ gets closer to a certain value, even if the function doesn’t exist at that exact spot (like $\frac{0}{0}$).
Rule: Always factorize first to remove the “zero” from the denominator before you find the limit.
Example: Evaluating a Limit#
Find $\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$
Direct substitution gives $\frac{0}{0}$ — undefined! But we can factorize:
$$ \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4 $$2. First Principles — The Definition of the Derivative#
This is the formal way we calculate the derivative using the limit logic:
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$The 4-Step Method#
- Write $f(x+h)$: Plug $(x+h)$ into the original equation wherever there’s an $x$.
- Subtract $f(x)$: Calculate $f(x+h) - f(x)$.
- Divide by $h$: Factor out $h$ from every term in the numerator and cancel.
- Apply the limit: Only NOW let $h = 0$ in whatever remains.
⚠️ THE #1 RULE: Never Drop the Limit#
You MUST write $\lim_{h \to 0}$ at the start of every single line of your working until the very last step where you finally substitute $h = 0$. The limit symbol is your “licence” to have $h$ in the expression — without it, you’re dividing by zero. Dropping it early loses marks every time.
3. Worked Examples#
Example 1: $f(x) = x^2$ — Full Solution with Correct Notation#
Notice how $\lim_{h \to 0}$ appears on every line until the final substitution:
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$$$ = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} $$$$ = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} $$$$ = \lim_{h \to 0} \frac{2xh + h^2}{h} $$$$ = \lim_{h \to 0} \frac{h(2x + h)}{h} $$$$ = \lim_{h \to 0} (2x + h) $$NOW — and only now — we substitute $h = 0$:
$$ = 2x + 0 $$$$\boxed{f'(x) = 2x}$$Why this matters: Every line above still has $h$ in it. The $\lim_{h \to 0}$ tells the reader “we haven’t set $h = 0$ yet — we’re still simplifying.” Only after we’ve cancelled the $h$ from the denominator is it safe to finally let $h = 0$. This is the entire point of limits.
Example 2: $f(x) = 3x^2 - 5x + 2$#
First, prepare: $f(x+h) = 3(x+h)^2 - 5(x+h) + 2 = 3x^2 + 6xh + 3h^2 - 5x - 5h + 2$
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$$$ = \lim_{h \to 0} \frac{(3x^2 + 6xh + 3h^2 - 5x - 5h + 2) - (3x^2 - 5x + 2)}{h} $$$$ = \lim_{h \to 0} \frac{6xh + 3h^2 - 5h}{h} $$$$ = \lim_{h \to 0} \frac{h(6x + 3h - 5)}{h} $$$$ = \lim_{h \to 0} (6x + 3h - 5) $$Now substitute $h = 0$:
$$\boxed{f'(x) = 6x - 5}$$Example 3: $f(x) = \frac{1}{x}$#
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$$$ = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} $$$$ = \lim_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h} = \lim_{h \to 0} \frac{-h}{h \cdot x(x+h)} $$$$ = \lim_{h \to 0} \frac{-1}{x(x+h)} $$Now substitute $h = 0$:
$$\boxed{f'(x) = -\frac{1}{x^2}}$$Example 4: $f(x) = x^3$ (The challenging one)#
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$$$ = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h} $$$$ = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} $$$$ = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} $$$$ = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h} $$$$ = \lim_{h \to 0} (3x^2 + 3xh + h^2) $$Key expansion: $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$. Memorize this!
Now substitute $h = 0$:
$$\boxed{f'(x) = 3x^2}$$Example 5: First Principles at a Specific Point#
Find the gradient of $f(x) = x^2 + 3x$ at $x = 2$.
Method: Use the “point” form of first principles:
$$ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} $$$f(2) = 4 + 6 = 10$
$f(2+h) = (2+h)^2 + 3(2+h) = 4 + 4h + h^2 + 6 + 3h = 10 + 7h + h^2$
$$ f'(2) = \lim_{h \to 0} \frac{(10 + 7h + h^2) - 10}{h} = \lim_{h \to 0} \frac{7h + h^2}{h} = \lim_{h \to 0}(7 + h) = 7 $$The gradient at $x = 2$ is 7.
4. The Key Algebraic Expansions#
You will need these repeatedly:
| Expression | Expansion |
|---|---|
| $(x+h)^2$ | $x^2 + 2xh + h^2$ |
| $(x+h)^3$ | $x^3 + 3x^2h + 3xh^2 + h^3$ |
| $\frac{1}{x+h} - \frac{1}{x}$ | $\frac{-h}{x(x+h)}$ |
🚨 Common Mistakes#
- Dropping $\lim_{h \to 0}$ too early: You must write $\lim_{h \to 0}$ in every line until the very last step when you actually substitute $h = 0$. Dropping it loses marks.
- Expansion errors with $(x+h)^2$: Students often write $x^2 + h^2$ and forget the $2xh$ middle term. This ruins the entire calculation.
- Setting $h = 0$ before cancelling: Never substitute $h = 0$ while $h$ is still in the denominator. You must factor out and cancel the $h$ first.
- Forgetting the negative sign: In $f(x+h) - f(x)$, every term of $f(x)$ must be subtracted. Use brackets: $-(3x^2 - 5x + 2) = -3x^2 + 5x - 2$.
💡 Pro Tip: The “Cheat-Code” Check#
Before starting a First Principles question, spend 2 seconds using the Power Rule (the shortcut from the next lesson) to find the answer. Now you know exactly what your final line should look like — this helps you catch errors mid-calculation!
🏠 Back to Calculus | ⏭️ Power Rule