Differentiating from first principles is powerful, but tedious. What if there were shortcuts? In this lesson, you will learn the three fundamental rules — power, constant multiple, and sum/difference — that let you differentiate polynomials in seconds instead of minutes.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Consider the function $f(x) = x^3 + 2x$. If you had to find its derivative using first principles, you would need to expand $(x+h)^3$, simplify, cancel $h$, and take a limit. That takes several minutes. Do you think there might be a faster way to differentiate terms like $x^3$ and $2x$ individually? What pattern do you notice from derivatives you have already calculated?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: (a + b)² = a² + b².
Right: (a + b)² = a² + 2ab + b²; the middle term 2ab is essential and commonly forgotten.
📚 Core Content
If $f(x) = x^n$, then:
$$f'(x) = nx^{n-1}$$
This works for any real number $n$ — positive integers, fractions, negatives, and zero. You simply bring the power down to the front and subtract one from the power.
Constants slide straight past the derivative:
$$\frac{d}{dx}\bigl(cf(x)\bigr) = c \cdot f'(x)$$
For example: $\frac{d}{dx}(7x^3) = 7 \cdot 3x^2 = 21x^2$.
You can differentiate term by term:
$$\frac{d}{dx}\bigl(f(x) \pm g(x)\bigr) = f'(x) \pm g'(x)$$
For example: if $y = 4x^3 - 2x^2 + 5x - 7$, then:
$$\frac{dy}{dx} = 12x^2 - 4x + 5$$
Notice that the derivative of the constant $-7$ is $0$.
Once you have the derivative, finding the equation of a tangent line is straightforward.
At $x = a$, the gradient of the tangent is $m = f'(a)$. Using point–gradient form with the point $(a, f(a))$:
$$y - f(a) = f'(a)(x - a)$$
The normal is perpendicular to the tangent. If the tangent has gradient $m$, the normal has gradient $-\frac{1}{m}$ (provided $m \neq 0$):
$$y - f(a) = -\frac{1}{f'(a)}(x - a)$$
🧮 Worked Examples
🧪 Activities
1 $f(x) = x^5 - 3x^2 + 4x - 1$
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2 $y = 4x^3 + \frac{2}{x}$
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3 $f(x) = \sqrt{x} + x^2$
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4 Find the equation of the tangent to $y = x^2 - 4x$ at $x = 3$.
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1 Explain why $\frac{d}{dx}(x^3 + x^2) = \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2)$. What property of differentiation makes this valid?
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2 A student differentiates $f(x) = \frac{1}{x^2}$ and gets $f'(x) = -\frac{1}{x}$. Identify the error and give the correct derivative.
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Earlier you thought about whether there might be a faster way to differentiate $f(x) = x^3 + 2x$ than using first principles.
There is: using the power rule, we differentiate each term separately. The derivative of $x^3$ is $3x^2$, and the derivative of $2x$ is $2$. So $f'(x) = 3x^2 + 2$ — a process that takes seconds rather than minutes. This is why the differentiation rules are so powerful: they preserve the rigour of calculus while making computation practical.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. (a) Find $f'(x)$ for $f(x) = 2x^4 - 3x^2 + 5x - 2$. 2 MARKS
(b) Find the equation of the tangent to $y = 2x^4 - 3x^2 + 5x - 2$ at $x = 1$. 2 MARKS
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7. A curve has equation $y = ax^2 + bx + c$. The tangent to the curve at $x = 1$ has gradient 1, and the curve passes through the point $(1, 5)$. The normal at $x = 1$ intersects the $y$-axis at $(0, 6)$. Find the values of $a$, $b$, and $c$. 4 MARKS
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8. The population of bacteria in a dish is modelled by $P(t) = 500 + 20t^2$, where $t$ is in hours. 3 MARKS
(a) Find the rate of change of the population at $t = 3$ hours.
(b) Explain what this value means in the context of the problem.
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1. $f'(x) = 5x^4 - 6x + 4$
2. $y = 4x^3 + 2x^{-1}$, so $\frac{dy}{dx} = 12x^2 - 2x^{-2} = 12x^2 - \frac{2}{x^2}$
3. $f(x) = x^{1/2} + x^2$, so $f'(x) = \frac{1}{2}x^{-1/2} + 2x = \frac{1}{2\sqrt{x}} + 2x$
4. $\frac{dy}{dx} = 2x - 4$, so at $x = 3$, $m = 2$. Point: $(3, -3)$. Tangent: $y + 3 = 2(x - 3)$, so $y = 2x - 9$.
1. This works because differentiation is a linear operation. The derivative of a sum is the sum of the derivatives (sum rule).
2. The student forgot to reduce the power by 1. Since $f(x) = x^{-2}$, the correct derivative is $f'(x) = -2x^{-3} = -\frac{2}{x^3}$.
1. B — Using the power rule: $\frac{d}{dx}(x^4) = 4x^3$.
2. A — $\frac{d}{dx}(3x^2 + 2x) = 6x + 2$.
3. A — Rewrite as $x^{1/2}$, then $\frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.
4. B — Rewrite as $x^{-1}$, then $\frac{d}{dx}(x^{-1}) = -x^{-2} = -\frac{1}{x^2}$.
5. B — At $x = 2$, $y = 4$ and $\frac{dy}{dx} = 4$, so the normal has gradient $-\frac{1}{4}$. Equation: $y - 4 = -\frac{1}{4}(x - 2)$, which simplifies to $y = -\frac{1}{4}x + \frac{9}{2}$.
6. C — The derivative of any constant is zero.
7. B — Rewrite as $3x^{-2}$, then $\frac{dy}{dx} = -6x^{-3} = -\frac{6}{x^3}$.
Q6 (4 marks): (a) $f'(x) = 8x^3 - 6x + 5$ [1]. (b) At $x = 1$, $y = 2(1)^4 - 3(1)^2 + 5(1) - 2 = 2$ [1]. The gradient is $m = f'(1) = 8 - 6 + 5 = 7$ [1]. Using point–gradient form: $y - 2 = 7(x - 1)$, so $y = 7x - 5$ [1].
Q7 (4 marks): First, $\frac{dy}{dx} = 2ax + b$ [1]. The tangent at $x = 1$ has gradient $1$, so $2a + b = 1$ [1]. The curve passes through $(1, 5)$, giving $a + b + c = 5$ [1]. The normal at $x = 1$ passes through $(1, 5)$ and $(0, 6)$, so its gradient is $\frac{6-5}{0-1} = -1$, confirming the tangent gradient is $1$. Solving simultaneously gives $a = 1$, $b = -1$, $c = 5$ [1].
Q8 (3 marks): (a) $P'(t) = 40t$, so at $t = 3$, $P'(3) = 120$ bacteria per hour [2]. (b) After 3 hours, the population is increasing at a rate of 120 bacteria per hour [1].
Differentiation Rules
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