Learn powerful shortcuts that make differentiation fast and easy. No more lengthy first principles calculations!
Prerequisites: Before starting this lesson, make sure you understand differentiation from first principles. These rules are derived from first principles!
In the previous lesson, you learned how to differentiate functions from first principles. While this method is fundamental and helps you understand what differentiation really means, it has a significant drawback: it's time-consuming and prone to errors.
Pros:
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Requirement:
Good News!
Mathematicians have used first principles to derive a set of rules that work for all standard functions. Once you learn these rules, you can differentiate most functions in seconds!
Use First Principles When:
Use Differentiation Rules When:
Before we formally state the rules, let's discover them ourselves! Look at these derivatives that were calculated using first principles and see if you can identify any patterns.
| Function: $f(x)$ | Derivative: $f'(x)$ |
|---|---|
| $x$ | $1$ |
| $-4x$ | $-4$ |
| $x^2$ | $2x$ |
| $3x^2$ | $6x$ |
| $-x^3$ | $-3x^2$ |
| $2x^3$ | $6x^2$ |
| $\frac{1}{x}$ or $x^{-1}$ | $-\frac{1}{x^2}$ or $-x^{-2}$ |
| $\sqrt{x}$ or $x^{1/2}$ | $\frac{1}{2\sqrt{x}}$ or $\frac{1}{2}x^{-1/2}$ |
These five rules will allow you to differentiate almost any polynomial or rational function quickly and accurately.
For any real number $n$:
In words:
"Multiply by the power, then reduce the power by 1"
Examples:
For any constant $k$:
Why is this true?
A constant function like $f(x) = 5$ is a horizontal line. Horizontal lines have a gradient of zero, so the derivative is zero!
Examples:
For any constant $k$ and function $f(x)$:
In words:
"Constants can be moved outside the derivative"
Differentiate the function part, then multiply by the constant.
Examples:
For any functions $f(x)$ and $g(x)$:
In words:
"The derivative of a sum equals the sum of the derivatives"
Differentiate each term separately, then add them up.
Examples:
$f(x) = x^3 + x^2$
$f'(x) = 3x^2 + 2x$
$y = 5x^4 + 3x^2 + 7$
$\frac{dy}{dx} = 20x^3 + 6x + 0 = 20x^3 + 6x$
$g(x) = x^5 + 2x^3 + 4x + 1$
$g'(x) = 5x^4 + 6x^2 + 4$
For any functions $f(x)$ and $g(x)$:
In words:
"The derivative of a difference equals the difference of the derivatives"
Just like the sum rule, but with subtraction!
Examples:
$f(x) = x^4 - x^2$
$f'(x) = 4x^3 - 2x$
$y = 3x^5 - 7x^3 + 2x$
$\frac{dy}{dx} = 15x^4 - 21x^2 + 2$
$g(x) = 10x^2 - 5x + 8$
$g'(x) = 20x - 5$
In practice, you'll often use multiple rules at once. The key is to apply them systematically, one term at a time.
Example: Differentiate $f(x) = 4x^5 - 3x^2 + 7x - 2$
Apply the rules to each term:
Term 1: $4x^5 \to 4 \cdot 5x^4 = 20x^4$ (constant multiple + power rule)
Term 2: $-3x^2 \to -3 \cdot 2x = -6x$ (constant multiple + power rule)
Term 3: $7x \to 7 \cdot 1 = 7$ (constant multiple + power rule)
Term 4: $-2 \to 0$ (constant rule)
Therefore: $f'(x) = 20x^4 - 6x + 7$
Use the Constant Multiple Rule and the Power Rule:
$\frac{dy}{dx} = 3 \cdot \frac{d}{dx}[x^2]$
$= 3 \cdot 2x$
$= 6x$
Answer: $\frac{dy}{dx} = 6x$
Compare this to first principles - it took us just one line instead of 5-6 steps!
The power rule works for negative exponents too!
$f'(x) = 16 \cdot (-2)x^{-2-1}$
$= -32x^{-3}$
$= -\frac{32}{x^3}$
Remember: Always write your final answer with positive exponents (unless the question specifies otherwise).
Differentiate each term separately:
$f'(x) = \frac{d}{dx}[x^4] - \frac{d}{dx}[6x^2] - \frac{d}{dx}[1]$
$= 4x^3 - 6(2x) - 0$
$= 4x^3 - 12x$
The power rule works for fractional exponents too!
$\frac{dy}{dx} = 3 \cdot \frac{3}{2}x^{3/2 - 1} - 4 \cdot \frac{1}{2}x^{1/2 - 1} + 0$
$= \frac{9}{2}x^{1/2} - 2x^{-1/2}$
$= \frac{9}{2}\sqrt{x} - \frac{2}{\sqrt{x}}$
You can leave your answer in exponential form or convert back to square root form - both are correct!
These examples require you to simplify or expand the function before differentiating.
Important: We haven't learned the product rule yet, so we must expand first!
$y = \frac{1}{4}(x - 1)(12 + x^2)$
$= \frac{1}{4}(12x + x^3 - 12 - x^2)$
$= \frac{1}{4}(x^3 - x^2 + 12x - 12)$
$= \frac{1}{4}x^3 - \frac{1}{4}x^2 + 3x - 3$
$\frac{dy}{dx} = \frac{1}{4}(3x^2) - \frac{1}{4}(2x) + 3(1) - 0$
$= \frac{3}{4}x^2 - \frac{1}{2}x + 3$
Answer: $\frac{dy}{dx} = \frac{3}{4}x^2 - \frac{1}{2}x + 3$
$f(y) = \sqrt{y} = y^{1/2}$
$f'(y) = \frac{1}{2}y^{1/2 - 1}$
$= \frac{1}{2}y^{-1/2}$
$= \frac{1}{2\sqrt{y}}$
Key Strategy: Always rewrite roots and fractions as powers before differentiating!
$f(z) = (z - 1)(z + 1) = z^2 - 1$
$f'(z) = 2z - 0$
$= 2z$
Important: We haven't learned the quotient rule yet, so we must simplify first!
$(t + 2)^3 = (t + 2)(t + 2)(t + 2)$
$= (t^2 + 4t + 4)(t + 2)$
$= t^3 + 2t^2 + 4t^2 + 8t + 4t + 8$
$= t^3 + 6t^2 + 12t + 8$
$p(t) = \frac{t^3 + 6t^2 + 12t + 8}{\sqrt{t}}$
$= \frac{t^3}{t^{1/2}} + \frac{6t^2}{t^{1/2}} + \frac{12t}{t^{1/2}} + \frac{8}{t^{1/2}}$
$= t^{3 - 1/2} + 6t^{2 - 1/2} + 12t^{1 - 1/2} + 8t^{-1/2}$
$= t^{5/2} + 6t^{3/2} + 12t^{1/2} + 8t^{-1/2}$
$\frac{dp}{dt} = \frac{5}{2}t^{3/2} + 6 \cdot \frac{3}{2}t^{1/2} + 12 \cdot \frac{1}{2}t^{-1/2} + 8 \cdot (-\frac{1}{2})t^{-3/2}$
$= \frac{5}{2}t^{3/2} + 9t^{1/2} + 6t^{-1/2} - 4t^{-3/2}$
$\frac{dp}{dt} = \frac{5}{2}t\sqrt{t} + 9\sqrt{t} + \frac{6}{\sqrt{t}} - \frac{4}{t\sqrt{t}}$
1. Power Rule:
$\frac{d}{dx}[x^n] = nx^{n-1}$
2. Constant Rule:
$\frac{d}{dx}[k] = 0$
3. Constant Multiple:
$\frac{d}{dx}[kf(x)] = kf'(x)$
4. Sum Rule:
$\frac{d}{dx}[f + g] = f' + g'$
5. Difference Rule:
$\frac{d}{dx}[f - g] = f' - g'$
$\frac{1}{x^2} = x^{-2}$
$\frac{1}{x^3} = x^{-3}$
$\sqrt{x} = x^{1/2}$
$\sqrt[3]{x} = x^{1/3}$
$\frac{1}{\sqrt{x}} = x^{-1/2}$
Forgetting to reduce the power by 1
Wrong: $\frac{d}{dx}[x^3] = 3x^3$ Correct: $\frac{d}{dx}[x^3] = 3x^2$
Forgetting to multiply by the power
Wrong: $\frac{d}{dx}[x^4] = x^3$ Correct: $\frac{d}{dx}[x^4] = 4x^3$
Treating constants as variables
Wrong: $\frac{d}{dx}[5] = 5$ Correct: $\frac{d}{dx}[5] = 0$
Not simplifying before differentiating
For $(x+2)(x-3)$, you must expand to $x^2 - x - 6$ before differentiating (until you learn the product rule)
Forgetting the negative sign with negative exponents
Wrong: $\frac{d}{dx}[x^{-2}] = 2x^{-3}$ Correct: $\frac{d}{dx}[x^{-2}] = -2x^{-3}$
Not converting roots to exponential form first
Always write $\sqrt{x}$ as $x^{1/2}$ before differentiating
Apply the differentiation rules to solve these problems. Remember: use the rules, not first principles!
Test your understanding of the differentiation rules with these multiple-choice questions.
Continue your calculus journey with the chain rule, product rule, and more advanced differentiation techniques.
CAPS Aligned
Grade 12 Mathematics