# The Rules of Differentiation

> Power rule, sums, product, quotient and chain rules, and the derivatives of eˣ, ln x, sin x, cos x and tan x.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/differentiation-rules/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The shortcut through the park
Every day, Marcus walks around the edge of the park to get to school. One day his friend shows him a path straight through the middle. Same destination, half the time.

First principles always works, but it is the long way around. Mathematicians spotted patterns in the answers and turned them into **rules**. With the rules, finding a derivative takes seconds. This chapter is that shortcut.

**In short:** The rules of differentiation find derivatives without first principles. The power rule says $\dfrac{d}{dx}x^n = nx^{n-1}$, and the product, quotient and chain rules handle functions that are multiplied, divided or put inside each other.

## Starting point: Years 9 and 10

The rules work on **powers of x**, so first rewrite everything as a power. This uses the index laws.

| Written as | Rewrite as a power |
|---|---|
| $\sqrt{x}$ | $x^{\frac{1}{2}}$ |
| $\dfrac{1}{x}$ | $x^{-1}$ |
| $\dfrac{1}{x^3}$ | $x^{-3}$ |
| $\dfrac{5}{\sqrt{x}}$ | $5x^{-\frac{1}{2}}$ |
| $x\sqrt{x}$ | $x^{\frac{3}{2}}$ |
| $\dfrac{x^2 + x}{x}$ | $x + 1$ |

You also need two facts about straight lines:
- The gradient of $y = 3x + 2$ is 3, everywhere.
- The gradient of $y = 7$ (a horizontal line) is 0.

These two facts are already derivatives: $\dfrac{d}{dx}(3x + 2) = 3$ and $\dfrac{d}{dx}(7) = 0$.

### Worked example: Get it ready
Rewrite $y = \dfrac{4}{x^2} - 2\sqrt{x} + \dfrac{x^3}{2}$ as powers.

$$y = 4x^{-2} - 2x^{\frac{1}{2}} + \frac{1}{2}x^3$$

### Watch out: A root on the bottom
$\dfrac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$, not $x^{\frac{1}{2}}$ or $x^{-2}$.

**Mind map**

```text
Differentiation rules
  Power rule
    Bring down, take one off
    Constants vanish
  Combining
    Sum: term by term
    Product: u'v + uv'
    Quotient: (u'v - uv')/v²
    Chain: outside times inside
  Special functions
    eˣ stays eˣ
    ln x gives 1/x
    sin gives cos
    cos gives -sin
    tan gives sec²
```

### Year 11: The power rule
$$\frac{d}{dx}x^n = nx^{n-1}$$

Bring the power down in front, then take one off the power. It works for any $n$: whole, negative or fraction.

- $\dfrac{d}{dx}x^5 = 5x^4$
- $\dfrac{d}{dx}x^{-2} = -2x^{-3}$
- $\dfrac{d}{dx}x^{\frac{1}{2}} = \dfrac{1}{2}x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}$

**Constants and sums.** A number in front stays: $\dfrac{d}{dx}(4x^3) = 12x^2$. A number on its own disappears. Differentiate a sum term by term.

$$\frac{d}{dx}\left(4x^3 - 5x + 9\right) = 12x^2 - 5$$

### Year 11: Product, quotient and chain rules
**Product rule.** If $y = uv$, then

$$\frac{dy}{dx} = u'v + uv'$$

$y = x^2(3x - 1)^4$: here $u = x^2$, $v = (3x - 1)^4$. You also need the chain rule for $v$.

**Chain rule.** For a function inside a function, differentiate the outside, keep the inside, then multiply by the derivative of the inside.

$$\frac{d}{dx}\big(f(x)\big)^n = n\big(f(x)\big)^{n-1} f'(x)$$

So $\dfrac{d}{dx}(3x - 1)^4 = 4(3x - 1)^3 \times 3 = 12(3x - 1)^3$.

**Quotient rule.** If $y = \dfrac{u}{v}$, then

$$\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$$

$y = \dfrac{x}{x + 1}$: $u' = 1$, $v' = 1$, so $\dfrac{dy}{dx} = \dfrac{(x + 1) - x}{(x + 1)^2} = \dfrac{1}{(x + 1)^2}$.

**Exponentials.** In Year 11, $\dfrac{d}{dx}e^x = e^x$.

### Year 12: Derivatives of special functions
| $f(x)$ | $f'(x)$ |
|---|---|
| $e^x$ | $e^x$ |
| $e^{f(x)}$ | $f'(x)e^{f(x)}$ |
| $\ln x$ | $\dfrac{1}{x}$ |
| $\ln f(x)$ | $\dfrac{f'(x)}{f(x)}$ |
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
| $a^x$ | $a^x \ln a$ |

With the chain rule: $\dfrac{d}{dx}\sin(3x) = 3\cos(3x)$ and $\dfrac{d}{dx}\cos(x^2) = -2x\sin(x^2)$.

**Combining rules.** $y = x^2 e^{-x}$ needs the product rule:

$$\frac{dy}{dx} = 2xe^{-x} + x^2(-e^{-x}) = xe^{-x}(2 - x)$$

Angles must be in radians for the trig derivatives.

**Use log laws first.** $y = \ln\dfrac{x^3}{x + 2} = 3\ln x - \ln(x + 2)$, so $\dfrac{dy}{dx} = \dfrac{3}{x} - \dfrac{1}{x + 2}$. Much easier than the quotient rule inside a log.

**Logs of any base.** $\log_a x = \dfrac{\ln x}{\ln a}$, so $\dfrac{d}{dx}\log_a x = \dfrac{1}{x\ln a}$. The tangent to $y = \log_2 x$ at $x = 2$ has gradient $\dfrac{1}{2\ln 2}$.

**The chain rule with function notation.** If $y = f(3x - 1)$, then $\dfrac{dy}{dx} = 3f'(3x - 1)$. If $g(x) = [f(x)]^2$, then $g'(x) = 2f(x)f'(x)$. So if $f(1) = 4$ and $f'(1) = -2$, then $g'(1) = 2(4)(-2) = -16$.

**The second derivative.** Differentiating twice gives $f''(x)$, also written $\dfrac{d^2y}{dx^2}$. For $y = e^{2x}$, $y' = 2e^{2x}$ and $y'' = 4e^{2x}$. It measures how the gradient is changing, and is used in curve sketching.

**A useful link.** $\dfrac{d}{dx}\tan x = \sec^2 x = 1 + \tan^2 x$, so $y = \tan x$ satisfies $\dfrac{dy}{dx} = 1 + y^2$.

## Quick summary

- Rewrite roots and fractions as powers of $x$ before differentiating.
- Power rule: $\dfrac{d}{dx}x^n = nx^{n-1}$. Constants on their own vanish.
- Product: $u'v + uv'$. Quotient: $\dfrac{u'v - uv'}{v^2}$. Chain: outside derivative times inside derivative.
- $e^x \to e^x$, $\ln x \to \dfrac{1}{x}$, $\sin x \to \cos x$, $\cos x \to -\sin x$, $\tan x \to \sec^2 x$.

## Quiz

Try these questions from the chapter quiz.

1. Write $\sqrt{x}$ as a power of $x$.
   - A. $x^2$
   - B. $x^{\frac{1}{2}}$
   - C. $x^{-\frac{1}{2}}$
   - D. $x^{-2}$
   - Answer: B. $x^{\frac{1}{2}}$. A square root is the power one half.

2. Write $\dfrac{1}{x^4}$ as a power of $x$.
   - A. $x^4$
   - B. $x^{\frac{1}{4}}$
   - C. $x^{-4}$
   - D. $-4x$
   - Answer: C. $x^{-4}$. A power on the bottom becomes negative.

3. Write $\dfrac{3}{\sqrt{x}}$ as a power of $x$.
   - A. $3x^{-\frac{1}{2}}$
   - B. $3x^{\frac{1}{2}}$
   - C. $\dfrac{1}{3}x^{-\frac{1}{2}}$
   - D. $3x^{-2}$
   - Answer: A. $3x^{-\frac{1}{2}}$. $\sqrt{x} = x^{\frac{1}{2}}$ on the bottom becomes $x^{-\frac{1}{2}}$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/differentiation-rules/#quiz