Rules for Differentiation Revision Notes for VCE SSCE Mathematical Methods

Rules for Differentiation

The power rule

When we differentiate using first principles, we discover a pattern. Looking at some simple examples:

For $f(x) = x$ , we get $f'(x) = 1$
For $f(x) = x^2$ , we get $f'(x) = 2x$
For $f(x) = x^3$ , we get $f'(x) = 3x^2$

This pattern reveals a fundamental relationship between the power and its derivative, leading us to one of the most powerful tools in calculus.

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Notice the pattern: in each case, the power becomes the coefficient, and the power itself decreases by 1. This consistent behavior is what makes the power rule so useful.

This pattern leads us to the power rule, one of the most important rules in differentiation:

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The Power Rule

For $f(x) = x^n$ , we have $f'(x) = nx^{n-1}$ , where $n = 1, 2, 3, ...$

In other words, to differentiate $x$ raised to a power:

Bring the power down as a coefficient
Reduce the power by 1

This result can be proven using the binomial theorem, but you don't need to memorize the proof for your exam.

Rules for differentiating polynomials

The following rules are essential tools when finding derivatives of polynomial functions. Understanding these will make differentiation much easier and allow you to tackle more complex problems with confidence.

Constant function rule

If $f(x) = c$ (where $c$ is a constant), then $f'(x) = 0$ .

This makes sense because a constant function is a horizontal line, which has zero gradient.

Multiple rule

If $f(x) = k \cdot g(x)$ (where $k$ is a constant), then $f'(x) = k \cdot g'(x)$ .

The derivative of a number multiple is simply the multiple of the derivative. Constants can be factored out of the differentiation process.

Example: If $f(x) = 5x^2$ , then $f'(x) = 5(2x) = 10x$

Sum rule

If $f(x) = g(x) + h(x)$ , then $f'(x) = g'(x) + h'(x)$ .

The derivative of a sum equals the sum of the derivatives. This means we can differentiate each term separately and then add the results.

Example: If $f(x) = x^2 + 2x$ , then $f'(x) = 2x + 2$

Difference rule

If $f(x) = g(x) - h(x)$ , then $f'(x) = g'(x) - h'(x)$ .

The derivative of a difference equals the difference of the derivatives. Like the sum rule, we can differentiate each term separately.

Example: If $f(x) = x^2 - 2x$ , then $f'(x) = 2x - 2$

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The process of finding the derivative function is called differentiation. These rules work together to let us differentiate any polynomial function quickly and efficiently.

Worked examples

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Worked Example: Differentiating a Polynomial

Differentiate $x^5 - 2x^3 + 2$ with respect to $x$ .

Solution:

Let $f(x) = x^5 - 2x^3 + 2$

Then $f'(x) = 5x^4 - 2(3x^2) + 0$

f'(x) = 5x^4 - 6x^2

Explanation: We apply the power rule to each term. The derivative of $x^5$ is $5x^4$ , the derivative of $-2x^3$ is $-6x^2$ (using the multiple rule), and the derivative of the constant 2 is 0.

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Worked Example: Finding a Specific Derivative Value

Find the derivative of $f(x) = 3x^3 - 6x^2 + 1$ and thus find $f'(1)$ .

Solution:

Let $f(x) = 3x^3 - 6x^2 + 1$

Then $f'(x) = 3(3x^2) - 6(2x) + 0$

f'(x) = 9x^2 - 12x

Therefore $f'(1) = 9 - 12 = -3$

Alternative notation systems

There are different ways to write derivatives. It's important to recognize and use both systems, as different textbooks and exams may use different notations. Being fluent in multiple notation systems will help you understand mathematics from various sources.

Leibniz notation

An alternative way to denote derivatives uses the symbol $\frac{dy}{dx}$ .

If $y = x^3$ , then the derivative can be written as $\frac{dy}{dx} = 3x^2$

In general, if $y$ is a function of $x$ , then the derivative of $y$ with respect to $x$ is denoted by $\frac{dy}{dx}$ .

Similarly, if $z$ is a function of $t$ , then the derivative of $z$ with respect to $t$ is denoted $\frac{dz}{dt}$ .

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Important Warning About Leibniz Notation

In Leibniz notation, the symbol $d$ is not a factor and cannot be cancelled. The notation $\frac{dy}{dx}$ represents a single entity—the derivative—not a fraction.

Do not treat $\frac{dy}{dx}$ as if it were a fraction with numerator $dy$ and denominator $dx$ .

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Historical Context

The notation originated in the eighteenth century when mathematicians used $\delta x$ to mean a small difference in $x$ and $\delta y$ to mean a small difference in $y$ , where $\delta$ (delta) is the lowercase Greek letter d. This historical connection helps explain why we use $d$ in the derivative notation.

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Worked Example: Using Leibniz Notation

a) If $y = t^2$ , find $\frac{dy}{dt}$

b) If $x = t^3 + t$ , find $\frac{dx}{dt}$

c) If $z = \frac{1}{3}x^3 + x^2$ , find $\frac{dz}{dx}$

Solution:

a) $y = t^2$

\frac{dy}{dt} = 2t

b) $x = t^3 + t$

\frac{dx}{dt} = 3t^2 + 1

c) $z = \frac{1}{3}x^3 + x^2$

\frac{dz}{dx} = x^2 + 2x

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Worked Example: Simplifying Before Differentiating

a) For $y = (x + 3)^2$ , find $\frac{dy}{dx}$

b) For $z = (2t - 1)^2(t + 2)$ , find $\frac{dz}{dt}$

c) For $y = \frac{x^2 + 3x}{x}$ , find $\frac{dy}{dx}$

d) Differentiate $y = 2x^3 - 1$ with respect to $x$

Solution:

a) First expand: $y = (x + 3)^2 = x^2 + 6x + 9$

Therefore $\frac{dy}{dx} = 2x + 6$

b) Expand the expression:

z = (4t^2 - 4t + 1)(t + 2)

z = 4t^3 - 4t^2 + t + 8t^2 - 8t + 2

z = 4t^3 + 4t^2 - 7t + 2

Therefore $\frac{dz}{dt} = 12t^2 + 8t - 7$

c) First simplify: $y = \frac{x^2 + 3x}{x} = x + 3$ (for $x \neq 0$ )

Therefore $\frac{dy}{dx} = 1$ (for $x \neq 0$ )

d) $y = 2x^3 - 1$

Therefore $\frac{dy}{dx} = 6x^2$

Operator notation

There's another way to express "find the derivative with respect to $x$ ". This notation emphasizes the operation of differentiation itself.

Instead of writing "find the derivative of $2x^2 - 4x$ with respect to $x$ ", we can write "find $\frac{d}{dx}(2x^2 - 4x)$ ".

In general: $\frac{d}{dx}(f(x)) = f'(x)$

Here, $\frac{d}{dx}$ acts as an operator that tells us to differentiate whatever follows it with respect to $x$ .

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Worked Example: Using Operator Notation

Find:

a) $\frac{d}{dx}(5x - 4x^3)$

b) $\frac{d}{dz}(5z^2 - 4z)$

c) $\frac{d}{dz}(6z^3 - 4z^2)$

Solution:

a) $\frac{d}{dx}(5x - 4x^3) = 5 - 12x^2$

b) $\frac{d}{dz}(5z^2 - 4z) = 10z - 4$

c) $\frac{d}{dz}(6z^3 - 4z^2) = 18z^2 - 8z$

Finding the gradient of a tangent line

Remember that the tangent line to the graph of function $f$ at point $(a, f(a))$ is the line through that point with gradient $f'(a)$ .

This connection between derivatives and tangent lines is fundamental to calculus. The derivative at a point gives us the exact slope of the tangent line at that point.

This means we can find the gradient of any tangent line by:

Finding the derivative $f'(x)$
Substituting the $x$ -coordinate of the point into $f'(x)$

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Worked Example: Finding the Gradient of a Tangent

For the curve determined by the rule $f(x) = 3x^3 - 6x^2 + 1$ , find the gradient of the tangent line to the curve at the point $(1, -2)$ .

Solution:

First find the derivative: $f'(x) = 9x^2 - 12x$

Then evaluate at $x = 1$ : $f'(1) = 9 - 12 = -3$

The gradient of the tangent line at the point $(1, -2)$ is $-3$ .

Finding coordinates from gradient conditions

Sometimes we need to find points on a curve where the tangent has a specific gradient. This is the reverse problem—instead of finding the gradient at a given point, we know the gradient and need to find the point(s).

This involves:

Finding $f'(x)$
Setting $f'(x)$ equal to the required gradient
Solving for $x$
Finding the corresponding $y$ -coordinate

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Worked Example: Finding Points with Given Gradient

For each curve, find the coordinates of points where the gradient of the tangent line has the given value:

a) $y = x^3$ , gradient = 8

b) $y = x^2 - 4x + 2$ , gradient = 0

c) $y = 4 - x^3$ , gradient = $-6$

Solution:

a) $y = x^3$ implies $\frac{dy}{dx} = 3x^2$

Set $3x^2 = 8$

x^2 = \frac{8}{3}

x = \pm\sqrt{\frac{8}{3}} = \pm\frac{2\sqrt{6}}{3}

When $x = \frac{2\sqrt{6}}{3}$ : $y = \left(\frac{2\sqrt{6}}{3}\right)^3 = \frac{16\sqrt{6}}{9}$

When $x = -\frac{2\sqrt{6}}{3}$ : $y = \left(-\frac{2\sqrt{6}}{3}\right)^3 = -\frac{16\sqrt{6}}{9}$

The points are $\left(\frac{2\sqrt{6}}{3}, \frac{16\sqrt{6}}{9}\right)$ and $\left(-\frac{2\sqrt{6}}{3}, -\frac{16\sqrt{6}}{9}\right)$

b) $y = x^2 - 4x + 2$ implies $\frac{dy}{dx} = 2x - 4$

Set $2x - 4 = 0$

x = 2

When $x = 2$ : $y = 4 - 8 + 2 = -2$

The only point is $(2, -2)$

c) $y = 4 - x^3$ implies $\frac{dy}{dx} = -3x^2$

Set $-3x^2 = -6$

x^2 = 2

x = \pm\sqrt{2}

When $x = \sqrt{2}$ : $y = 4 - (\sqrt{2})^3 = 4 - 2\sqrt{2}$

When $x = -\sqrt{2}$ : $y = 4 - (-\sqrt{2})^3 = 4 + 2\sqrt{2}$

The points are $(\sqrt{2}, 4 - 2\sqrt{2})$ and $(-\sqrt{2}, 4 + 2\sqrt{2})$

Angles and gradients

The gradient of a curve at a point is the gradient of the tangent at that point. Each point on a curve has an associated tangent line, and we can describe that tangent using either its gradient or the angle it makes with the horizontal axis.

If $\alpha$ is the angle a straight line makes with the positive direction of the $x$ -axis, then the gradient $m$ of the line equals $\tan \alpha$ . That is:

m = \tan \alpha

Example: If $\alpha = 135°$ , then $\tan \alpha = -1$ , so the gradient is $-1$ .

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Worked Example: Using Angles to Find Points

Find the coordinates of points on the curve $y = x^2 - 7x + 8$ at which the tangent line:

a) makes an angle of $45°$ with the positive direction of the $x$ -axis

b) is parallel to the line $y = -2x + 6$

Solution:

First find the derivative: $\frac{dy}{dx} = 2x - 7$

a) Since the angle is $45°$ , the gradient is $\tan 45° = 1$

Set $2x - 7 = 1$

2x = 8

x = 4

When $x = 4$ : $y = 16 - 28 + 8 = -4$

The coordinates are $(4, -4)$

b) The line $y = -2x + 6$ has gradient $-2$

Set $2x - 7 = -2$

2x = 5

x = \frac{5}{2}

When $x = \frac{5}{2}$ : $y = \frac{25}{4} - \frac{35}{2} + 8 = \frac{25 - 70 + 32}{4} = -\frac{13}{4}$

The coordinates are $\left(\frac{5}{2}, -\frac{13}{4}\right)$

Increasing and decreasing functions

A function's behavior—whether it's going up or down—is directly related to the sign of its derivative. This connection gives us a powerful tool for analyzing functions without having to draw their graphs.

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Key Definitions

A function $f$ is strictly increasing on an interval if $x_2 > x_1$ implies $f(x_2) > f(x_1)$
A function $f$ is strictly decreasing on an interval if $x_2 > x_1$ implies $f(x_2) < f(x_1)$

The derivative provides a direct test for whether a function is increasing or decreasing on an interval.

Important results:

If $f'(x) > 0$ for all $x$ in an interval, then the function is strictly increasing on that interval. Think of the tangents at each point—they each have positive gradient, meaning the function is rising.

If $f'(x) < 0$ for all $x$ in an interval, then the function is strictly decreasing on that interval. Think of the tangents at each point—they each have negative gradient, meaning the function is falling.

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Special Case

A function can be strictly increasing even if the derivative equals zero at some points. For example, $f(x) = x^3$ is strictly increasing everywhere, but $f'(0) = 0$ . What matters is that the derivative is non-negative throughout the interval and positive on most of it.

Sign of the derivative

The gradient of a tangent can be positive, negative, or zero, and this tells us important information about the function's behavior at each point.

Looking at a typical graph, we can identify several features based on the tangent lines:

Where the tangent slopes upward (positive gradient): $g'(x) > 0$
Where the tangent is horizontal (zero gradient): $g'(x) = 0$
Where the tangent slopes downward (negative gradient): $g'(x) < 0$

At points where the gradient is zero, the function often has a local maximum or minimum (turning point).

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Worked Example: Identifying Where Derivatives are Positive, Negative, or Zero

For the graph of $f: \mathbb{R} \to \mathbb{R}$ , find:

a) $\{x : f'(x) > 0\}$ (where the derivative is positive)

b) $\{x : f'(x) < 0\}$ (where the derivative is negative)

c) $\{x : f'(x) = 0\}$ (where the derivative is zero)

Solution:

a) $\{x : f'(x) > 0\} = \{x : -1 < x < 5\} = (-1, 5)$

The function is increasing (has positive gradient) between $x = -1$ and $x = 5$ .

b) $\{x : f'(x) < 0\} = \{x : x < -1\} \cup \{x : x > 5\} = (-\infty, -1) \cup (5, \infty)$

The function is decreasing (has negative gradient) before $x = -1$ and after $x = 5$ .

c) $\{x : f'(x) = 0\} = \{-1, 5\}$

The gradient is zero at the turning points $x = -1$ and $x = 5$ .

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Key Points to Remember:

Power rule: For $f(x) = x^n$ , we have $f'(x) = nx^{n-1}$ — bring the power down and reduce by 1
Basic rules: The derivative of a constant is 0; the derivative of a sum is the sum of derivatives; the derivative of a difference is the difference of derivatives; constants can be factored out
Notation: You can write derivatives as $f'(x)$ , $\frac{dy}{dx}$ , or $\frac{d}{dx}(f(x))$ — they all mean the same thing
Tangent gradients: The gradient of the tangent at point $(a, f(a))$ is found by evaluating $f'(a)$
Sign matters: If $f'(x) > 0$ , the function is increasing; if $f'(x) < 0$ , the function is decreasing; if $f'(x) = 0$ , there may be a turning point

Rules for Differentiation (VCE SSCE Mathematical Methods): Revision Notes