Differentiation

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Notes

Differentiation Basics

**Differentiation** changes a curve equation $y = \ldots$ into a **gradient function** $\frac{dy}{dx} = \ldots$ .
To differentiate $y = x^n$ , bring down the power and reduce it by one: $\frac{dy}{dx} = n x^{n-1}$ .
For $y = kx^n$ , multiply by the coefficient: $\frac{dy}{dx} = kn x^{n-1}$ .
Special cases: $y = kx$ gives $\frac{dy}{dx} = k$ ; $y = c$ (constant) gives $\frac{dy}{dx} = 0$ .
Differentiate each term separately when a curve has multiple terms.

Finding the Gradient at a Point

To find the gradient at a point, substitute the $x$ -coordinate into $\frac{dy}{dx}$ .
If given a gradient, set $\frac{dy}{dx}$ equal to that value and solve for $x$ .
The $y$ -coordinate is not needed to find the gradient.

Stationary Points & Turning Points

A **stationary point** occurs where the gradient is zero: $\frac{dy}{dx} = 0$ .
**Turning points** are stationary points where the curve changes direction (maxima or minima).
To find coordinates: (1) differentiate, (2) set $\frac{dy}{dx} = 0$ and solve for $x$ , (3) substitute $x$ into original equation to get $y$ .

Classifying Stationary Points Using Graphs

A positive quadratic ( $x^2$ positive) has a **minimum**; a negative quadratic has a **maximum**.
A positive cubic has a **maximum** on the left and a **minimum** on the right.
A negative cubic has a **minimum** on the left and a **maximum** on the right.

Classifying Using the First Derivative

Examine the sign of $\frac{dy}{dx}$ just before and after the stationary point.
If gradient changes from positive to zero to negative → **maximum**.
If gradient changes from negative to zero to positive → **minimum**.

Classifying Using the Second Derivative

The **second derivative** $\frac{d^2y}{dx^2}$ is the derivative of $\frac{dy}{dx}$ .
Substitute the $x$ -coordinate of the stationary point into $\frac{d^2y}{dx^2}$ .
If $\frac{d^2y}{dx^2} < 0$ → **maximum**; if $\frac{d^2y}{dx^2} > 0$ → **minimum**; if zero, test fails.

Problem Solving with Differentiation (Optimisation)

Use differentiation to find maximum or minimum values of quantities (e.g., area, volume).
Form an equation for the quantity in terms of one variable, then differentiate and set $\frac{dy}{dx} = 0$ .
Solve for the variable, then substitute back to find the optimum value.
Check if it is a max or min using second derivative or graph shape.

Gradient Function Concept

Stationary Points: Max and Min

Second Derivative Test

Optimisation Example

Practice questions

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1.What is the derivative of $y = x^{5}$ ?
Easy
- A $5x^{4}$
- B $5x^{5}$
- C $x^{4}$
- D $4x^{5}$
2.What is the derivative of $y = 2x^{3}$ ?
Easy
- A $6x^{2}$
- B $5x^{2}$
- C $2x^{2}$
- D $6x^{3}$
3.What is the derivative of $y = 4$ ?
Easy
- A0
- B4
- C1
- D4x
4.What is the derivative of $y = 6 + 4x - x^{2}$ ?
Easy
- A $4 - 2x$
- B $4 + 2x$
- C $6 + 4x - 2x$
- D $4x - 2x$
5.Find the gradient of $y = 24 + 5x - x^{2}$ at $x = -1.5$ .
Medium
- A8
- B2
- C-8
- D-2
6.Given $y = 2x^{k} + u x^{7}$ and $dy/dx = 18 x^{k-1} + 21 x^{6}$ , find k and u.
Medium
- A $k=9, u=3$
- B $k=9, u=7$
- C $k=18, u=21$
- D $k=2, u=3$
7.Find the x-coordinate of the turning point of $y = 6 + 4x - x^{2}$ .
Medium
- A2
- B-2
- C4
- D6
8.Find the coordinates of the turning point of $y = 2x^{2} + 8x - 9$ .
Medium
- A(-2, -17)
- B(2, 15)
- C(-2, 1)
- D(2, -17)

Summary

## Differentiation - Quick Revision ### Key Rules - **Power rule**: If $y = x^n$, then $\frac{dy}{dx} = n x^{n-1}$. - **Constant multiple**: If $y = k x^n$, then $\frac{dy}{dx} = k n x^{n-1}$. - **Linear term**: If $y = kx$, then $\frac{dy}{dx} = k$. - **Constant**: If $y = c$, then $\frac{dy}{dx} = 0$. - Differentiate term by term for sums/differences. ### Finding Gradient at a Point 1. Differentiate to get $\frac{dy}{dx}$. 2. Substitute the $x$-coordinate into $\frac{dy}{dx}$. ### Stationary Points (Turning Points) - Set $\frac{dy}{dx} = 0$ and solve for $x$. - Substitute $x$ into original equation to find $y$. ### Classifying Stationary Points #### Method 1: Second Derivative - Find $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$. - If $\frac{d^2y}{dx^2} < 0$ at the point → **maximum**. - If $\frac{d^2y}{dx^2} > 0$ at the point → **minimum**. - If $\frac{d^2y}{dx^2} = 0$, test fails; use another method. #### Method 2: Gradient Table - Check sign of $\frac{dy}{dx}$ just before and after the point. - Positive → zero → negative: maximum. - Negative → zero → positive: minimum. #### Method 3: Graph Shape - Positive quadratic: minimum. Negative quadratic: maximum. - Positive cubic: max left, min right. Negative cubic: min left, max right. ### Problem Solving (Optimisation) 1. Express quantity in one variable using given constraints. 2. Differentiate and set derivative = 0. 3. Solve for variable. 4. Substitute back to find max/min value. 5. Justify nature (e.g., negative quadratic → max). ### Common Mistakes - Forgetting to substitute $x$ back into original equation. - Mixing up curve equation and gradient function. - Using $y$-coordinate when only gradient is needed.

Key terms

Differentiation: Algebraic method to find the gradient function (derivative) of a curve.
Gradient function (derivative): A function, written as dy/dx, that gives the gradient of a curve at any point.
Stationary point: A point on a curve where the gradient is zero (dy/dx = 0).
Turning point: A stationary point where the curve changes direction (maximum or minimum).
Maximum point: A turning point where the curve reaches a peak; second derivative < 0.
Minimum point: A turning point where the curve reaches a trough; second derivative > 0.
Second derivative: The derivative of the derivative, denoted d²y/dx²; used to classify stationary points.
Optimisation: Using differentiation to find maximum or minimum values in real-world problems.

Slides

What is Differentiation?

Differentiation is an algebraic method that changes the equation of a curve, $y =$ ..., into a gradient function, $dy/dx =$ ...
The gradient function gives the gradient of the curve at any point when you substitute in its x-coordinate.
To differentiate a power of x, bring down the power and reduce the power by 1: if $y = x^{n}$ , then $dy/dx = n x^(n-1)$ .

What is the derivative of $y = x^{5}$ ?

$5x^{4}$
$5x^{5}$
$x^{4}$
$4x^{5}$

Differentiating Terms with Coefficients and Constants

If $y = k x^{n}$ , then $dy/dx = k n x^(n-1)$ . For example, $y = 2x^{5}$ gives $dy/dx = 10x^{4}$ .
If $y = k x (i.e$ . $x^{1})$ , then $dy/dx = k$ . For example, $y = 2x$ gives $dy/dx = 2$ .
If $y = c (a constant)$ , then $dy/dx = 0$ . For example, $y = 4$ gives $dy/dx = 0$ .

Differentiate $y = 3x^{4} + 2x - 5$ . $dy/dx = ____ x^{3} + ____$

Finding the Gradient at a Point

To find the gradient of a curve at a specific point, differentiate to get dy/dx, then substitute the x-coordinate of the point.
The y-coordinate is not needed for the gradient.
If given a gradient and asked to find the x-coordinate, set dy/dx equal to that gradient and solve for x.

Find the gradient of $y = x^{3} - 2x^{2} + 6$ at $x = 2$ .

Stationary Points and Turning Points

A stationary point is where the gradient is zero $(dy/dx = 0)$ .
Turning points are a type of stationary point where the curve changes direction (maximum or minimum).
To find turning points: (1) find dy/dx, (2) set $dy/dx = 0$ and solve for x, (3) substitute x into original equation to find y.

Find the turning point of $y = 2x^{2} + 8x - 9$ . $x-coordinate = ____$

Classifying Turning Points Using the Second Derivative

The second derivative, $d^{2}y/dx^{2}$ , is found by differentiating dy/dx.
If $d^{2}y/dx^{2} < 0$ at a stationary point, it is a maximum.
If $d^{2}y/dx^{2} > 0$ at a stationary point, it is a minimum.
If $d^{2}y/dx^{2} = 0$ , the test fails and another method is needed.

For a stationary point at $x = a$ , if $d^{2}y/dx^{2} = -5$ , what type is it?

Maximum
Minimum
Not enough info
Inflection

Classifying Using Gradients Before and After

Alternatively, examine the sign of dy/dx just before and after the stationary point.
If dy/dx changes from positive to zero to negative, it is a maximum.
If dy/dx changes from negative to zero to positive, it is a minimum.

Which of these are true about stationary points?

At a stationary point, $dy/dx = 0$ .
All stationary points are turning points.
A maximum point has $d^{2}y/dx^{2} < 0$ .
A minimum point has $d^{2}y/dx^{2} > 0$ .

Problem Solving with Differentiation

Differentiation can be used to find maximum or minimum values in real-world problems (optimisation).
Set the derivative equal to zero to find the x-value at the optimum.
Substitute that x back into the original formula to find the maximum or minimum value.
Use the shape of the graph (e.g., negative quadratic → maximum) to confirm the nature.

A farmer has 60 m of fencing for three sides. Area $A = 60x - 2x^{2}$ . Find $dA/dx = 60 - ____ x$

For $A = 60x - 2x^{2}$ , what x gives maximum area?

Summary of Differentiation

Differentiation turns y into dy/dx using the power rule: bring down power, reduce by 1.
Constants differentiate to zero; kx differentiates to k.
Stationary points occur when $dy/dx =$ 0; classify using second derivative or gradient table.
Differentiation helps solve optimisation problems by finding maximum/minimum values.

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