Differentiation

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Differentiation Basics

Differentiation changes a curve equation y = \ldots into a gradient function \frac{dy}{dx} = \ldots.
To differentiate y = xⁿ, bring down the power and reduce it by one: \frac{dy}{dx} = n x^n-1.
For y = kxⁿ, 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.

Gradient at a point via the tangent

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.

Turning points on a cubic curve

Classifying Stationary Points Using Graphs

A positive quadratic (x² 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 turning points on a cubic

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{d²y}{dx²} is the derivative of \frac{dy}{dx}.
Substitute the x-coordinate of the stationary point into \frac{d²y}{dx²}.
If \frac{d²y}{dx²} < 0 → maximum; if \frac{d²y}{dx²} > 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.

A cuboid with variable dimensions

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1.What is the derivative of y = x⁵?
Easy
- A5x⁴
- B5x⁵
- Cx⁴
- D4x⁵
2.What is the derivative of y = 2x³?
Easy
- A6x²
- B5x²
- C2x²
- D6x³
3.What is the derivative of y = 4?
Easy
- A0
- B4
- C1
- D4x
4.What is the derivative of y = 6 + 4x - x²?
Easy
- A4 - 2x
- B4 + 2x
- C6 + 4x - 2x
- D4x - 2x
5.Find the gradient of y = 24 + 5x - x² at x = -1.5.
Medium
- A8
- B2
- C-8
- D-2
6.Given y = 2x^k + u x⁷ and dy/dx = 18 x^k-1 + 21 x⁶, find k and u.
Medium
- Ak=9, u=3
- Bk=9, u=7
- Ck=18, u=21
- Dk=2, u=3
7.Find the x-coordinate of the turning point of y = 6 + 4x - x².
Medium
- A2
- B-2
- C4
- D6
8.Find the coordinates of the turning point of y = 2x² + 8x - 9.
Medium
- A(-2, -17)
- B(2, 15)
- C(-2, 1)
- D(2, -17)

Plus de questions

9.For the curve y = 2x² - 6x - 11, find the x-coordinate where the gradient is 2.
Medium
- A2
- B1
- C3
- D4
10.Find the coordinates of the stationary points of y = x³ - 6x² + 16.
Medium
- A(0,16) and (4,-16)
- B(0,16) and (4,0)
- C(2,0) and (4,-16)
- D(0,16) and (2,0)
11.Given y = 5x^m + p x¹² and dy/dx = 30 x^m-1 + 6 x¹¹, find m and p.
Medium
- Am=6, p=0.5
- Bm=6, p=6
- Cm=30, p=6
- Dm=5, p=0.5
12.Find the stationary points of y = 4x³ - 3x + 3.
Medium
- A(0.5,2) and (-0.5,4)
- B(0.5,4) and (-0.5,2)
- C(1,4) and (-1,2)
- D(0.5,3) and (-0.5,3)
13.The curve y = x³ - 3x + 4 has stationary points. Which is a maximum?
Hard
- A(-1,6)
- B(1,2)
- C(-1,2)
- D(1,6)
14.Find the stationary points of y = x⁴ - 4x³.
Hard
- A(0,0) and (3,-27)
- B(0,0) and (3,27)
- C(0,0) and (4,0)
- D(1,-3) and (3,-27)
15.For y = x³ - 6x² - 15x, find the x-coordinates of its stationary points.
Medium
- A5 and -1
- B5 and 1
- C-5 and 1
- D6 and -15
16.Differentiate y = x⁴.
Easy
- Ady/dx = 4x³
- Bdy/dx = 4x⁴
- Cdy/dx = x³
- Ddy/dx = 4x⁵
17.What is the derivative of y = x⁶?
Easy
- A6x⁵
- B5x⁶
- C6x⁶
- Dx⁵
18.Differentiate y = 7x.
Easy
- A7
- B7x
- C0
- Dx
19.What is the derivative of y = 9?
Easy
- A0
- B9
- C9x
- D1
20.Differentiate y = 3x⁴.
Easy
- A12x³
- B4x³
- C12x⁴
- D3x³
21.Differentiate y = 5 + 2x - x³.
Easy
- A2 - 3x²
- B2 - 3x
- C5 + 2 - 3x²
- D2x - 3x²
22.Find the gradient of y = x² + 3x - 4 at x = 1.
Easy
- A5
- B1
- C3
- D0
23.The curve y = x³ - 6x² + 16 has a turning point at x = 4. What is the y-coordinate of this turning point?
Medium
- A-16
- B16
- C0
- D4
24.Find the coordinates of the turning point of y = 6 + 4x - x².
Medium
- A(2, 10)
- B(2, 6)
- C(-2, 2)
- D(4, 6)
25.Differentiate y = 5x³.
Easy
- A15x²
- B5x²
- C15x³
- D5x⁴
26.Differentiate y = 7.
Easy
- A7
- B7x
- C0
- D1
27.Differentiate y = 3x.
Easy
- A3
- B3x
- C0
- D1
28.Find the gradient of y = x² + 4x - 5 at x = 2.
Easy
- A8
- B4
- C6
- D10
29.The curve y = 3x² - 12x + 7 has a turning point. What is the x-coordinate of this turning point?
Medium
- A2
- B-2
- C4
- D0
30.Find the coordinates of the turning point of y = x² - 6x + 10.
Medium
- A(3, 1)
- B(3, -1)
- C(-3, 1)
- D(3, 19)
31.For y = 2x³, find dy/dx.
Easy
- A6x²
- B6x³
- C2x²
- D8x²
32.Given y = x³ + 2x² - 5x + 1, find dy/dx.
Medium
- A3x² + 4x - 5
- B3x² + 2x - 5
- Cx² + 4x - 5
- D3x² + 4x + 5
33.The curve y = x³ - 3x² - 9x + 5 has two stationary points. Find the x-coordinate of the local maximum.
Hard
- A-1
- B3
- C1
- D-3
34.A curve has equation y = 2x³ - 9x² + 12x - 3. Find the x-coordinate of the local minimum.
Hard
- A2
- B1
- C3
- D0
35.The derivative of a constant term is always zero.
Easy
True or false?
36.What is the derivative of y = 5x?
Easy
- A5
- B0
- C5x
- D1
37.What is the derivative of y = 3x²?
Easy
- A6x
- B3x
- C2x
- D6x²
38.The curve y = x³ - 3x² + 2 has a stationary point at x = 0. What is the y-coordinate of this stationary point?
Medium
- A2
- B0
- C-2
- D1
39.For which value of x does the curve y = 2x² - 8x + 3 have a stationary point?
Medium
- A2
- B4
- C-2
- D1
40.The turning point of y = -x² + 4x - 1 is a
Medium
- Amaximum
- Bminimum
- Cpoint of inflection
- Dcannot be determined
41.Which of the following are true about the gradient function dy/dx? (Select all that apply)
Medium
- AIt gives the gradient of the curve at any point.
- BIt is also called the derivative.
- CIt is always positive.
- DIt can be found by differentiation.
42.Match each curve equation to its gradient function.
Medium
- y = x²
- y = 2x
- y = 5
- dy/dx = 0
- dy/dx = 2
- dy/dx = 2x
43.Arrange the steps in order to find the stationary point of y = x² - 4x + 3.
Medium
- Set dy/dx = 0 and solve for x
- Differentiate to find dy/dx
- Substitute x into original equation to find y
- Write the coordinates of the stationary point
44.Find the gradient of y = x² + 2x at x = 3.
Easy
45.A farmer uses 100 m of fencing to make a rectangular enclosure against a wall (so only three sides are fenced). If the width (perpendicular to the wall) is x m, the area A m² is given by A = x(100 - 2x). Find the maximum area.
Medium
46.Complete the sentence: The derivative of y = xⁿ is dy/dx = ___.
Easy
The derivative of y = xⁿ is dy/dx = ____.
47.A rectangular field is to be fenced on three sides using 100 m of fencing. The fourth side is a river. Find the maximum area of the field in square metres.
Medium
48.Find the gradient of y = x² + 2x at x = 3.
Easy
- A10
- B9
- C8
- D7
49.A farmer uses 100 m of fencing to make a rectangular enclosure against a wall (so only three sides are fenced). If the width (perpendicular to the wall) is x m, the area A m² is given by A = x(100 - 2x). Find the maximum area.
Medium
- A1250 m²
- B1375 m²
- C1125 m²
- D1500 m²
50.A rectangular field is to be fenced on three sides using 100 m of fencing. The fourth side is a river. Find the maximum area of the field in square metres.
Medium
- A1250 m²
- B1375 m²
- C1125 m²
- D1500 m²

Notes de révision

## 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.

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.

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