CIE IGCSE Mathematics (0580) Differentiation Study Notes - New Syllabus
CIE IGCSE Mathematics (0580) Differentiation Study Notes
LEARNING OBJECTIVE
- Estimating Gradients of Curves
Key Concepts:
- Estimating Gradients of Curves
- Basic Rules of Derivatives
- Gradients and Stationary Points
- Nature of Stationary Points
Estimating Gradients of Curves
Estimating Gradients of Curves
Differentiation is the process of finding the gradient (or slope) of a curve at a given point. Since curves do not have constant gradients, we estimate the gradient at a particular point by drawing a tangent to the curve at that point.
Key Idea:
- The gradient of a curve at a specific point is equal to the gradient of the tangent line drawn at that point.
- A tangent is a straight line that touches the curve at exactly one point and has the same direction as the curve at that point.
- You can estimate the gradient of the curve by measuring the rise over run of the tangent line: \( \text{Gradient} = \dfrac{\text{Vertical Change}}{\text{Horizontal Change}} \)
Example:
Estimate the gradient of the curve \( y = x^2 \) at the point \( x = 2 \).
▶️ Answer/Explanation
Step 1: Draw the curve \( y = x^2 \) and mark the point where \( x = 2 \). The corresponding y-value is \( y = 4 \), so the point is \( (2, 4) \).
Step 2: Using a ruler, draw a tangent to the curve at the point \( (2, 4) \). The line should just touch the curve at that point and follow its direction.
Step 3: Choose two points on the tangent line, as far apart as possible, for accuracy. Suppose these are:
- Point A: \( (2, 4) \)
- Point B: \( (0, -4) \)
Step 4: Use the gradient formula:
$\text{Gradient} = \frac{-4 – 4}{0 – 2} = \frac{8}{2} = 4 $
Answer: The estimated gradient of the curve at \( x = 2 \) is 4.
Basic Rules of Derivatives
Basic Rules of Derivatives
Differentiation helps us find the gradient of a function at any point. For algebraic functions of the form \( ax^n \), the following rule is used:
If \( f(x) = ax^n \), then its derivative is:
$ f'(x) = \frac{d}{dx}(ax^n) = a \cdot n \cdot x^{n – 1} $
This rule also applies to expressions that are sums of such terms.
Key Properties:
- Constants remain unchanged under differentiation.
- The derivative of a constant term (e.g., 7) is zero.
- Apply the power rule to each term separately and add the results.
Example:
Find the derivative of \( f(x) = 5x^3 \).
▶️ Answer/Explanation
Use the rule: \( \frac{d}{dx}(ax^n) = a \cdot n \cdot x^{n – 1} \)
$ f'(x) = 5 \cdot 3 \cdot x^{2} = 15x^2 $
Example:
Differentiate \( f(x) = 3x^4 – 2x^2 + 7 \)
▶️ Answer/Explanation
Differentiating each term:
- \( \frac{d}{dx}(3x^4) = 12x^3 \)
- \( \frac{d}{dx}(-2x^2) = -4x \)
- \( \frac{d}{dx}(7) = 0 \)
Answer: \( f'(x) = 12x^3 – 4x \)
Example:
Find the derivative of \( f(x) = -\frac{1}{2}x^3 + 4x \)
▶️ Answer/Explanation
$ f'(x) = -\frac{1}{2} \cdot 3 \cdot x^2 + 4 = -\frac{3}{2}x^2 + 4 $
Gradients and Stationary Points
Gradients and Stationary Points
Differentiation can be used to find the gradient of a curve at any point and to determine the location and nature of stationary points (also called turning points).
Gradient at a Point:
The derivative of a function \( f'(x) \) gives the slope (gradient) of the curve at any value of \( x \).
To find the gradient at a specific point: $ \text{Gradient} = f'(x_0) \quad \text{where } x_0 \text{ is the x-value of the point.} $
Stationary Points:
A stationary point occurs where the gradient is zero: $ f'(x) = 0 $ At these points, the curve is flat — it may be a maximum, minimum, or point of inflection.
To Find Stationary Points:
- Differentiate the function \( f(x) \)
- Solve \( f'(x) = 0 \) to find the x-values
- Substitute into the original function to get the y-values
Example:
Find the gradient of the curve \( f(x) = 3x^2 + 2x – 5 \) at \( x = 1 \).
▶️ Answer/Explanation
Step 1: Differentiate: \( f'(x) = 6x + 2 \)
Step 2: Substitute \( x = 1 \):
$ f'(1) = 6(1) + 2 = 8 $
Answer: The gradient at \( x = 1 \) is 8.
Example:
Find the stationary points of \( f(x) = x^2 – 4x + 6 \).
▶️ Answer/Explanation
Step 1: Differentiate: \( f'(x) = 2x – 4 \)
Step 2: Solve \( f'(x) = 0 \):
$ 2x – 4 = 0 \Rightarrow x = 2 $
Step 3: Find y-value:
$ f(2) = (2)^2 – 4(2) + 6 = 4 – 8 + 6 = 2 $
Stationary point: \( (2, 2) \)
Identifying Maximum and Minimum Points
Identifying Maximum and Minimum Points
Maximum and minimum points (also called turning or stationary points) occur where the gradient of a curve is zero — i.e., where the first derivative is zero:
$ f'(x) = 0 $
Once the stationary points are found, we need to determine whether each point is a maximum or a minimum. This can be done using any of the following methods:
1. Accurate Sketch
- Sketch the curve using key features: roots, y-intercept, vertex (turning point), and shape.
- A peak (top of the hill) indicates a maximum.
- A dip (bottom of a valley) indicates a minimum.
2. Second Derivative Test
- Find \( f”(x) \), the second derivative.
- If \( f”(x) > 0 \), the curve is concave up → Minimum point.
- If \( f”(x) < 0 \), the curve is concave down → Maximum point.
3. Gradient Sign Test (First Derivative Sign Change)
- Check the sign of \( f'(x) \) just before and just after the stationary point.
- If gradient changes from positive to negative → Maximum.
- If gradient changes from negative to positive → Minimum.
Example:
Use the second derivative to classify the turning point of \( f(x) = x^2 – 6x + 7 \).
▶️ Answer/Explanation
Step 1: First derivative: \( f'(x) = 2x – 6 \)
Set \( f'(x) = 0 \Rightarrow x = 3 \)
Step 2: Second derivative: \( f”(x) = 2 \)
\( f”(3) = 2 > 0 \Rightarrow \text{Minimum point} \)
Substitute into original: \( f(3) = 9 – 18 + 7 = -2 \)
Answer: Minimum point at \( (3, -2) \)
Example:
Determine the nature of the turning point of \( f(x) = -x^2 + 4x – 1 \) using the gradient sign method.
▶️ Answer/Explanation
Step 1: First derivative: \( f'(x) = -2x + 4 \)
Set \( f'(x) = 0 \Rightarrow x = 2 \)
Check gradient either side:
- \( x = 1.9 \Rightarrow f'(1.9) = -2(1.9) + 4 = 0.2 \) (positive)
- \( x = 2.1 \Rightarrow f'(2.1) = -2(2.1) + 4 = -0.2 \) (negative)
Gradient changes from positive to negative → Maximum
Find y-value: \( f(2) = -4 + 8 – 1 = 3 \)
Answer: Maximum point at \( (2, 3) \)
Example:
The sketch of a curve shows a turning point at \( (1, -5) \), and the curve opens upward. What type of turning point is it?
▶️ Answer/Explanation
If the curve opens upwards (U-shape), the turning point is at the lowest point.
Answer: Minimum point at \( (1, -5) \)