SKETCHING GRADIENT FUNCTIONS
Given y = f(x), sketch the graph of y = f'(x)
SPEC 7.1 • DIFFERENTIATION
The Big Idea
The gradient function f'(x) tells us the gradient of f(x) at every point.
Key Insight
By analysing the shape of y = f(x), we can deduce what y = f'(x) looks like — without any algebra!
The Process
Look at f(x) and ask: Where is the gradient zero? Positive? Negative? Large? Small?
Key Features: f(x) → f'(x)
Feature on y = f(x)Feature on y = f'(x)
Maximum (turning point)Root — crosses from + to −
Minimum (turning point)Root — crosses from − to +
Point of inflectionTouches x-axis but doesn't cross
f(x) increasingf'(x) > 0 (above x-axis)
f(x) decreasingf'(x) < 0 (below x-axis)
Vertical asymptoteVertical asymptote (same x-value)
Horizontal asymptotef'(x) → 0 (approaches x-axis)
Curve Type Relationships
Differentiation reduces the degree of polynomials by 1:
If f(x) is...Then f'(x) is...
Linear (straight line)Constant (horizontal line)
Quadratic (parabola)Linear (straight line)
CubicQuadratic (parabola)
QuarticCubic
sin xcos x
eˣeˣ (same shape!)
Tip
Count the turning points on f(x) — that's how many roots f'(x) will have.
01Worked Example
Sketch y = f'(x) for the quadratic y = f(x) shown with minimum at (2, −1).
y = f(x)y = f'(x)
Step 1: Minimum at x = 2 → root on f'(x) at x = 2
Step 2: x < 2: decreasing → f'(x) < 0; x > 2: increasing → f'(x) > 0
Step 3: Quadratic → Linear
f'(x) = 2x − 4 (straight line crossing at x = 2)
02Worked Example
Sketch y = f'(x) for the cubic with local max at (−1, 4) and local min at (2, −3).
y = f(x)y = f'(x)
Step 1: Turning points at x = −1 and x = 2 → roots on f'(x)
Step 2: x < −1: increasing (+); −1 < x < 2: decreasing (−); x > 2: increasing (+)
Step 3: Cubic → Quadratic (U-shape)
f'(x) is a U-shaped parabola with roots at x = −1 and x = 2
03Worked Example
Sketch y = f'(x) for y = sin x on 0 ≤ x ≤ 2π.
y = sin xy = f'(x) = cos x
Step 1: Max at x = π/2, min at x = 3π/2 → roots at these x-values
Step 2: Positive (0 to π/2), negative (π/2 to 3π/2), positive (3π/2 to 2π)
f'(x) = cos x
04Worked Example
Sketch y = f'(x) for y = x³ with a point of inflection at the origin.
y = x³y = f'(x) = 3x²
Key insight: NO turning points — always increasing!
At x = 0, gradient is zero but f(x) continues increasing.
f'(x) = 3x² (touches but doesn't cross at origin)
05Worked Example
Sketch y = f'(x) for this curve with turning points at A, B, and C.
y = f(x)y = f'(x)
Step 1: A = max, B = min, C = max → 3 roots on f'(x)
Step 2: Sign pattern: + before A, − (A to B), + (B to C), − after C
f'(x) is a cubic crossing at A, B, C
06Worked Example
Sketch y = f'(x) for y = 1/x (vertical asymptote x = 0).
y = 1/xy = f'(x) = −1/x²
Asymptote rules:
•
Vertical at x = 0 → f'(x) also has vertical asymptote at x = 0
•
1/x is always decreasing → f'(x) always negative
f'(x) = −1/x² (always negative)
07Worked Example
Sketch y = f'(x) for this curve with horizontal asymptote y = 2.
y = f(x)y = f'(x)
•
No turning points → f'(x) never crosses x-axis
•
Always increasing → f'(x) always positive
•
Horizontal asymptote → f'(x) → 0 as x → ∞
f'(x) is positive and decreases towards 0
Practice
Question 1:
Cubic → Quadratic f'(x) (U-shaped parabola with roots at x = −2 and x = 3)
U-shaped parabola with roots at x = −2 and x = 3
Practice
Question 2:
3 turning points → cubic f'(x) with 3 roots
Cubic crossing at P, Q, R with pattern + − + −
Practice
Question 3:
Vertical asymptote at x = 2 → f'(x) also has asymptote at x = 2
Positive left of x=2, negative right of x=2
Summary
5 Steps to Sketch f'(x) from f(x)
StepWhat to Do
1Mark turning points → roots on f'(x)
2f(x) increasing → f'(x) above x-axis
3f(x) decreasing → f'(x) below x-axis
4Vertical asymptote stays; horizontal → f'(x) → 0
5Determine curve type (cubic → quadratic, etc.)
Key Points
Maximum → f'(x) crosses from + to −
Minimum → f'(x) crosses from − to +
Inflection → f'(x) touches but doesn't cross
Next: The Big Idea