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Higher tier essential: use the power rule to find gradients, locate turning points and solve kinematics problems.
Differentiation is the branch of calculus that finds the rate of change of a function. In IGCSE Edexcel Maths (Higher tier, Topic 3.7), students must differentiate polynomial expressions using the power rule, find the gradient of a curve at a specific point, locate and classify turning points using the second derivative test, and apply calculus to kinematics (displacement, velocity, acceleration). This topic typically appears in the later questions of Paper 2 and is worth 4–8 marks.
Source: Edexcel IGCSE Mathematics (9-1) Specification 4MA1
The power rule is the only differentiation rule tested at IGCSE level. It states:
If y = axn, then dy/dx = anxn−1
Multiply the coefficient by the power, then reduce the power by one. This works for any value of n, including negative and fractional powers (once you rewrite the expression using index laws).
When differentiating a polynomial such as y = 3x4 − 2x3 + 5x − 8, differentiate each term separately: dy/dx = 12x3 − 6x2 + 5.
The derivative dy/dx gives a general gradient function. To find the gradient at a particular point, substitute the x-coordinate of that point into the derivative.
For example, if y = x3 − 4x and you need the gradient at x = 2:
The gradient of the curve at x = 2 is 8. This value can then be used to write the equation of the tangent line at that point using y − y1 = m(x − x1).
Turning points (also called stationary points) occur where the gradient is zero. The process has three stages:
The second derivative d2y/dx2 tells you the nature of each turning point:
Kinematics is the study of motion. In IGCSE calculus, the three quantities are linked by differentiation:
Displacement s → Velocity v = ds/dt → Acceleration a = dv/dt
Each quantity is the derivative of the one before it with respect to time (t). Key exam scenarios include:
Differentiate y = (x3 + 4x) / x.
Step 1 — Simplify first: Divide each term by x: y = x3/x + 4x/x = x2 + 4x−1
Step 2 — Differentiate term by term:
d/dx(x2) = 2x
d/dx(4x−1) = 4 × (−1) × x−2 = −4x−2
Step 3 — Combine: dy/dx = 2x − 4x−2
The key skill is rewriting the fraction using index laws before differentiating. You cannot differentiate a fraction directly.
Find the turning points of y = x3 − 3x2 − 9x + 1 and classify each one.
Step 1 — Differentiate: dy/dx = 3x2 − 6x − 9
Step 2 — Set dy/dx = 0: 3x2 − 6x − 9 = 0 → Divide by 3: x2 − 2x − 3 = 0 → (x − 3)(x + 1) = 0 → x = 3 or x = −1
Step 3 — Find y-coordinates (substitute into ORIGINAL):
When x = 3: y = 27 − 27 − 27 + 1 = −26 → Point (3, −26)
When x = −1: y = −1 − 3 + 9 + 1 = 6 → Point (−1, 6)
Step 4 — Classify using d2y/dx2: d2y/dx2 = 6x − 6
At x = 3: d2y/dx2 = 18 − 6 = 12 > 0 → Minimum at (3, −26)
At x = −1: d2y/dx2 = −6 − 6 = −12 < 0 → Maximum at (−1, 6)
Always substitute x back into the original equation for y. Substituting into dy/dx would give 0 (which is the gradient, not the y-coordinate).
A particle moves along a straight line. Its displacement at time t seconds is s = 2t3 − 15t2 + 24t metres. Find the acceleration when the particle is stationary.
Step 1 — Find velocity: v = ds/dt = 6t2 − 30t + 24
Step 2 — Find when stationary (v = 0): 6t2 − 30t + 24 = 0 → Divide by 6: t2 − 5t + 4 = 0 → (t − 1)(t − 4) = 0 → t = 1 or t = 4
Step 3 — Find acceleration: a = dv/dt = 12t − 30
When t = 1: a = 12 − 30 = −18 m/s2 (decelerating)
When t = 4: a = 48 − 30 = 18 m/s2 (accelerating)
This question chains all three kinematics steps: differentiate s to get v, set v = 0 for stationary, then differentiate v to get a and substitute.
A farmer has 200 m of fencing and wants to enclose a rectangular area against a wall (so only 3 sides need fencing). If the width is x metres, the area is A = 100x − 2x2. Differentiating: dA/dx = 100 − 4x. Setting dA/dx = 0 gives x = 25, and substituting back: A = 100(25) − 2(25)2 = 2500 − 1250 = 1250 m2. The second derivative d2A/dx2 = −4 < 0, confirming this is a maximum. This is the largest possible area.
The constant trap: Differentiating a constant like 5 and writing 5 instead of 0. A constant has no x term, so its derivative is always zero.
Substituting into the derivative for y-coordinates: When finding turning point coordinates, students solve dy/dx = 0 correctly but then substitute x back into dy/dx (which gives 0) instead of into the original equation y. Always return to the original equation for the y-value.
Notation confusion: dy/dx, f'(x), ds/dt and dv/dt all mean "differentiate" but with respect to different variables. In kinematics, differentiate with respect to t (time), not x.
Forgetting to rewrite before differentiating: Expressions like 4/x or √x must be rewritten as 4x−1 or x1/2 before applying the power rule. You cannot differentiate fractions or roots directly.
Always simplify first: If the expression contains fractions or roots, rewrite every term using index notation before touching the power rule. This avoids errors and scores method marks.
State the second derivative test result clearly: Write "d2y/dx2 = 12 > 0 therefore minimum." Examiners award a mark for the explicit comparison and conclusion.
Label kinematics steps: Write "v = ds/dt" and "a = dv/dt" as headings before each differentiation. This makes your working clear and helps the examiner follow your method.
Check units in kinematics: If s is in metres and t in seconds, then v is in m/s and a is in m/s2. Including units in your final answer demonstrates understanding and avoids losing marks.
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Last updated: March 2026
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