OCR A Level Mathematics A: Differentiation fundamentals: gradients, second derivatives, first principles and standard derivatives (H240) – Revision Notes

These are condensed notes for OCR A Level Mathematics A (H240), content references 1.07a to 1.07l, specification Version 3.1 (August 2026). For full explanations read the study guide, then test yourself with the practice questions. Items marked A Level only are Stage 2 content. See also the course hub, the printable checklist, the pure mathematics revision notes and the free diagnostics.

Definitions to know

Term	Meaning	Ref
Derivative f’(x), dy/dx	Gradient of the tangent at (x, y); rate of change of y with respect to x	1.07a, 1.07b
Chord gradient	[f(a + h) − f(a)] / h	1.07b
Second derivative f’’(x), d²y/dx²	Derivative of the derivative; rate of change of gradient	1.07d, 1.07e
Convex on an interval	f’’(x) > 0 there	1.07f, A Level only
Concave on an interval	f’’(x) < 0 there	1.07f, A Level only
Point of inflection	f’’(x) = 0 and the curve changes between concave and convex	1.07f, A Level only

Derivatives to learn

f(x)	f’(x)	Stage
xⁿ (n rational)	nxⁿ⁻¹	1
constant	0	1
e^kx	k e^kx	A Level only
a^kx	k ln a · a^kx	A Level only
ln x	1/x	A Level only
sin kx	k cos kx	A Level only
cos kx	−k sin kx	A Level only
tan kx	k sec² kx	A Level only

Constant multiples come out front. Sums and differences go term by term. Trigonometric results need radians.

Method boxes

Differentiating a power function

1. Rewrite roots and reciprocals as powers.
2. Split any fraction with a one-term denominator into separate terms.
3. Bring the power down as a multiplier, then take 1 off the power.
4. Differentiate again for f’’(x) if asked.

First principles for xⁿ (1.07g)

1. Write [f(x + h) − f(x)] / h with the lim (h → 0) notation.
2. Expand (x + h)ⁿ. The powers 2, 3 and 4 are the ones you may be asked for.
3. Cancel the xⁿ term and divide every remaining term by h.
4. Let h → 0 and state the result.

First principles for sin x and cos x (1.07h, A Level only)

1. Use the addition formula for sin(x + h) or cos(x + h).
2. Group terms to show (cos h − 1)/h and sin h / h.
3. Quote the limits 0 and 1 (radians) and read off the result.

Convexity and inflection (1.07f, A Level only)

1. Find f’’(x) and factorise.
2. Test the sign of f’’ on each side of every root.
3. State the intervals, then name a point of inflection only where the sign changes.

Notation and the gradient sketch

You may see	It means
f’(x) or dy/dx	First derivative
f’’(x) or d²y/dx²	Second derivative
dA/dt	Rate of change of A with respect to t
lim (h → 0)	The value the expression approaches as h gets smaller

Method to sketch y = f’(x) from y = f(x) (1.07c)

1. Mark every turning point of f. Plot a root of f’ directly below each one.
2. Where f rises, put f’ above the axis. Where f falls, put it below.
3. Where f is steepest, put a turning point of f’.
4. Where f is a straight line, draw f’ as a horizontal line at that gradient.
5. Check the degree: f’ should look one degree lower than f.

Small worked reminders

First principles for x²: [(x + h)² − x²] / h = 2x + h, and the limit is 2x.

Chord for y = x³ from x = 2: gradient 12 + 6h + h², so the tangent gradient is 12.

Cosine from first principles: [cos(x + h) − cos x] / h = cos x · (cos h − 1)/h − sin x · (sin h / h), which tends to −sin x.

Rewrite before differentiating: y = 8/√x = 8x^(−1/2), so dy/dx = −4x^(−3/2).

Exponential with a general base: y = 3^(2x) gives dy/dx = 2 ln 3 · 3^(2x).

Second derivative of a trigonometric function: y = sin 3x gives y’ = 3 cos 3x and y’’ = −9 sin 3x.

Reading a gradient sketch: if f has a maximum then a minimum, f’ is below the axis between them. For a cubic with a positive x³ coefficient, f’ is a U-shaped parabola.

Three fuller examples

Powers and a second derivative. Let y = 2x^(3/2) − 12√x + 5/x. Find dy/dx and d²y/dx² when x = 4.

y = 2x^(3/2) − 12x^(1/2) + 5x^(−1)
dy/dx = 3x^(1/2) − 6x^(−1/2) − 5x^(−2)
d²y/dx² = (3/2)x^(−1/2) + 3x^(−3/2) + 10x^(−3)
at x = 4: dy/dx = 6 − 3 − 5/16 = 43/16
          d²y/dx² = 3/4 + 3/8 + 10/64 = 41/32

Inflection. Let f(x) = x³ − 6x² + 9x + 1 (A Level only). Then f’‘(x) = 6x − 12 = 6(x − 2). It is negative for x < 2 (concave) and positive for x > 2 (convex), so the sign changes at x = 2. The point of inflection is (2, 3), and the gradient there is f’(2) = −3, so it is not a stationary point.

Rate of change in context. A balance of £500 grows as A = 500 × 1.04^t after t years (A Level only). Then dA/dt = 500 ln 1.04 × 1.04^t. At t = 10, dA/dt = 29.03 to 2 decimal places, so the balance is growing at about £29 per year at that moment. The derivative is a rate, so give it units.

Must-know distinctions

• f’(x) versus f’’(x). The first is the gradient. The second tells you how the gradient is changing.
• d²y/dx² versus (dy/dx)². The first is the second derivative. The second is the first derivative squared.
• xⁿ versus a^kx. Use the power rule when the variable is in the base. When the variable is in the power, use the exponential rule with ln a.
• f’‘(x) = 0 versus an inflection. The first is necessary. You also need a sign change in f’’.
• Convex versus concave. Convex means f’’ > 0 and the curve bends upwards. Concave means f’’ < 0.
• Graph of f versus graph of f’. A turning point on f is a root of f’. A section where f rises is a section where f’ is positive.

Quick self-test

1. Differentiate 7x⁻².
2. Differentiate x^(3/2).
3. Differentiate 4e^(−x/2).
4. Differentiate 3^(2x).
5. Differentiate tan 5x.
6. Find d²y/dx² when y = sin 3x.
7. What is the value of lim (h → 0) of (cos h − 1)/h?
8. For f(x) = x⁵ − 10x³, find the values of x where the curve is concave.
9. Find the gradient of y = ln x at x = e².
10. A cubic has a maximum at x = 2 and a minimum at x = 6. State the x-coordinates where f’(x) = 0 and describe the shape of the graph of f’.
11. In first principles for x², what does the quotient simplify to before the limit?

Answers

1. −14x⁻³.
2. (3/2)x^(1/2), which is (3/2)√x.
3. −2e^(−x/2).
4. 2 ln 3 · 3^(2x).
5. 5 sec² 5x.
6. −9 sin 3x.
8. f’’(x) = 20x³ − 60x = 20x(x² − 3). This is negative for x < −√3 and for 0 < x < √3.
9. 1/e², which is e⁻².
10. f’(x) = 0 at x = 2 and x = 6. A positive-cubic maximum then minimum gives f’ as a U-shaped parabola with roots 2 and 6. It dips below the axis between them, with its lowest point at x = 4.
11. 2x + h.

Where marks are usually lost

• Leaving √x or 1/x² in place and trying to differentiate it directly.
• Dropping the ln a factor for a^kx, or the k factor for e^kx and the trigonometric functions.
• A sign slip on cos kx, whose derivative is −k sin kx.
• Writing the limit too early, or not at all, in a first-principles answer.
• Expanding (x + h)³ or (x + h)⁴ with a wrong coefficient.
• Calling a point an inflection from f’’ = 0 without testing the sign on both sides.
• Mixing up concave and convex, or stating them without the interval.
• Sketching f’ with the roots in the wrong places; they must sit under the turning points of f.
• Differentiating a trigonometric function with the angle in degrees, when the standard results need radians.
• Leaving exact values untidy, for example 3 − 3/16 instead of 45/16.

Official syllabus

OCR Level 3 Advanced GCE in Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR: section 1.07, Differentiation, references 1.07a to 1.07l.