Differentiation and Gradients of Curves

Differentiation gives you a formula for the gradient of a curve, so you can find the steepness at any point without drawing a tangent. Tutors described these as “easy three marks” — the rule is mechanical, and most lost marks come from stopping too early rather than from the calculus.

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1. The power rule

If y = axⁿ, then dy/dx = anx⁽ⁿ⁻¹⁾

In words: multiply by the power, then reduce the power by one.

y	dy/dx
x³	3x²
5x⁴	20x³
x	1
7x	7
9	0
x⁻²	−2x⁻³

Differentiating term by term:

y = 3x³ + 4x² − 5x + 7

• 3x³ → 9x²
• 4x² → 8x
• −5x → −5
• 7 → 0
• dy/dx = 9x² + 8x − 5

A constant differentiates to zero. The +7 disappears entirely, because a constant doesn’t change the gradient — it only shifts the curve up or down. Students both forgot this and doubted it; it is worth understanding rather than memorising.

x on its own differentiates to 1, and 7x to 7. There is no x left, because x¹ → 1x⁰ = 1.

Reduce the power, don’t just drop it. x³ becomes 3x², not 3x. Writing x² where x³ was needed was a recorded error.

With negative powers: rewrite fractions first.

y = 1 − 15/x → write as y = 1 − 15x⁻¹

• dy/dx = −15 × (−1)x⁻² = 15x⁻², i.e. 15/x²

Convert fractions to negative powers before differentiating. You cannot apply the power rule to something written as a fraction.

Notation: dy/dx and f′(x) mean the same thing — the derivative, the gradient function.

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2. Finding the gradient at a point

Differentiate first, then substitute.

Example: find the gradient of y = x³ − 4x at the point where x = 2.

1. dy/dx = 3x² − 4
2. At x = 2: 3(4) − 4 = 8

Never substitute before differentiating. Putting x = 2 into the original equation gives you the y-coordinate, not the gradient — a recorded error that produces a confident wrong answer.

Related uses:

• Gradient of the tangent at a point = the derivative there
• Gradient of the normal = −1 ÷ (gradient of tangent) — the negative reciprocal
• To find the equation of the tangent: get the gradient from dy/dx, get the point from the original equation, then use y = mx + c

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3. Stationary points (turning points)

At a stationary point, the curve is momentarily flat: dy/dx = 0

Method:

1. Differentiate
2. Set dy/dx = 0 and solve for x
3. Substitute x back into the ORIGINAL equation to find y
4. State the coordinates

Example: find the turning points of y = x³ − 3x² + 4

1. dy/dx = 3x² − 6x
2. 3x² − 6x = 0 → 3x(x − 2) = 0 → x = 0 or x = 2
3. x = 0 → y = 4; x = 2 → y = 8 − 12 + 4 = 0
4. Turning points: (0, 4) and (2, 0)

Step 3 is the most-missed step in the topic. Finding x is not the answer — a point needs both coordinates. And you must substitute into the original equation, not the derivative. The derivative gives 0 at a turning point by definition, so using it produces y = 0 every time.

A cubic usually has two turning points; a quadratic has one. If your numbers give a different count, check your factorising.

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4. Maximum or minimum?

Three acceptable methods — use whichever you find most reliable:

1. The shape of the curve. For a positive cubic, the first stationary point (smaller x) is the maximum and the second is the minimum. For a positive quadratic, the single stationary point is a minimum.

2. The sign of the gradient either side. Test a value just below and just above:

• gradient goes + → 0 → − : maximum
• gradient goes − → 0 → + : minimum

3. The second derivative. Differentiate again:

• d²y/dx² < 0 → maximum
• d²y/dx² > 0 → minimum

For y = x³ − 3x² + 4: d²y/dx² = 6x − 6. At x = 0 it is −6 (< 0) → maximum; at x = 2 it is +6 (> 0) → minimum. This matches the coordinates found above.

A minimum is the bottom of a dip, not necessarily the lowest point of the whole curve. These are local maxima and minima.

Justify your answer. “Maximum, because d²y/dx² = −6 < 0” earns the mark; an unexplained label may not.

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5. Drawing a tangent — the non-calculus method

You may still be asked to estimate a gradient from a graph:

1. Draw a tangent at the point with a ruler
2. Take two widely spaced points on the tangent
3. Gradient = rise ÷ run

This gives an estimate; differentiation gives the exact value. If a question says “by drawing a tangent”, you must do it graphically even if you could differentiate.

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6. Scope note

Some lessons went beyond 0580 into:

• Related rates of change — dy/dt, dx/dt and “flipping the derivative”
• Differentiating logarithmic and other non-polynomial functions
• The chain, product and quotient rules

None of these is on 0580. The syllabus covers differentiating simple polynomials (terms of the form axⁿ), finding gradients, and locating and classifying stationary points. Related rates and log differentiation belong to Additional Mathematics (0606) and A-level — check your own syllabus document before revising them.

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7. Mistakes that cost marks

Substituting before differentiating.

Forgetting that a constant differentiates to 0 — or thinking it stays.

Dropping the power instead of reducing it.

Differentiating x as 0 instead of 1.

Not rewriting fractions as negative powers first.

Sign errors with negative powers.

Stopping at x without finding the y-coordinate.

Substituting into the derivative rather than the original equation.

Not saying whether a point is a maximum or minimum, or giving no justification.

Using calculus when the question said “by drawing a tangent”.

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Frequently asked questions

What does differentiation give you? A formula for the gradient of a curve at any point.

What is the power rule? y = axⁿ → dy/dx = anx⁽ⁿ⁻¹⁾ — multiply by the power, reduce the power by one.

What happens to a constant? It differentiates to 0.

What does x differentiate to? 1.

How do I find the gradient at a point? Differentiate first, then substitute the x-value.

What is a stationary point? A point where the gradient is zero — the curve is momentarily flat.

How do I find turning points? Set dy/dx = 0, solve for x, then substitute into the original equation for y.

How do I tell a maximum from a minimum? Use the second derivative (negative → maximum, positive → minimum), the sign of the gradient either side, or the shape of the curve.

What is the gradient of the normal? The negative reciprocal of the tangent’s gradient.

Do I need the chain rule for 0580? No — 0580 covers simple polynomials only.

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Quick revision checklist

• I can apply the power rule to any term axⁿ
• I know constants differentiate to 0 and x to 1
• I reduce the power rather than dropping it
• I rewrite fractions as negative powers first
• I differentiate before substituting
• I can find the gradient of a tangent at a given point
• I can find the gradient of the normal
• I can find the equation of a tangent
• I set dy/dx = 0 to find stationary points
• I substitute back into the original equation for the y-coordinate
• I give turning points as coordinates
• I can classify maxima and minima, with justification
• I can estimate a gradient by drawing a tangent when asked
• I know which differentiation material is beyond 0580

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These notes cover differentiation and gradients of curves in the Cambridge IGCSE Mathematics (0580) syllabus, and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Maths lessons, with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus and formula list for your own exam series.