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The derivative from first principles questions
A curve has no single gradient, but each point on it does, and catching that number takes a genuinely new idea: measure the slope of a chord, then let the chord shrink. What survives the shrinking is the derivative, and everything else in calculus is built on top of this one move.
6 original questions · 22 marks · the the derivative from first principles notes · Differentiation
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Write down the first-principles definition of the derivative f′(x), and explain in one sentence what the limit as h → 0 achieves.
Worked answer
f′(x) = lim as h → 0 of [f(x + h) − f(x)]/h. The fraction is the gradient of the chord joining x to x + h, and the limit slides the far end of that chord into the near end, turning a chord into a tangent. B1 for the definition, B1 for the chord becoming a tangent. Writing the quotient without the limit sign scores nothing, since the limit is the definition.For f(x) = x2, calculate the gradient of the chord from x = 3 with h = 0.1, and again with h = 0.01. State what the answers suggest about the gradient of the curve at x = 3.
Worked answer
With h = 0.1 the chord gradient is (3.12 − 9)/0.1 = 6.1. With h = 0.01 it is 6.01. The values are settling towards 6, which the derivative 2x confirms at x = 3. M1 for a chord gradient, A1 for 6.1, A1 for 6.01. Numerical chords are evidence, never proof, so a question asking you to prove the result still needs the algebra.Prove from first principles that the derivative of f(x) = x2 is 2x.
Worked answer
The chord gradient is [(x + h)2 − x2]/h = (2xh + h2)/h. For h ≠ 0 this cancels to 2x + h, and as h → 0 what survives is 2x. Three moves earn the marks: expand, divide, let h tend to zero. M1 for the chord gradient, A1 for the expansion, A1 for cancelling to 2x + h, A1 for the limit. The middle move requires h ≠ 0 to be stated, and the words as h → 0 are the ones doing the work, rather than a bare substitution of h = 0.Prove from first principles that the derivative of f(x) = 3x2 + 2x is 6x + 2.
Worked answer
The chord gradient is [3(x + h)2 + 2(x + h) − 3x2 − 2x]/h = (6xh + 3h2 + 2h)/h, which is 6x + 3h + 2 for h ≠ 0. As h → 0 it becomes 6x + 2. M1 for the chord gradient, A1 for 6xh + 3h2 + 2h, A1 for 6x + 3h + 2, A1 for 6x + 2. Every term of f contributes its own derivative, and that is the termwise rule being born. Forgetting to expand 3(x + h)2 fully is the usual source of a lost 6xh.Find, from first principles, the gradient of the curve y = x3 at the point (2, 8).
Worked answer
The chord gradient from x = 2 is [(2 + h)3 − 8]/h = (12h + 6h2 + h3)/h = 12 + 6h + h2. As h → 0 the gradient is 12. M1 for the chord gradient from x = 2, A1 for the expansion, A1 for 12 + 6h + h2, A1 for 12. Working at the specific point keeps the binomial expansion short, since the 8 cancels and only the h terms remain.Prove from first principles that the derivative of f(x) = 1/x, x ≠ 0, is −1/x2.
Worked answer
The chord gradient is [1/(x + h) − 1/x]/h. Combine the two fractions over the common denominator x(x + h), giving [x − (x + h)]/[x(x + h)h] = −h/[x(x + h)h]. For h ≠ 0 the h cancels to leave −1/[x(x + h)], and as h → 0 that tends to −1/x2. M1 for the chord gradient, M1 for combining over x(x + h), A1 for −h/[x(x + h)h], A1 for cancelling, A1 for the limit. The difficulty is entirely algebraic. The subtraction has to happen inside one fraction before the division by h can cancel anything, and candidates who write 1/(x + h) − 1/x = 1/h lose every mark. Note also that the h in the denominator only cancels because h ≠ 0 throughout the working.
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