Questions on the derivative from first principles

Question 1

Write down the first-principles definition of the derivative f′(x), and explain in one sentence what the limit as h → 0 achieves.

Accepted 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. The limit sign is part of the definition, so the quotient alone does not define the derivative.

Question 2

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.

Accepted 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.

Question 3

Use the definition of the derivative to show that the derivative of f(x) = 5x − 2 is 5, and state what your working shows about the gradients of the chords of this graph.

Accepted Answer

The chord gradient is [5(x + h) − 2 − (5x − 2)]/h = 5h/h, which is 5 for every h ≠ 0, so the limit as h → 0 is also 5. B1 for the chord gradient simplifying to 5. The quotient reached its answer before the limit was taken, so every chord of this graph has gradient 5 whatever h is, however far apart its two ends are. B1 for that observation. The graph is a straight line and a chord of a straight line is a piece of the line itself, which is why the gradient of a straight line needs no limit.

Question 4

The curve C has equation y = 4 − x². Use the definition of the derivative to show, from first principles, that dy/dx = −2x. The gradient function is itself a straight line: write down its gradient, and state what that value says about the way the gradient of C changes as x increases.

Accepted Answer

The chord gradient from x to x + h is [4 − (x + h)² − (4 − x²)]/h. The two 4s cancel, and expanding the square leaves (−2xh − h²)/h, which is −2x − h for h ≠ 0. As h → 0 what survives is −2x. M1 for the chord gradient, A1 for −2x with the limit taken. The gradient function y = −2x is a straight line of gradient −2, so d²y/dx² = −2: the gradient of C falls by 2 for every unit x increases, at the same steady rate everywhere on the curve. B1 for −2 and what it says about the gradient. The second derivative is the rate of change of the gradient and not of y, so a constant −2 does not mean the curve itself falls steadily; at x = 0 the gradient is zero and the curve is momentarily level, while the gradient is still falling through zero there.

Question 5

Prove from first principles that the derivative of f(x) = x² is 2x.

Accepted 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 are needed: 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.

Question 6

Prove from first principles that the derivative of f(x) = 3x² + 2x is 6x + 2.

Accepted 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. Forgetting to expand 3(x + h)² fully is the usual source of a lost 6xh.

Question 7

Find, from first principles, the gradient of the curve y = x³ at the point (2, 8).

Accepted 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.

Question 8

For f(x) = x² − 5, write down the gradient of the chord joining the points on the curve where x = 2 and where x = 2 + h, and simplify it. Hence state the value of f′(2). Explain why the chord gradient you obtained is the same as the gradient of the matching chord of g(x) = x².

Accepted Answer

The chord joins (2, −1) to (2 + h, (2 + h)² − 5), so its gradient is [(2 + h)² − 5 − (−1)]/h. M1 for the chord gradient. The numerator is 4 + 4h + h² − 5 + 1 = 4h + h², so for h ≠ 0 the chord gradient is 4 + h. A1 for 4 + h. As h → 0 it tends to 4, so f′(2) = 4. A1 for f′(2) = 4. For g the numerator is g(2 + h) − g(2) = (2 + h)² − 4, which is the same 4h + h²: the −5 sits in both f(2 + h) and f(2) and subtracts away, so the two chords have equal gradients for every h. B1 for that reason. Subtracting a constant slides the whole curve down the y-axis without tilting it, and the definition reads the difference f(x + h) − f(x) rather than f itself, which is why no chord gradient and no derivative can notice the constant.

Question 9

A curve y = f(x) is decreasing for x < −1, has a minimum point at (−1, −4), is increasing for −1 < x < 2, has a maximum point at (2, 5), and is decreasing for x > 2. Sketch the graph of y = f′(x), giving the coordinates of the points where your sketch crosses the x-axis, and state the sign of f′(x) on each of the three intervals.

Accepted Answer

The zeros of the gradient function sit directly below the stationary points of the curve, so the sketch crosses the x-axis at (−1, 0) and (2, 0). M1 for putting the zeros of f′ under the stationary points of f, A1 for both coordinates. Between those values f is increasing, so f′(x) > 0 for −1 < x < 2, and outside them f is decreasing, so f′(x) < 0 for x < −1 and for x > 2. B1 for all three signs. The sketch is therefore one curve lying below the axis on the left, rising through (−1, 0), staying above the axis as far as (2, 0), and lying below the axis afterwards. B1 for a sketch that crosses the axis at each zero rather than touching it, since f′ changes sign at both. The heights −4 and 5 belong to f, not to f′. Nothing in the description says how steep f becomes, so the vertical scale of the sketch is not fixed; the zeros and the signs are what the information settles.

Question 10

The volume V litres of water in a tank t seconds after a valve is opened is modelled by V = 60 − 0.5t², for 0 ≤ t ≤ 10. Find the average rate of change of V over the first 4 seconds. Then use the definition of the derivative to find the rate at which V is changing at the instant t = 4. Give both answers with their units.

Accepted Answer

An average rate of change across an interval is the gradient of the chord across it. Here V(0) = 60 and V(4) = 60 − 8 = 52, so the average rate is (52 − 60)/4 = −2. M1 for the chord gradient across the four seconds, A1 for −2. For the instant, take the chord from t = 4 to t = 4 + h: [60 − 0.5(4 + h)² − 52]/h = (−8 − 4h − 0.5h² + 8)/h, which is −4 − 0.5h for h ≠ 0, and as h → 0 that tends to −4. M1 for the chord gradient from t = 4, A1 for −4 − 0.5h and its limit. A volume in litres divided by a time in seconds gives litres per second, so the two answers are −2 litres per second and −4 litres per second. B1 for the units on both. The two numbers differ because the tank is emptying faster as time goes on: −2 describes the four seconds taken together, and −4 describes the single instant t = 4. Only the second is a derivative.

Question 11

The gradient function y = f′(x) of a curve y = f(x) is a parabola with a minimum point at (1, −4), crossing the x-axis at x = −1 and at x = 3. State the values of x at which f has stationary points, state the interval on which f is decreasing, and state the value of x at which f″(x) = 0, giving a reason for that last answer.

Accepted Answer

A stationary point of f is a value of x at which f′ is zero, so f has stationary points at x = −1 and x = 3. B1 for both. Between those zeros the parabola dips below the axis, so f′(x) < 0 there and f is decreasing on −1 < x < 3. B1 for the interval. The second derivative f″ is the gradient of the graph of f′, and that graph has gradient zero at its own minimum point, so f″(x) = 0 at x = 1. B1 for x = 1, B1 for the reason. The value −4 is the least gradient the curve y = f(x) ever has, reached at x = 1.

Question 12

Prove from first principles that the derivative of f(x) = 1/x, x ≠ 0, is −1/x².

Accepted 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 subtraction has to happen inside one fraction before the division by h can cancel anything, and 1/(x + h) − 1/x is not 1/h.

Question 13

Prove from first principles that the derivative of f(x) = 1/x², x ≠ 0, is −2/x³.

Accepted Answer

The chord gradient is [1/(x + h)² − 1/x²]/h. Combine the two fractions over the common denominator x²(x + h)², giving [x² − (x + h)²]/[x²(x + h)²h]. The numerator is a difference of two squares, and factorising it as [x − (x + h)][x + (x + h)] gives −h(2x + h), so the whole quotient is −h(2x + h)/[x²(x + h)²h]. For h ≠ 0 the h cancels to leave −(2x + h)/[x²(x + h)²], and as h → 0 that tends to −2x/x⁴ = −2/x³. M1 for the chord gradient, M1 for combining over x²(x + h)², A1 for the numerator −h(2x + h), A1 for cancelling to −(2x + h)/[x²(x + h)²], A1 for the limit. As with 1/x, the subtraction has to happen inside one fraction before the division by h can cancel anything, and the point of the factorising is to manufacture the single factor of h that the cancellation needs. Both x² and (x + h)² must be kept in the denominator to the end: replacing (x + h)² by x² before the limit is taken assumes what the limit is there to decide. The result agrees with the power rule applied to x⁻², which the specification lets you quote elsewhere, but not in a proof from first principles.

Question 14

Prove from first principles that the derivative of f(x) = √x, x > 0, is 1/(2√x).

Accepted Answer

The chord gradient is [√(x + h) − √x]/h. Nothing cancels while the surds are subtracted, so multiply top and bottom by the conjugate √(x + h) + √x. The numerator becomes the difference of two squares, (x + h) − x = h, and the quotient is h/[h(√(x + h) + √x)]. For h ≠ 0 the h cancels, leaving 1/(√(x + h) + √x), and as h → 0 the first surd tends to √x, so the limit is 1/(2√x). M1 for the chord gradient, M1 for multiplying by the conjugate, A1 for the numerator h, A1 for cancelling to 1/(√(x + h) + √x), A1 for 1/(2√x). The conjugate is doing the same job the common denominator does for 1/x: it manufactures the single factor of h that the division by h can cancel against. The identity √(x + h) = √x + √h is false. The result agrees with the power rule applied to x^1/2, which the specification lets you quote elsewhere, but not in a proof from first principles.

Question 15

For f(x) = x², show that the gradient of the chord joining the points where x = a and x = b, with a ≠ b, is equal to f′(m), where m = (a + b)/2 is the midpoint of a and b. Show that the corresponding statement for g(x) = x³ is false, by finding the chord gradient and g′(m) in terms of a and b and giving one pair of values of a and b for which the two differ.

Accepted Answer

The chord gradient for f is (b² − a²)/(b − a), and factorising the difference of two squares gives (b − a)(b + a)/(b − a), which cancels to a + b because a ≠ b. M1 for the chord gradient, A1 for a + b. Since f′(x) = 2x, f′(m) = 2 × (a + b)/2 = a + b as well, so the two agree for every such pair. A1 for the comparison, which completes the proof. For g the chord gradient is (b³ − a³)/(b − a) = a² + ab + b², while g′(m) = 3(a + b)²/4. M1 for both expressions. Take a = 0 and b = 2: the chord gradient is 8/2 = 4 and g′(1) = 3, and 4 ≠ 3, so the statement fails. A1 for a pair of values with both numbers, B1 for the conclusion. Subtracting the two general expressions leaves (a − b)²/4, which is zero only when a = b, so for the cube the two never agree on an admissible pair; one counter-example is all that the question asks for.