Questions on the product, quotient and chain rules

Question 1

Differentiate y = (2x + 1)⁴.

Accepted Answer

Differentiate the outside first, leaving the inside untouched, to get 4(2x + 1)³, then multiply by the inside's derivative, 2. So dy/dx = 8(2x + 1)³. M1 A1.

Question 2

Differentiate y = e^x².

Accepted Answer

The outside is e to something, which is its own derivative, and the inside x² contributes 2x. So dy/dx = 2x e^x². M1 A1.

Question 3

Using the derivatives given in the formulae booklet, differentiate y = 3 sec x, and y = cot 2x.

Accepted Answer

The booklet gives the derivative of sec x as sec x tan x, and the constant 3 is carried through unchanged, so 3 sec x differentiates to 3 sec x tan x. B1. It gives the derivative of cot x as −cosec² x, and the inner function 2x contributes a factor of 2 through the chain rule, so cot 2x differentiates to −2 cosec² 2x. B1. Each result holds where the function is defined: sec x and tan x need cos x ≠ 0, and cot 2x and cosec 2x need sin 2x ≠ 0.

Question 4

Differentiate y = ln (4x + 1) for x > −¼.

Accepted Answer

The outer function is a logarithm, whose derivative is 1 over whatever is inside, and the inner function 4x + 1 contributes 4. Multiplying the two rates gives dy/dx = 4/(4x + 1). M1 for the chain rule with both stages, A1 for the answer. The restriction x > −¼ keeps 4x + 1 positive, which is where ln is defined. An answer of 1/(4x + 1) has left out the inner derivative, and one of 4/x has differentiated the wrong function as the outer one.

Question 5

Differentiate y = (x² − 3)⁶, and find the coordinates of the stationary points of the curve.

Accepted Answer

With u = x² − 3, dy/du = 6u⁵ and du/dx = 2x, so dy/dx = 12x(x² − 3)⁵. Setting this to zero gives x = 0 or x² = 3, so x = 0, √3 and −√3. The corresponding y values are (−3)⁶ = 729 at x = 0 and 0 at both x = ±√3, so the stationary points are (0, 729), (√3, 0) and (−√3, 0). M1 for the chain rule, A1 for the derivative, M1 for solving, A1 for all three x values, A1 for the points. Setting only the bracket to zero misses x = 0 entirely, and a product is zero when either factor is.

Question 6

Differentiate y = x³e^2x.

Accepted Answer

Product rule with u = x³ and v = e^2x: dy/dx = 3x²e^2x + 2x³e^2x, which factorises as x²e^2x(3 + 2x). M1 for the rule, A1 for the two terms, A1 for the factorised form.

Question 7

Differentiate y = x sin x, and find the gradient of the curve at x = π.

Accepted Answer

Product rule: dy/dx = 1 × sin x + x × cos x = sin x + x cos x. At x = π, sin π = 0 and cos π = −1, so the gradient is 0 + π(−1) = −π. M1 for the rule, A1 for the derivative, A1 for −π. Differentiating each factor and multiplying the results would give cos x on its own, and that is wrong.

Question 8

Differentiate y = (2x + 1)/(x² + 1), and find the gradient of the curve at x = 0.

Accepted Answer

Quotient rule with u = 2x + 1 and v = x² + 1: dy/dx = [2(x² + 1) − (2x + 1)(2x)]/(x² + 1)², which simplifies to (−2x² − 2x + 2)/(x² + 1)². At x = 0 the gradient is 2/1 = 2. M1 for the rule, A1 for the numerator, A1 for the simplification, A1 for the value. Swapping the two products in u′v − uv′ flips the sign of the answer.

Question 9

Differentiate y = sin³ x.

Accepted Answer

Read the expression as (sin x)³. The outside gives 3(sin x)² and the inside gives cos x, so dy/dx = 3 sin² x cos x. M1 A1.

Question 10

Differentiate y = cosec 3x, and find the exact gradient of the curve at the point where x = π/4.

Accepted Answer

The booklet gives the derivative of cosec x as −cosec x cot x, and the chain rule supplies a factor of 3 for the inner 3x, so dy/dx = −3 cosec 3x cot 3x. M1 for the booklet derivative with the chain factor, A1 for the derivative. At x = π/4 the angle 3x is 3π/4, where sin (3π/4) = √2/2 and cos (3π/4) = −√2/2, so cosec (3π/4) = 2/√2 = √2 and cot (3π/4) = −1. M1 for both exact values. The gradient is therefore −3 × √2 × (−1) = 3√2. A1.

Question 11

Differentiate y = x²√(2x − 1) for x > ½, giving your answer as a single fraction.

Accepted Answer

The expression is a product whose second factor is itself a composite, so both rules are needed. Product rule with u = x² and v = (2x − 1)^1/2. M1 for the product rule. The chain rule gives v′ = ½(2x − 1)^−1/2 × 2 = (2x − 1)^−1/2, the inner derivative 2 cancelling the ½. M1 for the chain rule on the root. So dy/dx = 2x(2x − 1)^1/2 + x²(2x − 1)^−1/2. A1 for both terms. Put them over the common denominator √(2x − 1): the first becomes 2x(2x − 1)/√(2x − 1), so dy/dx = [2x(2x − 1) + x²]/√(2x − 1) = (4x² − 2x + x²)/√(2x − 1). M1 for combining over one denominator. Collecting the numerator gives x(5x − 2)/√(2x − 1). A1. The restriction x > ½ keeps 2x − 1 positive, which both the square root and the negative power require.

Question 12

Given that y = x² ln 3x for x > 0, show that dy/dx = x(2 ln 3x + 1). Given also that z = (ln 3x)/x² for x > 0, use the quotient rule to show that dz/dx = (1 − 2 ln 3x)/x³.

Accepted Answer

Product rule with u = x² and v = ln 3x. Note that ln 3x = ln 3 + ln x, so its derivative is 1/x. Then dy/dx = 2x ln 3x + x² × 1/x = 2x ln 3x + x = x(2 ln 3x + 1), as required. M1 for the product rule, A1 for its two terms, A1 for the printed form. For z the same two functions change places: with u = ln 3x and v = x², the quotient rule gives dz/dx = [x² × 1/x − ln 3x × 2x]/(x²)². M1 for the quotient rule with u and v the right way round. The numerator is x − 2x ln 3x = x(1 − 2 ln 3x), and the denominator is x⁴. A1 for the numerator. Cancelling one factor of x, which x > 0 permits, gives dz/dx = (1 − 2 ln 3x)/x³. A1 for the printed form. The derivative of ln 3x is 1/x rather than 1/(3x). The chain rule gives 3/(3x), and the threes cancel. In the quotient rule the numerator is v du/dx − u dv/dx, in that order: writing it the other way round turns the answer's sign over, and here would print 2 ln 3x − 1 on top.

Question 13

By writing tan x as sin x/cos x, use the quotient rule to show that the derivative of tan x is sec² x.

Accepted Answer

With u = sin x and v = cos x, the quotient rule gives [cos x × cos x − sin x × (−sin x)]/cos² x = (cos² x + sin² x)/cos² x. The numerator is 1 by the Pythagorean identity, so the derivative is 1/cos² x = sec² x. M1 for the quotient rule, A1 for the collected numerator, A1 for the conclusion. The double negative in −sin x × (−sin x) is what makes the numerator add rather than cancel, and losing it produces (cos² x − sin² x)/cos² x, which is cos 2x sec² x and goes nowhere.

Question 14

Given that u and v are differentiable functions of x, with v ≠ 0, use the product rule and the chain rule to prove that the derivative of u/v is (u′v − uv′)/v².

Accepted Answer

Write the quotient as a product with a negative power: u/v = uv⁻¹, which needs v ≠ 0 before it can be written at all. M1 for the rewrite. The product rule then gives the derivative as u′v⁻¹ + u × (the derivative of v⁻¹). M1 for the product rule. That second piece is a composite: the outer function is a reciprocal, and t⁻¹ differentiates to −t⁻², while the inner function v contributes v′, so v⁻¹ differentiates to −v⁻²v′. M1 for the chain rule on v⁻¹. Substituting, the derivative is u′v⁻¹ − uv′v⁻², that is u′/v − uv′/v². A1. Putting both terms over v² gives (u′v − uv′)/v². A1, which completes the proof.

Question 15

Given that y = (3x + 1)⁵(2x − 7)³, show that dy/dx = 3(3x + 1)⁴(2x − 7)²(16x − 33), and find the three values of x at which the tangent to the curve is horizontal.

Accepted Answer

Product rule with u = (3x + 1)⁵ and v = (2x − 7)³, each factor needing a chain rule of its own: u′ = 15(3x + 1)⁴ and v′ = 6(2x − 7)². M1 for the product rule with the chain rule inside both factors. So dy/dx = 15(3x + 1)⁴(2x − 7)³ + 6(3x + 1)⁵(2x − 7)². A1 for both terms. Both carry (3x + 1)⁴, (2x − 7)² and a factor of 3, so take those out: dy/dx = 3(3x + 1)⁴(2x − 7)²[5(2x − 7) + 2(3x + 1)]. M1 for the common factor. The bracket collects to 10x − 35 + 6x + 2 = 16x − 33, which is the printed form. A1. A horizontal tangent needs dy/dx = 0, and a product vanishes when one of its factors does, so solve 3x + 1 = 0, 2x − 7 = 0 and 16x − 33 = 0. M1 for setting each factor to zero. The three values are x = −1/3, x = 7/2 and x = 33/16. A1. Take the lower power of each repeated bracket outside, together with the numerical factor; expanding instead gives a polynomial of degree seven with all three roots buried in it.