Differentiation

Differentiate y = 3x⁴ − 5x² + 7 with respect to x.

• A12x³ − 10x✓
• B7x³ − 10x
• C12x³ − 5x
• D12x³ − 10x + 7

Answer: Differentiating term by term: d/dx(3x⁴) = 12x³, d/dx(−5x²) = −10x, and the constant vanishes, giving 12x³ − 10x.

Find the gradient of the curve y = x³ − 4x at the point where x = 2.

• A8✓
• B0
• C12
• D4

Answer: dy/dx = 3x² − 4, so at x = 2 the gradient is 3(4) − 4 = 12 − 4 = 8.

The curve y = x² − 8x + 3 has a stationary point. Find its x-coordinate.

• A4✓
• B−4
• C2
• D8

Answer: Setting dy/dx = 2x − 8 = 0 gives x = 4 at the stationary point.

Given y = 6/x = 6x⁻¹, find dy/dx.

• A−6x⁻²✓
• B−6x²
• C6x⁻²
• D6 ln x

Answer: Writing y = 6x⁻¹ and applying the power rule gives dy/dx = −6x⁻², i.e. −6/x².

Differentiate y = 4√x with respect to x. (Write √x as x^(1/2).)

• A4x^(−1/2)
• B2x^(1/2)
• C8x^(−1/2)
• D2x^(−1/2)✓

Answer: Write y = 4x^(1/2). By the power rule dy/dx = 4 × (1/2)x^(1/2 − 1) = 2x^(−1/2), i.e. 2/√x. The 4x^(−1/2) distractor forgets the 1/2 factor; 2x^(1/2) subtracts the index wrongly; 8x^(−1/2) doubles the coefficient.

Given y = 5x³ − 2/x², find dy/dx. (Write 2/x² as 2x^(−2).)

• A15x² − 4x^(−3)
• B15x² + 4x^(−3)✓
• C15x² − 4x^(−1)
• D15x² + 4x^(−3) + c

Answer: y = 5x³ − 2x^(−2). Then dy/dx = 15x² − 2(−2)x^(−3) = 15x² + 4x^(−3), i.e. 15x² + 4/x³. Writing 15x² − 4x^(−3) keeps the minus sign and misses that multiplying by the negative power makes the term positive; 15x² − 4x^(−1) reduces the index to −1 instead of −3; and adding + c belongs to integration, not differentiation.

Differentiate y = (3x − 1)⁵ using the chain rule.

• A5(3x − 1)⁴
• B3(3x − 1)⁴
• C15(3x − 1)⁴✓
• D15(3x − 1)⁵

Answer: Let u = 3x − 1, so y = u⁵ and dy/du = 5u⁴, du/dx = 3. By the chain rule dy/dx = 5(3x − 1)⁴ × 3 = 15(3x − 1)⁴. The 5(3x − 1)⁴ distractor forgets to multiply by du/dx = 3; 3(3x − 1)⁴ drops the power-5 factor; 15(3x − 1)⁵ fails to reduce the power.

Differentiate y = x² e^x using the product rule.

• A2x e^x
• Bx² e^x + 2x
• Ce^x(x² + 2x)✓
• D2x e^x + x²

Answer: With u = x² (u′ = 2x) and v = e^x (v′ = e^x), the product rule gives dy/dx = u′v + uv′ = 2x e^x + x² e^x = e^x(x² + 2x). Stopping at 2x e^x differentiates x² but drops the second product-rule term altogether, while 2x e^x + x² and x² e^x + 2x each lose the e^x from one of the two terms, treating it as 1.

Differentiate y = x/(x + 1) using the quotient rule.

• A1/(x + 1)²✓
• B−1/(x + 1)²
• C(2x + 1)/(x + 1)²
• D1/(x + 1)

Answer: With u = x (u′ = 1) and v = x + 1 (v′ = 1), the quotient rule gives (u′v − uv′)/v² = [1·(x + 1) − x·1]/(x + 1)² = [x + 1 − x]/(x + 1)² = 1/(x + 1)². Subtracting the numerator terms the wrong way round gives −1/(x + 1)²; adding rather than subtracting gives (2x + 1)/(x + 1)²; and leaving the denominator unsquared gives 1/(x + 1).

The curve y = x³ − 3x² + 4 has stationary points. Using the second derivative, classify the stationary point at x = 2.

• AMaximum
• BPoint of inflection
• CMinimum✓
• DCannot be determined

Answer: dy/dx = 3x² − 6x = 3x(x − 2), so stationary points are at x = 0 and x = 2. The second derivative is d²y/dx² = 6x − 6. At x = 2, d²y/dx² = 12 − 6 = 6 > 0, so the point is a minimum.

Find the equation of the tangent to the curve y = x² − 4x + 5 at the point (3, 2).

• Ay = 2x + 2
• By = −2x + 8
• Cy = 2x − 2
• Dy = 2x − 4✓

Answer: dy/dx = 2x − 4, so at x = 3 the gradient is 2(3) − 4 = 2. The tangent through (3, 2) is y − 2 = 2(x − 3), giving y = 2x − 4. The y = −2x + 8 distractor uses the wrong sign of gradient; y = 2x + 2 and y = 2x − 2 use a wrong intercept.

Find the equation of the normal to the curve y = x² at the point (1, 1).

• Ay = 2x − 1
• By = (1/2)x + 1/2
• Cy = −2x + 3
• Dy = −(1/2)x + 3/2✓

Answer: dy/dx = 2x gives a tangent gradient of 2 at x = 1, so the normal gradient is −1/2. The normal through (1, 1) is y − 1 = −(1/2)(x − 1), i.e. y = −(1/2)x + 3/2. The y = 2x − 1 distractor is the tangent; y = −2x + 3 uses gradient −2; y = (1/2)x + 1/2 uses the wrong sign for the perpendicular gradient.

For the function f(x) = x³ − 12x, determine the interval on which f is decreasing.

• Ax < −2
• B−2 < x < 2✓
• Cx > 2
• Dx < −2 or x > 2

Answer: f′(x) = 3x² − 12 = 3(x² − 4) = 3(x − 2)(x + 2). f is decreasing where f′(x) < 0, i.e. between the roots −2 and 2, so −2 < x < 2. The intervals x < −2, x > 2 and their union are where f′(x) > 0, so they describe where f is increasing.

The radius r of a circle increases at a rate of 0.2 cm/s. Find the rate of increase of the area A when r = 5 cm. (Use A = πr².)

• A2π cm²/s✓
• Bπ cm²/s
• C5π cm²/s
• D10π cm²/s

Answer: A = πr² gives dA/dr = 2πr. By the chain rule dA/dt = dA/dr × dr/dt = 2πr × 0.2. At r = 5, dA/dt = 2π(5)(0.2) = 2π cm²/s. Common errors are omitting the 0.2 factor, halving the result, or multiplying the area itself by dr/dt instead of differentiating first.

Differentiate y = 3x⁴ with respect to x.

• A12x³✓
• B3x⁵/5
• C36x²
• D3x⁴

Answer: Multiply by the power and reduce the power by one: d/dx(3x⁴) = 12x³. Giving the integral rather than the derivative — powers going up instead of down — is the usual slip.

Differentiate y = 5x³ − 2x with respect to x.

• A30x
• B5x⁴/4 − x²
• C15x² − 2✓
• D5x³ − 2x

Answer: Multiply by the power and reduce the power by one: d/dx(5x³ − 2x) = 15x² − 2. Giving the integral rather than the derivative — powers going up instead of down — is the usual slip.

Differentiate y = x⁵ + 7x² with respect to x.

• A20x³ + 14
• Bx⁶/6 + 7x³/3
• C5x⁴ + 14x✓
• Dx⁵ + 7x²

Answer: Multiply by the power and reduce the power by one: d/dx(x⁵ + 7x²) = 5x⁴ + 14x. Giving the integral rather than the derivative — powers going up instead of down — is the usual slip.

Differentiate y = (3x + 2)⁴.

• A4(3x + 2)³
• B12(3x + 2)³✓
• C3(3x + 2)³
• D12(3x + 2)⁴

Answer: Chain rule: bring the 4 down, reduce the power to 3, then multiply by the derivative of the bracket, which is 3. That gives 12(3x + 2)³. Forgetting to multiply by the derivative of the bracket is the single most common chain-rule error.

Differentiate y = (5x + 4)⁴.

• A20(5x + 4)⁴
• B4(5x + 4)³
• C5(5x + 4)³
• D20(5x + 4)³✓

Answer: Chain rule: bring the 4 down, reduce the power to 3, then multiply by the derivative of the bracket, which is 5. That gives 20(5x + 4)³. Forgetting to multiply by the derivative of the bracket is the single most common chain-rule error.

Differentiate y = (2x + 7)⁴.

• A2(2x + 7)³
• B4(2x + 7)³
• C8(2x + 7)³✓
• D8(2x + 7)⁴

Answer: Chain rule: bring the 4 down, reduce the power to 3, then multiply by the derivative of the bracket, which is 2. That gives 8(2x + 7)³. Forgetting to multiply by the derivative of the bracket is the single most common chain-rule error.

Find the x-coordinate of the stationary point of y = x² − 4x + 3.

• Ax = 2✓
• Bx = −2
• Cx = 4
• Dx = 1

Answer: Stationary points occur where dy/dx = 0. Here dy/dx = 2x − 4, so 2x = 4 and x = 2. Setting y rather than dy/dx to zero is a different question — that would find the roots, not the turning point.

Find the x-coordinate of the stationary point of y = x² − 9x + 3.

• Ax = 9
• Bx = −9/2
• Cx = 9/2✓
• Dx = 9/4

Answer: Stationary points occur where dy/dx = 0. Here dy/dx = 2x − 9, so 2x = 9 and x = 9/2. Setting y rather than dy/dx to zero is a different question — that would find the roots, not the turning point.

Find the x-coordinate of the stationary point of y = x² − 16x + 3.

• Ax = 16
• Bx = −8
• Cx = 8✓
• Dx = 4

Answer: Stationary points occur where dy/dx = 0. Here dy/dx = 2x − 16, so 2x = 16 and x = 8. Setting y rather than dy/dx to zero is a different question — that would find the roots, not the turning point.

Differentiate y = 3xe^(2x).

• A32e^(2x)
• B3e^(2x)(2x + 1)✓
• C3e^(2x)
• D3xe^(2x)(2)

Answer: Product rule with u = 3x and v = e^(2x): u′v + uv′ = 3e^(2x) + 3x·2e^(2x) = 3e^(2x)(2x + 1). Differentiating the two factors and multiplying the results is the classic misuse of the product rule — it is a sum, not a product.

Differentiate y = 5xeˣ.

• A5eˣ(x + 1)✓
• B5eˣ
• C5xeˣ
• D5eˣ(x − 1)

Answer: Product rule with u = 5x and v = eˣ: u′v + uv′ = 5eˣ + 5xeˣ = 5eˣ(x + 1). Differentiating the two factors and multiplying the results is the classic misuse of the product rule — it is a sum, not a product.

y = x³ − 6x has stationary points at x = −√(2) and x = √(2). Determine the nature of the point at x = √(2).

• AMinimum, because d²y/dx² > 0 there✓
• BMaximum, because d²y/dx² < 0 there
• CPoint of inflection, because d²y/dx² = 0 there
• DCannot be determined without a sketch

Answer: d²y/dx² = 6x. At x = √(2) this is 6√(2), which is positive, so the curve is concave up and the point is a minimum. A positive second derivative always means a minimum; the negative root gives a maximum by the same test.

y = x³ − 12x has stationary points at x = −2 and x = 2. Determine the nature of the point at x = 2.

• APoint of inflection, because d²y/dx² = 0 there
• BMaximum, because d²y/dx² < 0 there
• CMinimum, because d²y/dx² > 0 there✓
• DCannot be determined without a sketch

Answer: d²y/dx² = 6x. At x = 2 this is 12, which is positive, so the curve is concave up and the point is a minimum. A positive second derivative always means a minimum; the negative root gives a maximum by the same test.

What does dy/dx represent geometrically?

• AThe vertical distance from the curve down to the x-axis at that point
• BThe total area lying between the curve and the horizontal axis
• CThe gradient of the tangent at a point✓
• DThe average rate of change measured across the whole of the curve

Answer: The derivative is the instantaneous rate of change, which on a graph is the gradient of the tangent at that point. Area under the curve is integration, and the AVERAGE rate of change is a chord gradient, not a tangent.

If f′(a) = 0 and f″(a) = 0, what can be concluded about x = a?

• AIt is definitely a minimum point on the curve
• BIt is definitely a point of inflection where the curve changes its concavity
• CThe test is inconclusive✓
• DIt is definitely a maximum point on the curve

Answer: A zero second derivative means the test fails to decide, not that you have found an inflection. y = x⁴ has f′(0) = f″(0) = 0 and a minimum at the origin. You must examine the sign of f′ either side.