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. A keeps the minus sign (forgets the negative power gives +); C mishandles the index; D wrongly adds +c (this is 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). A differentiates only x²; B and D differentiate e^x incorrectly 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. B has a sign slip in the numerator; C adds instead of subtracts; D forgets to square the denominator.

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.