Differentiating e^x

1.1 — y = e^4x + e^−x (2 marks)

Sample response. dy/dx = 4 e^4x + (−1) e^−x = 4 e^4x − e^−x.

Marking notes. 1 mark — chain-rule factor 4 on the first term. 1 mark — chain-rule factor −1 on the second term (correct sign). A response that gives "4 e^4x + e^−x" (missing the negative) scores 1/2.

1.2 — y = x³ e^2x (3 marks)

Sample response. Product rule with u = x³, v = e^2x: u' = 3x², v' = 2 e^2x.   dy/dx = 3x² e^2x + x³ · 2 e^2x = e^2x(3x² + 2x³) = x² e^2x(3 + 2x).

Marking notes. 1 mark — product rule with correct u', v' (including chain-rule factor 2 on v'). 1 mark — substitution into u'v + uv'. 1 mark — factorising to x² e^2x(3 + 2x). Common error: students forget the factor 2 on v' — costs 1 mark.

1.3 — Tangent to y = e^x² at x = 1 (4 marks)

Sample response. y at x = 1: y = e¹ = e.   dy/dx = 2x e^x² (chain rule).   At x = 1: dy/dx = 2(1) e¹ = 2e.   Tangent: y − e = 2e(x − 1) ⇒ y = 2ex − 2e + e ⇒ y = 2ex − e (m = 2e, c = −e).

Marking notes. 1 mark — y-value at x = 1 is e. 1 mark — derivative 2x e^x² via chain rule. 1 mark — gradient 2e at x = 1. 1 mark — final equation y = 2ex − e in exact form. Approximate values (5.44x − 2.72) score 3/4 — exact form is required.

2.1 — Extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

Part (a) — derivatives. Product rule on f(x) = x e^−x (u = x, v = e^−x; u' = 1, v' = −e^−x):

f'(x) = 1 · e^−x + x · (−e^−x) = e^−x(1 − x).   [1 mark.]

Differentiate again (product rule on e^−x · (1 − x)):

f''(x) = −e^−x(1 − x) + e^−x(−1) = e^−x[−(1 − x) − 1] = e^−x(x − 2).   [1 mark.]

Part (b) — stationary points and inflection.

Stationary point: f'(x) = 0 ⇒ e^−x(1 − x) = 0. Since e^−x > 0, we need 1 − x = 0 ⇒ x = 1.   y = 1 · e⁻¹ = 1/e. So stationary point at (1, 1/e).

Second-derivative test at x = 1: f''(1) = e⁻¹(1 − 2) = −1/e < 0 ⇒ local maximum.   [1 mark.]

Inflection point: f''(x) = 0 ⇒ e^−x(x − 2) = 0 ⇒ x = 2.   y = 2 · e⁻² = 2/e². For x < 2, f''(x) < 0 (concave down); for x > 2, f''(x) > 0 (concave up). Sign change confirmed, so inflection at (2, 2/e²).   [1 mark.]

Part (c) — limits.

As x → −∞: x → −∞ and e^−x = e^|x| → ∞. The product of a large negative number and a large positive number is a large negative number, so f(x) → −∞.   [1 mark.]

As x → +∞: e^−x decays exponentially, which beats any polynomial growth (x → ∞). Formally, x · e^−x = x/e^x → 0 since e^x grows faster than x. So f(x) → 0⁺, with horizontal asymptote y = 0.   [1 mark.]

Part (d) — sketch description. The graph rises from −∞ on the far left (steeply), passes through the origin (the only x-intercept, since f(x) = 0 ⇔ x = 0), continues upward to the local maximum at (1, 1/e), bends from concave down to concave up at the inflection (2, 2/e²), then decays toward the horizontal asymptote y = 0 as x → ∞. Labels on sketch: origin (0, 0); maximum (1, 1/e); inflection (2, 2/e²); asymptote y = 0.   [1 mark.]

Total: 7/7.

Band descriptors for marker.

Band 3: f'(x) found but not factored; finds x = 1 but does not classify; no second derivative. ≈ 2-3 marks.

Band 4: f' and f'' both correct; finds local maximum at x = 1 with second-derivative test; misses or mislocates the inflection. ≈ 4-5 marks.

Band 5: All algebra correct, both critical points classified; limits stated but no justification using "exponential beats polynomial". ≈ 5-6 marks.

Band 6: Both derivatives factorised; max and inflection both found and classified using second-derivative sign analysis; both limits justified with explicit "x · e^−x → 0" argument and horizontal asymptote y = 0 named; sketch labels all five features. 7/7.