Mathematics Advanced • Year 11 • Module 4 • Lesson 11
Differentiating ln x
Apply derivatives of ln x to optimisation, tangent and rate-of-change problems in real contexts.
Problem 1 — Sound level (logarithmic scale)
The sound level in decibels for an intensity I (in W/m²) above the threshold I₀ is modelled by
L(I) = 10 · ln(I / I₀) / ln 10, for I > 0
Set up: What are we solving for?
(i) Show that L can be rewritten as L(I) = (10 / ln 10) · ( ln I − ln I₀ ). 1 mark
(ii) Differentiate L with respect to I and simplify dL/dI as a single fraction. 2 marks
(iii) Use your derivative to explain in one sentence why doubling I produces the same increase in decibels, no matter what value I started at. 2 marks
Stuck? Use ln(2I) − ln(I) = ln 2 to find the change in L.Problem 2 — pH and rate of change
For a solution with hydrogen-ion concentration H (in mol/L), pH is defined by
pH(H) = − ln H / ln 10, for H > 0
Set up: What are we solving for?
(i) Find d(pH)/dH and simplify. 2 marks
(ii) Evaluate d(pH)/dH at H = 10⁻⁴ mol/L (the rough acidity of orange juice). Give your answer to 2 significant figures. 2 marks
(iii) Interpret the sign of d(pH)/dH in one sentence (what happens to pH if H increases slightly?). 1 mark
Problem 3 — Minimum of y = x ln x
A simple model for entropy per unit mass uses S(x) = x ln x for x > 0, where x is the fraction of a substance in one of two states.
Set up: What are we solving for?
(i) Differentiate S(x) using the product rule. 2 marks
(ii) Find the x-value at which S is stationary, giving an exact answer. 2 marks
(iii) Compute S at the stationary point and use the sign of S'(x) on either side to classify the point as a minimum or maximum. 3 marks
Stuck? Revisit lesson § Worked example 3 — stationary point of y = x ln x.Problem 4 — Tangent to y = ln x
A surveyor wants the tangent line to the curve y = ln x at the point where x = e².
Set up: What are we solving for?
(i) State the y-coordinate at x = e² and find dy/dx at that point. 2 marks
(ii) Write the equation of the tangent in point-gradient form, then simplify to y = mx + c. 3 marks
(iii) Find the x-intercept of the tangent line and comment on its position relative to the original point. 2 marks
Problem 5 — Maximum information density
In information theory the function I(x) = (ln x) / x, for x > 0, measures a normalised information score. The publisher wants the x that maximises I.
Set up: What are we solving for?
(i) Differentiate I(x) using the quotient rule and simplify the numerator. 2 marks
(ii) Solve I'(x) = 0 to find the stationary x-value (exact form). 2 marks
(iii) Use the sign of I'(x) on either side of the stationary point to confirm it is a maximum, and give the exact maximum value of I. 3 marks
Stuck? Revisit lesson § SAQ 3 — stationary point of y = (ln x)/x.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Sound level
Set up. We test how decibels change with intensity, then use the rate to explain a "constant gain per doubling" property.
(i) L(I) = (10 / ln 10) · ln(I/I₀) = (10 / ln 10) · ( ln I − ln I₀ ) by the log-quotient law. ✓
(ii) Only the ln I term depends on I, so dL/dI = (10 / ln 10) · (1/I) = 10 / (I · ln 10).
(iii) Doubling I to 2I changes L by L(2I) − L(I) = (10/ln 10) · ( ln(2I) − ln I ) = (10/ln 10) · ln 2 ≈ 3.01 dB, independent of the starting I. The rate dL/dI shrinks as I grows, but the integral over a doubling is the constant 10 ln 2 / ln 10.
Problem 2 — pH
Set up. We differentiate the pH–concentration relation and interpret the sign physically.
(i) d(pH)/dH = −(1/ln 10) · (1/H) = −1 / (H · ln 10).
(ii) At H = 10⁻⁴: d(pH)/dH = −1 / (10⁻⁴ · ln 10) = −10⁴ / 2.3026 ≈ −4 300 (per mol/L) (≈ −4.3 × 10³).
(iii) The derivative is negative, so increasing H slightly decreases pH (the solution gets more acidic).
Problem 3 — Minimum of S(x) = x ln x
Set up. We find the stationary x of S and classify it.
(i) Product rule with u = x, v = ln x: S'(x) = 1·ln x + x·(1/x) = ln x + 1.
(ii) S'(x) = 0 ⇒ ln x = −1 ⇒ x = 1/e (≈ 0.368).
(iii) S(1/e) = (1/e)·ln(1/e) = (1/e)·(−1) = −1/e (≈ −0.368). For 0 < x < 1/e, ln x < −1, so S'(x) < 0 (decreasing). For x > 1/e, ln x > −1, so S'(x) > 0 (increasing). The point is a minimum.
Problem 4 — Tangent at x = e²
Set up. We need the point, the gradient, the tangent equation, and its x-intercept.
(i) y = ln(e²) = 2. dy/dx = 1/x, so at x = e² the gradient is 1/e² (≈ 0.135).
(ii) Point-gradient: y − 2 = (1/e²)(x − e²). Expand: y = x/e² − 1 + 2 = x/e² + 1.
(iii) x-intercept: 0 = x/e² + 1 ⇒ x = −e² (≈ −7.39). The intercept sits to the left of the y-axis, even though the original point (e², 2) is well to the right — the tangent is nearly flat, so it travels a long way before reaching y = 0.
Problem 5 — Maximum of I(x) = (ln x)/x
Set up. We use the quotient rule, locate the stationary point, and verify it is a maximum.
(i) Quotient rule with u = ln x, v = x: I'(x) = [(1/x)·x − ln x · 1] / x² = (1 − ln x) / x².
(ii) x² ≠ 0, so I'(x) = 0 ⇔ 1 − ln x = 0 ⇔ ln x = 1 ⇔ x = e.
(iii) For 0 < x < e, ln x < 1, so I'(x) > 0 (increasing). For x > e, ln x > 1, so I'(x) < 0 (decreasing). Hence x = e is a maximum. Max value: I(e) = (ln e)/e = 1/e (≈ 0.368).