Tasks 04 - Differentiation

Mathematics for Master Students

These problems accompany Session 04 of the tutorial. Work through them at your own pace before the next session. We start Session 05 with your questions. On the website, every problem has a fully worked solution you can expand to check yourself. The PDF version contains only the problems, so you can print it and use it as a worksheet.

The problems are staged: the first two are warm-ups, the middle ones practice the core techniques, and the last two are at the level of the mock exam. Throughout, name the structure first (sum, product, quotient or chain) and the right rule follows automatically.

Problem 1: Power Rule Warm-Up

Differentiate each function. For (c) and (d), rewrite as a power first, differentiate, then translate back.

  1. \(f(x) = x^6\)
  2. \(g(x) = 4x^3 - 5x^2 + 2x - 11\)
  3. \(h(x) = \dfrac{3}{x}\)
  4. \(k(x) = \sqrt{x}\)

The power rule is \(\left(x^n\right)' = n x^{n-1}\): exponent down front, then reduce the exponent by one. Sums differentiate term by term, and constant multiples tag along.

  1. A single power: \[f'(x) = 6x^5\]

  2. Term by term; the constant \(-11\) has derivative \(0\): \[g'(x) = 4 \cdot 3x^2 - 5 \cdot 2x + 2 = 12x^2 - 10x + 2\]

  3. Rewrite \(\frac{3}{x} = 3x^{-1}\), then apply the power rule with \(n = -1\): \[h'(x) = 3 \cdot (-1) x^{-2} = -3x^{-2} = -\frac{3}{x^2}\]

  4. Rewrite \(\sqrt{x} = x^{1/2}\), then apply the power rule with \(n = \frac{1}{2}\): \[k'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}\] The exponent \(\frac{1}{2} - 1 = -\frac{1}{2}\) lands back under a root. Session 1’s exponent rules are what make roots and fractions ordinary powers.

Problem 2: Product, Quotient and Chain

Differentiate each function using the named rule.

  1. \(u(x) = (x^2 + 3)(2x - 1)\): product rule (then check by expanding first)
  2. \(v(x) = \dfrac{x - 1}{x + 2}\): quotient rule
  3. \(w(x) = (3x + 1)^5\): chain rule

  1. Product rule \((fg)' = f'g + fg'\) with \(f = x^2 + 3\) (so \(f' = 2x\)) and \(g = 2x - 1\) (so \(g' = 2\)): \[u'(x) = 2x \cdot (2x - 1) + (x^2 + 3) \cdot 2 = 4x^2 - 2x + 2x^2 + 6 = 6x^2 - 2x + 6\] Check by expanding first: \(u(x) = 2x^3 - x^2 + 6x - 3\), so \(u'(x) = 6x^2 - 2x + 6\). ✓ Same answer.

  2. Quotient rule \(\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}\) with \(f = x - 1\) (so \(f' = 1\)) and \(g = x + 2\) (so \(g' = 1\)): \[v'(x) = \frac{1 \cdot (x + 2) - (x - 1) \cdot 1}{(x + 2)^2} = \frac{x + 2 - x + 1}{(x + 2)^2} = \frac{3}{(x + 2)^2}\] Watch the minus sign: it acts on the whole second product, which is why the \(-x\) and \(+1\) appear.

  3. Chain rule \(\left(f(g(x))\right)' = f'(g(x)) \cdot g'(x)\) with outer \((\cdot)^5\) and inner \(3x + 1\): \[w'(x) = 5(3x + 1)^4 \cdot 3 = 15(3x + 1)^4\]

WarningCommon Mistake

The chain rule’s inner derivative is not optional. Stopping at \(5(3x+1)^4\) and forgetting the \(\cdot\,3\) is the classic slip. Always finish with “times the derivative of the inside”.

Problem 3: Exponentials and Logarithms

Differentiate. Remember \(\left(e^x\right)' = e^x\) and \(\left(\ln x\right)' = \frac{1}{x}\), combined with the chain rule where needed.

  1. \(f(x) = e^{4x}\)
  2. \(g(x) = \ln(x^2 + 1)\)
  3. \(h(x) = x^2 e^x\)
  4. \(m(x) = 5\ln x - 2e^x\)

  1. Chain rule with outer \(e^{(\cdot)}\) and inner \(4x\); the exponential is its own derivative: \[f'(x) = e^{4x} \cdot 4 = 4e^{4x}\]

  2. Chain rule with outer \(\ln(\cdot)\) (derivative \(\frac{1}{\cdot}\)) and inner \(x^2 + 1\) (derivative \(2x\)): \[g'(x) = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}\]

  3. Product rule with \(f = x^2\) (so \(f' = 2x\)) and \(g = e^x\) (so \(g' = e^x\)): \[h'(x) = 2x \cdot e^x + x^2 \cdot e^x = e^x\left(2x + x^2\right) = x e^x (2 + x)\]

  4. Term by term, constants tag along: \[m'(x) = 5 \cdot \frac{1}{x} - 2 e^x = \frac{5}{x} - 2e^x\]

Problem 4: Tangent Lines

The tangent line to \(f\) at \(x = a\) has slope \(f'(a)\) and passes through \((a, f(a))\), so its equation is \(\;y = f(a) + f'(a)(x - a)\). Consider \(f(x) = x^2 - 6x + 10\).

  1. Find \(f'(x)\), and the slope of the tangent at \(x = 1\) and at \(x = 4\).
  2. Find the equation of the tangent line at \(x = 4\), simplified to the form \(y = mx + b\).
  3. At which \(x\) is the tangent horizontal? What is special about that point?

  1. Differentiate the polynomial: \[f'(x) = 2x - 6\] Then \(f'(1) = 2 \cdot 1 - 6 = -4\) (the graph falls here) and \(f'(4) = 2 \cdot 4 - 6 = 2\) (the graph rises here).

  2. Two ingredients: a point and a slope. The point: \(f(4) = 16 - 24 + 10 = 2\), so \((4, 2)\). The slope: \(f'(4) = 2\). Plug into the tangent formula: \[y = f(4) + f'(4)(x - 4) = 2 + 2(x - 4) = 2x - 6\] Check: at \(x = 4\) the line gives \(y = 2 \cdot 4 - 6 = 2 = f(4)\). ✓ It touches the curve at the right height.

  3. The tangent is horizontal where its slope is zero: \[f'(x) = 2x - 6 = 0 \quad \Rightarrow \quad x = 3\] Here \(f(3) = 9 - 18 + 10 = 1\). This is the vertex: the lowest point of the upward parabola. A flat tangent marks a minimum (or maximum); Session 5 turns this observation into an optimization recipe.

Problem 5: Marginal Cost

A workshop’s monthly cost of producing \(q\) units is \[C(q) = 0.1q^2 + 15q + 5000 \quad \text{(in €)}.\]

  1. Find the marginal cost function \(C'(q)\).
  2. Compute \(C'(100)\) and state its units. Interpret the number in one sentence.
  3. Compute the exact cost of the 101st unit, \(C(101) - C(100)\), and compare it with your answer from (b).
  4. Is marginal cost increasing or decreasing? Justify with \(C''(q)\).

  1. Differentiate term by term; the fixed cost \(5000\) vanishes because it does not react to producing one unit more: \[C'(q) = 0.2q + 15\]

  2. Substitute \(q = 100\): \[C'(100) = 0.2 \cdot 100 + 15 = 20 + 15 = 35\] The units are € per unit, a rate, not a total cost. Reading: at an output of 100 units, cost is growing at about €35 per additional unit, so the 101st unit costs approximately €35 to produce.

  3. The exact extra cost of moving from 100 to 101 units: \[C(100) = 0.1 \cdot 100^2 + 15 \cdot 100 + 5000 = 1000 + 1500 + 5000 = 7500\] \[C(101) = 0.1 \cdot 101^2 + 15 \cdot 101 + 5000 = 1020.1 + 1515 + 5000 = 7535.1\] \[C(101) - C(100) = 35.1\] The marginal cost €35 approximates the true €35.10 to within 10 cents. One unit is a tiny nudge next to 100 produced, so the tangent slope is an excellent stand-in.

  4. The second derivative is \[C''(q) = 0.2 > 0,\] so marginal cost is increasing: each further unit is dearer than the one before, and the cost curve is convex.

WarningCommon Mistake

\(C'(100) = 35\) does not mean “producing 100 units costs €35”. That total is \(C(100) = 7500\), i.e. €7500. The derivative is a rate in € per unit: the cost of the next unit, not the whole batch.

Problem 6: Elasticity of Demand

Weekly demand for a courier’s express service is \(D(p) = 400 - 8p\), where \(p\) is the price in €. Recall the elasticity \(\;\varepsilon = \dfrac{p}{D(p)} \cdot D'(p)\).

  1. Compute \(D'(p)\), then the elasticity at \(p = 20\). Classify it as elastic or inelastic, and say what a small price increase does to revenue.
  2. Repeat at \(p = 35\).
  3. Find the price at which \(\varepsilon = -1\) (unit elastic). Confirm that revenue \(R(p) = p \cdot D(p)\) peaks there.

The demand function is linear, so \(D'(p) = -8\) (constant). Demand falls in price, so we expect a negative elasticity throughout.

  1. At \(p = 20\): \(D(20) = 400 - 160 = 240\), so \[\varepsilon = \frac{20}{240} \cdot (-8) = \frac{-160}{240} = -\frac{2}{3} \approx -0.67\] Since \(|\varepsilon| = 0.67 < 1\), demand is inelastic: customers barely react. A 1% price increase loses only about 0.67% of demand, so revenue rises.

  2. At \(p = 35\): \(D(35) = 400 - 280 = 120\), so \[\varepsilon = \frac{35}{120} \cdot (-8) = \frac{-280}{120} = -\frac{7}{3} \approx -2.33\] Since \(|\varepsilon| = 2.33 > 1\), demand is elastic: demand overreacts. A price increase now loses more than 2% of demand per 1% of price, so revenue falls: raising the price backfires.

  3. Set the elasticity to \(-1\) and solve:

    Step 1 - Set up: \(\dfrac{p}{400 - 8p} \cdot (-8) = -1\); multiplying both sides by \(-1\) gives \(\dfrac{8p}{400 - 8p} = 1\).

    Step 2 - Cross-multiply: \(8p = 400 - 8p\).

    Step 3 - Collect and solve: \(16p = 400\), so \(p = 25\). Confirm the revenue peak: \(R(p) = p(400 - 8p) = 400p - 8p^2\), so \(R'(p) = 400 - 16p = 0\) gives \(p = 25\), exactly the unit-elastic price. There \(D(25) = 200\) and revenue is \(R(25) = 25 \cdot 200 = 5000\), i.e. €5000. Below \(p = 25\) demand is inelastic and raising the price helps; above it demand is elastic and raising the price hurts; the switch at \(\varepsilon = -1\) is precisely where revenue is largest.

Problem 7: Name the Structure

These are exam-level: each needs two rules. Name the outer structure first, then differentiate.

  1. \(f(x) = x^3 e^{2x}\)
  2. \(g(x) = \sqrt{x^2 + 4}\)
  3. \(h(x) = \dfrac{\ln x}{x^2}\)

  1. Product of \(x^3\) and \(e^{2x}\); the second factor needs the chain rule, \(\left(e^{2x}\right)' = 2e^{2x}\): \[f'(x) = 3x^2 \cdot e^{2x} + x^3 \cdot 2e^{2x} = e^{2x}\left(3x^2 + 2x^3\right) = x^2 e^{2x}(3 + 2x)\]

  2. Chain rule: rewrite \(\sqrt{x^2 + 4} = (x^2 + 4)^{1/2}\); outer \((\cdot)^{1/2}\), inner \(x^2 + 4\) (derivative \(2x\)): \[g'(x) = \frac{1}{2}(x^2 + 4)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 4}}\]

  3. Quotient rule with \(f = \ln x\) (so \(f' = \frac{1}{x}\)) and \(g = x^2\) (so \(g' = 2x\)): \[h'(x) = \frac{\frac{1}{x} \cdot x^2 - \ln x \cdot 2x}{\left(x^2\right)^2} = \frac{x - 2x\ln x}{x^4} = \frac{1 - 2\ln x}{x^3}\] The last step cancels one factor of \(x\) from top and bottom.

    Check: at \(x = 1\) (where \(\ln 1 = 0\)) the unsimplified form gives \(\frac{1 - 0}{1} = 1\) and the simplified form gives \(\frac{1 - 0}{1} = 1\): the cancellation was safe.

Problem 8: Growth and Curvature

A warehouse’s stored volume grows continuously as \(V(t) = 200\,e^{0.1t}\) (in m³ after \(t\) weeks).

  1. Find \(V'(t)\) and \(V''(t)\).
  2. Show that \(V'(t) = 0.1 \cdot V(t)\), and say what this means in words.
  3. Compute the growth rate at \(t = 0\) and at \(t = 10\) (you may leave the second answer as a multiple of \(e\), then give a decimal).
  4. Is \(V\) convex or concave? What does that say about how the volume grows?

  1. Each derivative uses the chain rule on \(e^{0.1t}\) (inner derivative \(0.1\)): \[V'(t) = 200 \cdot 0.1\,e^{0.1t} = 20\,e^{0.1t}, \qquad V''(t) = 20 \cdot 0.1\,e^{0.1t} = 2\,e^{0.1t}\]

  2. Factor \(V'\) back toward \(V\): \[V'(t) = 20\,e^{0.1t} = 0.1 \cdot \left(200\,e^{0.1t}\right) = 0.1 \cdot V(t)\] The growth is proportional to the current level: at every instant the volume grows at 10% of itself per week. That is the defining property of exponential growth: this is why \(e\) rules growth models.

  3. Substitute: \[V'(0) = 20\,e^{0} = 20 \ \text{m³ per week}, \qquad V'(10) = 20\,e^{1} = 20e \approx 54.4 \ \text{m³ per week}\] The stock is filling more than twice as fast at \(t = 10\) as at the start.

    Check: part (b) predicts the rate as 10% of the current level, and indeed \(0.1 \cdot V(10) = 0.1 \cdot 200e \approx 54.4\). ✓

  4. Since \(V''(t) = 2\,e^{0.1t} > 0\) for every \(t\), the function is convex everywhere. The volume does not just grow, its growth accelerates: each week adds more than the last, and the warehouse fills ever faster.