Derivatives, differentiation rules and the chain rule in Mathematik (CBK) at WU

Differential calculus is the third of the nine topics in the syllabus of Mathematik in the CBK at WU Vienna; the syllabus lists no subtopics for it.¹ The textbook by Birgit Rudloff, who leads the course, and Achim Zeileis has a third chapter titled "Differentialrechnung" (differential calculus). Its first two sections explain what a derivative is and give the rules for calculating one. Marginal cost, elasticity and optimisation are all built on them.

The typical mistakes in differentiating are small omissions: a forgotten inner derivative, a missing term of the product rule, a lost minus sign. The syllabus points you to the forum on Canvas for questions about exercises, and to the walk-in tutorial, where you work through problems with a tutor on hand.¹ German terms are given in brackets.

What a derivative is: Mathematik (CBK)

The average change of a function $f$ per unit between $x_0$ and $x_0+h$ is the difference quotient (Differenzenquotient) $\frac{f(x_0+h)-f(x_0)}{h}$. Geometrically it is the slope of the secant, the straight line through the two points on the curve. As $h$ goes to 0, the second point slides towards the first and the secant turns into the tangent. The limit as $h$ goes to 0, written ${\lim }_{h\to 0}$, is the derivative (Ableitung) at $x_0$:

$$f^{\prime }(x_0)=\underset{h\to 0}{\lim }\frac{f(x_0+h)-f(x_0)}{h}$$

Take $f(x)=4\sqrt{x}$ at $x_0=4$, where $f(4)=8$. With $h=5$ the secant reaches $f(9)=12$ and has slope $(12-8)/5=0.8$. With $h=1$ the slope is $4\sqrt{5}-8\approx 0.944$, with $h=0.1$ about $0.994$ and with $h=0.01$ about $0.999$. The smaller $h$ gets, the closer the secant lies to the tangent at the point $(4,8)$, and its slope approaches 1.

Our calculation

Numbers from the article’s example (our own numbers, not an exam question): f(x) = 4 · √x at x₀ = 4, so f(4) = 4 · 2 = 8. The second point is Q = (4 + h, f(4 + h)).

The slope of the secant through P and Q is the difference quotient (f(4 + h) − 8) / h, rounded here to three decimals:

h	5	1	0.1	0.01
slope	0.8	0.944	0.994	0.999

For instance (12 − 8) / 5 = 0.8 and (4 · √5 − 8) / 1 ≈ 0.944. The secant for h = 5 ends at Q = (9, 12).

Tangent: f′(x) = 4 · 1 / (2√x) = 2 / √x, so f′(4) = 2 / 2 = 1. The tangent at P is y = 8 + 1 · (x − 4) = x + 4. The x axis runs to 9, the y axis to 13.

That 1 is the derivative $f^{\prime }(4)$. Two rules from the next section, the power rule and the constant-factor rule, give it without any limit: $f^{\prime }(x)=4\cdot \frac{1}{2\sqrt{x}}=\frac{2}{\sqrt{x}}$, so $f^{\prime }(4)=2/2=1$. The limit tells you what the number means: if $x$ grows by a small $\Delta x$, $f$ changes by roughly $f^{\prime }(x_0)\cdot \Delta x$, so with $f^{\prime }(4)=1$ by about as much as $x$. The sign of the derivative also shows whether $f$ rises or falls (Monotonie): where $f^{\prime }(x)>0$ on an interval, $f$ is increasing there; where $f^{\prime }(x)<0$, it is decreasing. So $4\sqrt{x}$ increases for every $x>0$, because $2/\sqrt{x}$ is positive there.

Power, sum and product rules: Mathematik (CBK)

The power rule $(x^n{)}^{\prime }=n\cdot x^{n-1}$ holds for any real exponent (for negative exponents where $x\ne 0$, for non-integer ones where $x>0$), so it also covers roots and reciprocals such as $1/x$ once you write them as powers: $\sqrt{x}=x^{1/2}$ has the derivative $\frac{1}{2}{x}^{-1/2}=\frac{1}{2\sqrt{x}}$, and $\frac{1}{x}=x^{-1}$ has the derivative $-x^{-2}=-\frac{1}{x^2}$. The minus sign comes from the exponent. If it gets lost in the rewriting, the result is a positive derivative, as if $1/x$ were increasing, although it falls.

A constant factor stays where it is, $(c\cdot f{)}^{\prime }=c\cdot f^{\prime }$, and a sum is differentiated term by term. A constant term disappears, because it does not change. So $f(x)=3x^4-2x+7$ has the derivative $f^{\prime }(x)=12x^3-2$.

For a product of two functions $u$ and $v$ the product rule (Produktregel) applies:

$$(u\cdot v{)}^{\prime }=u^{\prime }\cdot v+u\cdot v^{\prime }$$

Simply multiplying the derivatives, $u^{\prime }\cdot v^{\prime }$, cannot be right, as $x^2\cdot x^3=x^5$ shows: the derivative is $5x^4$, but the product of the derivatives is $2x\cdot 3x^2=6x^3$. Just as easily, one of the two terms goes missing. Before calculating, write $u$, $u^{\prime }$, $v$ and $v^{\prime }$ in a column, and only then put them together. Fractions have their own quotient rule, but you can also write $u/v$ as the product $u\cdot v^{-1}$ and use the product and chain rules.

Chain rule, e^x and ln: Mathematik (CBK)

When one function sits inside another, as in $(3x+1{)}^5$ or $e^{0.2x}$, you need the chain rule (Kettenregel). Differentiate the outer function $u$, put the inner function $g(x)$ into it, and multiply by $g^{\prime }(x)$, the inner derivative (innere Ableitung):

$$(u(g(x)){)}^{\prime }=u^{\prime }(g(x))\cdot g^{\prime }(x)$$

So $((3x+1{)}^5{)}^{\prime }=5\cdot (3x+1{)}^4\cdot 3$, and the $3$ at the end is the inner derivative.

The exponential function is its own derivative, $(e^x{)}^{\prime }=e^x$, and the natural logarithm has the derivative $(\ln x{)}^{\prime }=1/x$ for $x>0$. With the chain rule these give the two forms you use in questions:

$$(e^{g(x)}{)}^{\prime }=e^{g(x)}\cdot g^{\prime }(x),(\ln g(x){)}^{\prime }=\frac{g^{\prime }(x)}{g(x)}$$

In the derivative of $e^{0.3x^2-2x}$ the exponent stays up top unchanged, and its derivative joins as a factor: $e^{0.3x^2-2x}\cdot (0.6x-2)$. Writing the exponent itself as the factor would mean using the inner function where the inner derivative belongs. With a negative exponent, the minus sign comes along: $(e^{-0.2x}{)}^{\prime }=-0.2\cdot e^{-0.2x}$. Drop it, and a decreasing function gets a positive derivative; monotonicity is therefore a quick check on the sign. For the logarithm, the inner derivative goes on top and the inner function below. You can test this on $(\ln (5x){)}^{\prime }=5/(5x)=1/x$: since $\ln (5x)=\ln 5+\ln x$, the constant $\ln 5$ disappears when you differentiate.

One task with product and chain rule: Mathematik (CBK)

Find the derivative of

$$f(x)=(4x-3)\cdot e^{0.25x-0.5}$$

at $x=2$. This is a product with $u=4x-3$ and $v=e^{0.25x-0.5}$. So $u^{\prime }=4$, and $v^{\prime }$ needs the chain rule: $v^{\prime }=e^{0.25x-0.5}\cdot 0.25$. The product rule gives

$$f^{\prime }(x)=4\cdot e^{0.25x-0.5}+(4x-3)\cdot 0.25\cdot e^{0.25x-0.5}$$

Only now, with the derivative complete, do you substitute the point; a number put in earlier would be a constant with derivative 0. At $x=2$ the exponent is $0.5-0.5=0$, so $e^0=1$, and $4x-3=5$. That makes $f^{\prime }(2)=4\cdot 1+5\cdot 0.25\cdot 1=5.25$.

Without the inner derivative $0.25$ the result is $4+5=9$, with the product of the derivatives $4\cdot 0.25=1$, and with only one term of the product rule 4 or $1.25$. To check, factor out $e^{0.25x-0.5}$: $f^{\prime }(x)=e^{0.25x-0.5}\cdot (x+3.5)$, and the bracket gives $5.25$ again at $x=2$.

Practice questions on derivatives: Mathematik (CBK)

The exam consists of 20 multiple-choice questions, each with 5 options of which exactly one is correct. Each correct answer is worth 1 point; of the wrong ones, the first 3 are free and each further one costs 0.25 points. You need 11 points to pass.¹

We wrote the three questions for this article in that format. The functions, numbers and wording are our own; none of them comes from a WU exam or from the textbook. Every wrong option is the result of a typical mistake, most of them the ones described above. Each solution shows the working, the typical mistake and how to rule options out.

Practice question 1Differential calculus

For f(x) = 5 · e^(0.2x² + x − 4), the derivative f′(3) (rounded) equals

1. a)4.90
2. b)8.90
3. c)11.13
4. d)13.35
5. e)24.48

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. Inner function g(x) = 0.2x² + x − 4, inner derivative g′(x) = 0.4x + 1. Chain rule: f′(x) = 5 · e^(g(x)) · g′(x).
2. At x = 3: g(3) = 1.8 + 3 − 4 = 0.8 and g′(3) = 1.2 + 1 = 2.2.
3. e^0.8 ≈ 2.2255409, so f′(3) = 5 · 2.2255409 · 2.2 ≈ 24.48.

a)

Wrong. The constant factor 5 forgotten: e^0.8 · 2.2 ≈ 4.90.

0 or −0.25

b)

Wrong. Multiplied by the inner function instead of the inner derivative: 5 · e^0.8 · 0.8 ≈ 8.90.

0 or −0.25

c)

Wrong. The inner derivative forgotten: 5 · e^0.8 ≈ 11.13. That is exactly the function value f(3).

0 or −0.25

d)

Wrong. The inner derivative taken without the term +1, so g′(x) = 0.4x instead of 0.4x + 1: 5 · e^0.8 · 1.2 ≈ 13.35.

0 or −0.25

e)

Right. Chain rule: f′(3) = 5 · e^0.8 · 2.2 ≈ 24.48.

+1

–

Left blank.

0

Typical mistake. The inner derivative. For e^(g(x)) the factor g′(x) always comes in; only when g′(x) = 1, as for e^(x + 5), does the derivative look like the function. Here g′(3) = 2.2. Without that factor you get 5 · e^0.8 ≈ 11.13, which is just the function value f(3).

Ruling out before you finish. By the chain rule f′(x) = f(x) · g′(x), with the exponent g(x) = 0.2x² + x − 4 and g′(x) = 0.4x + 1. For every positive x, g′(x) is greater than 1, so f′(3) is greater than f(3) = 5 · e^0.8 ≈ 11.13. That rules out options a, b and c and leaves d and e.

A guess among the 2 left has a 1-in-2 chance. It is worth 0.5 points on average while your three free wrong answers last, and 0.375 after that.

Practice question 2Differential calculus

For f(x) = (x² + 3) · e^(0.5x), the derivative f′(2) (rounded) equals

1. a)5.44
2. b)9.51
3. c)19.03
4. d)20.39
5. e)29.90

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. A product with u = x² + 3, u′ = 2x, and v = e^(0.5x), v′ = 0.5 · e^(0.5x) (chain rule, inner derivative 0.5).
2. f′(x) = 2x · e^(0.5x) + (x² + 3) · 0.5 · e^(0.5x). At x = 2, u = 7 and e^(0.5 · 2) = e ≈ 2.7182818.
3. f′(2) = 4e + 7 · 0.5 · e = 7.5e = 7.5 · 2.7182818 ≈ 20.39.

a)

Wrong. The derivatives multiplied instead of using the product rule: 2x · 0.5 · e^(0.5x) at x = 2, so 2e ≈ 5.44.

0 or −0.25

b)

Wrong. The first term of the product rule lost, only (x² + 3) · 0.5 · e^(0.5x) calculated: 7 · 0.5 · e = 3.5e ≈ 9.51.

0 or −0.25

c)

Wrong. The function value calculated instead of the derivative: f(2) = 7 · e ≈ 19.03.

0 or −0.25

d)

Right. Product rule with the chain rule: f′(2) = 4e + 7 · 0.5 · e = 7.5e ≈ 20.39.

+1

e)

Wrong. The inner derivative 0.5 forgotten in the second term: 4e + 7e = 11e ≈ 29.90.

0 or −0.25

–

Left blank.

0

Typical mistake. The inner derivative in the second term. Without the factor 0.5 in v′ you get 4e + 7e = 11e ≈ 29.90 instead of 20.39. Lose the first term altogether and 3.5e ≈ 9.51 is left; so write u, u′, v and v′ in a column before you substitute.

Ruling out before you finish. Both terms of the product rule, 2x · e^(0.5x) and (x² + 3) · 0.5 · e^(0.5x), are positive at x = 2. So the derivative is greater than the first term alone, 4e ≈ 10.87. That rules out options a and b and leaves three.

A guess among the 3 left has a 1-in-3 chance. It is worth ≈ 0.33 points on average while your three free wrong answers last, and ≈ 0.17 after that.

Practice question 3Differential calculus

For f(x) = ln(0.1x² − x + 4), the derivative f′(8) (rounded) equals

1. a)0.25
2. b)0.42
3. c)0.67
4. d)0.88
5. e)4.00

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. Inner function g(x) = 0.1x² − x + 4, inner derivative g′(x) = 0.2x − 1. Chain rule: f′(x) = g′(x) / g(x).
2. At x = 8: g(8) = 6.4 − 8 + 4 = 2.4 and g′(8) = 1.6 − 1 = 0.6.
3. f′(8) = 0.6 / 2.4 = 0.25.

a)

Right. Chain rule: f′(8) = g′(8) / g(8) = 0.6 / 2.4 = 0.25.

+1

b)

Wrong. The inner derivative forgotten, so only (ln x)′ = 1/x applied: 1 / 2.4 ≈ 0.42.

0 or −0.25

c)

Wrong. The inner derivative taken without the term −1, so g′(x) = 0.2x instead of 0.2x − 1: 1.6 / 2.4 ≈ 0.67.

0 or −0.25

d)

Wrong. The function value calculated instead of the derivative: f(8) = ln 2.4 ≈ 0.88.

0 or −0.25

e)

Wrong. The fraction written upside down, inner function over inner derivative: 2.4 / 0.6 = 4.

0 or −0.25

–

Left blank.

0

Typical mistake. The inner derivative. The derivative of ln(g(x)) is always g′(x) / g(x); only when g′(x) = 1, as for ln(x + 5), is the numerator 1. Here g′(8) = 0.6. Without that factor you get 1 / 2.4 ≈ 0.42. And when differentiating g, the −1 is easily lost: the numerator becomes 1.6 instead of 0.6, and 0.25 becomes 0.67.

Ruling out before you finish. The derivative is the fraction g′(8) / g(8). The numerator g′(8) = 0.2 · 8 − 1 = 0.6 is less than 1 and the denominator g(8) = 6.4 − 8 + 4 = 2.4 is greater than 2, so the fraction is less than 1/2 = 0.5. That rules out 0.67, 0.88 and 4.00, options c, d and e, and leaves a and b.

A guess among the 2 left has a 1-in-2 chance. It is worth 0.5 points on average while your three free wrong answers last, and 0.375 after that.

Differentiating by hand in Mathematik (CBK)

How the 20 questions spread over the chapters is something WU does not publish, so this article follows the structure of the textbook. Its third chapter opens with the section "Die Ableitung einer Funktion" (the derivative of a function), which starts from "Differenzenquotienten" (difference quotients). Then comes "Differenzieren" (differentiation), with the subsections "Elementare Ableitungsregeln", "Die Kettenregel" and "Exponentialfunktion und Logarithmus".²

A calculator is allowed as long as it has no extra functions for differential calculus, integration or matrices; models that solve linear systems or have a text memory are banned as well.¹ So every derivative is done by hand, and the calculator only supplies numerical values such as $e^{0.5}$.

Unless the syllabus provides otherwise, WU's exam guideline rules out formula sheets and watches of any kind.³ So you do not bring your own formula sheet, and whether one is handed out in the exam is an open question. Learn the power rule, the product rule, the chain rule and the derivatives of $e^x$ and $\ln x$ by heart. The forms $e^{g(x)}$ and $\ln g(x)$ follow from them with the chain rule.

The exam plan reserves two hours for the whole exam; the actual writing time is not in the syllabus.⁴ With 20 questions, that leaves at most six minutes for each. If the rules are automatic, that leaves time to check your answer by factoring out or by the sign.

Next step after the derivative rules: Mathematik (CBK)

1. Do the three questions above before opening the solutions, and for each one write down the inner function and its derivative first.
2. If you got one wrong, find the mistake in the solution that matches your answer. If it was the inner derivative, from now on box it before you go on. For more practice, see the third chapter of the free textbook by Birgit Rudloff and Achim Zeileis.⁵
3. How the derivative turns into a statement about costs and demand is the subject of the article on marginal cost, elasticity and optimisation. Exam dates and registration are in the overview of the Mathematik (CBK) exam.

Sources

1. 1Syllabus 0003 Mathematik (LVP), Wintersemester 2026/27WU Wien, Vorlesungsverzeichnis · undated page · checked on 27 September 2026
2. 2Rudloff, Zeileis: Mathematik für Wirtschaftswissenschaften, Kapitel 3: Differentialrechnungmathe4wiwi.org · undated page · checked on 27 September 2026
3. 3Richtlinie zur Abhaltung von Präsenzprüfungen, 2026 (PDF)WU Wien · document dated 2 March 2026 · checked on 27 September 2026
4. 4Prüfungsplan Semestermitte WiSe 2026/27, Prüfungswoche November 2026 (PDF)WU Wien, Prüfungsorganisation · document dated 30 March 2026 · checked on 27 September 2026
5. 5Rudloff, Zeileis: Mathematik für Wirtschaftswissenschaften (Online-Buch, Version 2023)mathe4wiwi.org · undated page · checked on 26 September 2026