Derivatives explained for high school

Derivatives are one of those chapters where many students learn to calculate before they have understood what they are calculating. The result is predictable: faced with a problem different from the ones in the book, they freeze. Yet, once the underlying idea is understood, derivatives are one of the cleanest tools of calculus.

The test almost always falls between November and January of Year 5 (the final year of Italian high school), right after the one on limits, and asks for three things: calculating derivatives, writing the tangent line, and discussing the points where the derivative doesn’t exist. The first two can be trained; the third is the one that separates a 6 from an 8, because it requires having understood that the derivative is a limit.

Here you will find the definition, the rules needed for 90% of problems, the points of non-differentiability, the theorems the test asks for, three worked exercises and the mistakes I see most often when I prepare Year 5 students.

The derivative as the limit of the difference quotient

The formal definition is this:

f^{'} (x_{0}) = h \to 0 lim \frac{f ( x _{0} + h ) - f ( x _{0} )}{h}

In words: the derivative at a point is the limit, as $h$ tends to zero, of the ratio between the change in the function and the change in $x$ . Geometrically it is the slope of the tangent line to the graph at the point $x_{0}$ ; in physics it is the instantaneous velocity, if the function is a law of motion.

The simplest example: take $f (x) = x^{2}$ at the point $x_{0} = 1$ :

f (1 + h) f (1 + h) - f (1) \frac{f ( 1 + h ) - f ( 1 )}{h} = (1 + h)^{2} = 1 + 2 h + h^{2} = 2 h + h^{2} = 2 + h

As $h$ tends to 0, the limit is 2.

So $f^{'} (1) = 2$ . You have found the slope of the tangent to the graph of $x^{2}$ at the point $(1, 1)$ . The whole rest of the chapter is a shortcut to avoid having to do this calculation each time — but the Year 5 test almost always asks for at least one derivative calculated with the definition, precisely to check that the shortcut hasn’t erased the idea.

If you want to see that limit instead of imagining it, the interactive figure of the secant becoming a tangent starts from exactly $x^{2}$ at $x_{0} = 1$ : you move the second point closer to the first and watch the slope of the secant fall from 3 to 2.5, then 2.1 and 2.01, while the line settles on the tangent.

If limits are still shaky, it is better to put them right first: they are the prerequisite for this whole chapter, and I wrote about them in limits, continuity and asymptotes.

The basic rules you really need

For typical high school exercises a few rules are enough. If you can apply them with confidence, you have already covered most problems:

Power: $D [x^{n}] = n \cdot x^{n - 1}$ . It also holds for negative and rational exponents.
Constant: $D [k] = 0$ . A constant multiplier is taken out: $D [k \cdot f (x)] = k \cdot f^{'} (x)$ .
Sum: $D [f + g] = f^{'} + g^{'}$ . Derivatives “pass” inside the sum.
Product: $D [f \cdot g] = f^{'} \cdot g + f \cdot g^{'}$ . Careful: it is not $f^{'} \cdot g^{'}$ .
Quotient:
$D [\frac{f}{g}] = \frac{f ^{'} \cdot g - f \cdot g ^{'}}{g ^{2}}$
Elementary functions: $D [sin x] = cos x$ , $D [cos x] = - sin x$ , $D [e^{x}] = e^{x}$ , $D [ln x] = \frac{1}{x}$ .

These six lines cover practically everything. Add the chain rule and you are at 95%.

The remaining 5% — every fundamental derivative with its domain of validity, ready-to-use composites and the derivative of the inverse function — is in the complete formula sheet of derivatives (available in Italian): here I explain how to reason, there you find them all in a table.

The chain rule: where everything is decided

The chain rule is needed when you have a composite function, that is, one function inside another. The formula:

D [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)

Example: $D [sin (x^{2})]$ . Here $f$ is sine, $g$ is $x^{2}$ . Then $f^{'} (g (x)) = cos (x^{2})$ and $g^{'} (x) = 2 x$ , so:

D [sin (x^{2})] = cos (x^{2}) \cdot 2 x = 2 x cos (x^{2})

Another example: $D [(3 x + 1)^{5}]$ . The outer function is $t^{5}$ , the inner function is $3 x + 1$ , so:

5 (3 x + 1)^{4} \cdot 3 = 15 (3 x + 1)^{4}

The problem is that the chain rule applies even when you don’t notice. $ln (2 x)$ is not “logarithm of 2x”, it is a composite function. The result:

\frac{1}{2 x} \cdot 2 = \frac{1}{x}

The same result as $ln x$ (because $ln (2 x) = ln 2 + ln x$ ), but the formally correct step goes through the chain rule.

Points of non-differentiability

This is the part worth the most in the test, and also the one almost nobody practises enough. A function can be continuous at a point and not differentiable there: continuity and differentiability are not the same thing. Only one implication holds, in one direction: if it is differentiable then it is continuous, never the other way round.

The method is always the same: you calculate the two one-sided derivatives at the point and compare them.

Case	One-sided derivatives	Example
Corner point	finite but different	$f (x) = ∣ x ∣$ at $x = 0$
Cusp	infinite with opposite sign	$f (x) = 3 x^{2}$ at $x = 0$
Inflection with vertical tangent	infinite with the same sign	$f (x) = 3 x$ at $x = 0$

In the case of $∣ x ∣$ the left derivative is $- 1$ and the right one $+ 1$ : two different slopes, the graph makes a corner. In $3 x^{2}$ the derivative is $\frac{2}{3 3 x}$ , which tends to $+ \infty$ from the right and to $- \infty$ from the left: the tip of the cusp. In $3 x$ the derivative is $\frac{1}{3 3 x ^{2}}$ , which tends to $+ \infty$ from both sides: the tangent is vertical but the graph passes through.

The signal that an exercise contains a point of non-differentiability is almost always one of these three: an absolute value, a root (of even or odd index) with a zero under it, or a piecewise-defined function.

The theorems the test asks for

There are three, and they are asked mostly as a statement with a check of the hypotheses — not as a proof.

Rolle. If $f$ is continuous on $[a, b]$ , differentiable on $(a, b)$ and $f (a) = f (b)$ , then there exists at least one interior point $c$ where $f^{'} (c) = 0$ . In words: if you start and arrive at the same height, somewhere in between the tangent is horizontal.

Lagrange (mean value). With the same hypotheses but without requiring $f (a) = f (b)$ , there exists an interior $c$ such that

f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}

that is, at some point the instantaneous slope coincides with the average slope between the endpoints.

De l’Hôpital. For an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ ,

x \to x_{0} lim \frac{f ( x )}{g ( x )} = x \to x_{0} lim \frac{f ^{'} ( x )}{g ^{'} ( x )}

provided the second limit exists. Two warnings that are worth marks: you differentiate numerator and denominator separately, not with the quotient rule; and it must be applied only after checking that the form really is indeterminate.

In theory questions the point is not to recite the statement but to check the hypotheses: is the function continuous on the closed interval? Is it differentiable on the open one? If one fails, the theorem doesn’t apply and the correct answer is to say so. Full statements, examples and three worked exercises are in Rolle, Lagrange and De L’Hôpital: worked exercises.

Three worked exercises

1. Tangent line at a point

Find the tangent line to the graph of $f (x) = x^{3} - 3 x$ at the point with abscissa $x_{0} = 2$ .

You need two numbers: the point and the slope.

f (2) = 8 - 6 = 2, f^{'} (x) = 3 x^{2} - 3, f^{'} (2) = 12 - 3 = 9

The formula of the line through a point with known slope:

y - y_{0} y - 2 = f^{'} (x_{0}) (x - x_{0}) = 9 (x - 2) ⟹ y = 9 x - 16

Check: for $x = 2$ the line gives $18 - 16 = 2$ , which is exactly $f (2)$ . It works.

2. Point of non-differentiability

Study the differentiability of $f (x) = ∣ x^{2} - 4 ∣$ at $x = 2$ .

Near $x = 2$ the function is written piecewise: for $x > 2$ it equals $x^{2} - 4$ , for $x < 2$ it equals $4 - x^{2}$ . I differentiate separately and calculate the limit at $2$ :

f_{+}^{'} (2) = x \to 2^{+} lim 2 x = 4, f_{-}^{'} (2) = x \to 2^{-} lim (- 2 x) = - 4

The two one-sided derivatives are finite but different: $x = 2$ is a corner point. The function is continuous there (both branches tend to 0), but not differentiable — the cleanest example of the fact that the two properties don’t coincide.

3. An optimisation problem

A square-based container, without a lid, must have a volume of $32 d m^{3}$ . Which dimensions make the surface area a minimum?

I call $x$ the side of the base and $h$ the height. The volume constraint links the two variables:

x^{2} h = 32 ⟹ h = \frac{32}{x ^{2}}

The surface to minimise (base plus four walls, no lid) becomes a function of $x$ alone:

S (x) = x^{2} + 4 x h = x^{2} + \frac{128}{x}, x > 0

I differentiate and set to zero:

S^{'} (x) = 2 x - \frac{128}{x ^{2}} = 0 ⟹ 2 x^{3} = 128 ⟹ x = 4

The sign of $S^{'}$ confirms that it is a minimum: negative for $x < 4$ , positive for $x > 4$ . The dimensions are a side of $4$ dm and a height $h = \frac{32}{16} = 2$ dm, for a surface area of $48 d m^{2}$ .

The decisive step is not the derivative: it is using the constraint to reduce to a single variable. That is where marks are lost in optimisation problems, not in the calculation.

Typical mistakes (and how to avoid them)

After hundreds of lessons in Year 5, mistakes on derivatives are almost always these five.

1. Forgetting the chain rule. The student sees $cos (2 x)$ and writes $- sin (2 x)$ instead of $- 2 sin (2 x)$ . It is the most frequent mistake of all. Practical rule: every time inside a function “there is something that is not $x$ on its own”, you must multiply by the derivative of that something.

2. Getting the derivative of the product wrong. Very many write $D [x \cdot sin x] = 1 \cdot cos x = cos x$ , which is wrong. The correct form is

D [x \cdot sin x] = 1 \cdot sin x + x \cdot cos x = sin x + x cos x

Memorise the formula with examples, not with symbols.

3. Not simplifying before differentiating. $D [\frac{x ^{2} - 1}{x - 1}]$ with the quotient rule is a pain. But

\frac{x ^{2} - 1}{x - 1} = x + 1 (x \neq = 1)

so the derivative is simply 1. Always simplify first.

4. Confusing continuity and differentiability. They are different things: $∣ x ∣$ is continuous everywhere and not differentiable at zero. In theory questions this distinction is asked almost every year.

5. Applying De l’Hôpital without checking the form. It should be used only on $\frac{0}{0}$ and $\frac{\infty}{\infty}$ , and you differentiate top and bottom separately. Using it on a determinate form gives a wrong number with all the steps in order, which is the worst way to go wrong.

How derivatives are used in maturità problems

Typical problems in the maturità (the school-leaving exam) ask for:

Function study: first derivative to find maxima and minima, second derivative for concavity and inflections. The complete path is in function study step by step.
Tangent at a point: as in exercise 1, with $y - y_{0} = f^{'} (x_{0}) (x - x_{0})$ .
Optimisation problems: write the quantity to be maximised as a function of a single variable, differentiate, set equal to zero.

The method is always the same: turn a problem “in words” into a function, differentiate it, reason about the signs. If the problem is not the method but your head in the first minutes of the test, also take a look at studying maths without stress.

When derivatives enter the function study, the series I wrote uses them like this: monotonicity and maxima and minima are in part 4: monotonicity, maxima and minima, concavity and inflections in part 5: concavity and inflections, and for the foundations needed first there are domain and symmetries, intercepts and sign, limits and asymptotes and, to close, the graph with worked examples.

How we work on this in lessons

On derivatives the first thing I check is not the calculation but the definition: I ask the student to differentiate a simple function with the difference quotient, by hand. Those who manage have the whole chapter in hand, points of non-differentiability included; those who don’t are applying rules by memory, and will freeze at the first exercise written differently from those in the book.

Then we work on recognising composites, which is the real skill of the test: ten functions in a row where you only say “outer, inner” without differentiating anything. In a quarter of an hour you build the automatic response that is worth half the mark. Lessons are online and one to one: the package (rates on the pricing page) is worth it if you want to cover limits, derivatives and function study as a single path towards the maturità.

The quickest way to understand where you stand is to try the practice test on derivatives: fifteen questions in fifty minutes, with a breakdown of weak points by topic at the end. The correct answer to each question is free and needs no registration; the guided step-by-step solution unlocks with your first lesson. If after the test you see that the calculation holds but the points of non-differentiability don’t (or vice versa), write to me for a free introductory call: we look at the programme of your test and I tell you honestly how many lessons are needed. The full picture of how I set up the path is on the maths tutoring page.

Derivatives explained properly, for high school