Differentiation of Special Functions

Take the function f(x) = |x| \cdot x. At first glance, it involves the absolute value — which has a corner at the origin and is not differentiable there. So is f also not differentiable at x = 0?

Try the limit definition directly:

The derivative exists and equals 0. The multiplication by x smoothed out the corner. The absolute value alone has a sharp V at the origin, but |x| \cdot x has a smooth cusp that just barely kisses the axis — differentiable, with slope zero.

This is the kind of surprise that makes special functions interesting. The standard rules (power rule, product rule, chain rule) assume the function is built from smooth pieces. When your function has corners, jumps, or sudden switches in formula, you need a different approach: split, check, and differentiate each piece.

Piecewise functions: the general strategy

A piecewise function is defined by different formulas on different intervals. The derivative on each interval is straightforward — just differentiate the formula that applies there. The only question is what happens at the boundary points where the formula changes.

The strategy is always the same:

1. Differentiate each piece separately on its open interval.
2. At each boundary point, compute the left-hand derivative and the right-hand derivative.
3. If they match (and the function is continuous there), the function is differentiable at that point and the derivative equals the common value.
4. If they don't match, the function is not differentiable at that point.

Continuity is a prerequisite — if the function has a jump at the boundary, it cannot be differentiable there. But continuity alone is not enough. The function |x| is continuous at x = 0, but its left derivative (-1) and right derivative (+1) disagree, so the derivative does not exist.

The absolute value function

The absolute value function is the most important piecewise function in calculus. Write it out:

Away from zero: for x > 0, \frac{d}{dx}|x| = 1. For x < 0, \frac{d}{dx}|x| = -1.

At zero: the left derivative is -1 and the right derivative is +1. They disagree. So |x| is not differentiable at x = 0.

You can write the derivative compactly using the signum function:

For a composite |u(x)| where u is differentiable:

At points where u(x) = 0, you need to check the left and right derivatives directly.

A general absolute-value derivative

Here is a pattern that covers many textbook problems. For f(x) = |x - a|:

The corner shifts to x = a, but the story is the same. The slope is +1 on one side, -1 on the other, and undefined at the corner itself.

The floor and ceiling functions

The greatest integer function (floor function) \lfloor x \rfloor is the largest integer \le x. Its graph is a staircase: flat segments at integer heights, with jumps at every integer.

Between integers, the function is constant: \lfloor x \rfloor = n when n \le x < n + 1. The derivative of a constant is 0. So for non-integer x:

At integers, the function has a jump discontinuity — it leaps from one step to the next. A function with a jump discontinuity is not continuous, and a function that is not continuous is certainly not differentiable. So \lfloor x \rfloor is not differentiable at any integer.

The ceiling function \lceil x \rceil (the smallest integer \ge x) has the same story with the jumps flipped. Its derivative is 0 between integers and does not exist at integers.

The fractional part function \{x\} = x - \lfloor x \rfloor has a different character. Between integers, \{x\} = x - n for the appropriate integer n, so its derivative is 1. At integers, the fractional part jumps from just below 1 back to 0, so the function is discontinuous and not differentiable there.

Composing the floor with smooth functions

A common exam pattern: differentiate f(x) = \lfloor x^2 \rfloor or f(x) = x \lfloor x \rfloor.

The rule is always the same: identify where the inner expression crosses an integer, because those are the points of potential non-differentiability. Between those points, the floor is constant, and you differentiate normally.

For f(x) = x \cdot \lfloor x \rfloor on the interval 1 \le x < 2: here \lfloor x \rfloor = 1, so f(x) = x, and f'(x) = 1. On the interval 2 \le x < 3: \lfloor x \rfloor = 2, so f(x) = 2x, and f'(x) = 2. At x = 2, check the left and right derivatives: f'(2^-) = 1 and f'(2^+) = 2. They disagree, so f is not differentiable at x = 2.

Max and min functions

Functions defined using \max or \min are piecewise functions in disguise. The function f(x) = \max(x, x^2) switches between x and x^2 depending on which is larger.

To differentiate such a function:

1. Find where the two expressions are equal — that is where the switch happens.
2. On each interval, differentiate whichever expression is the max (or min).
3. At the switch point, check the left and right derivatives.

For f(x) = \max(x, x^2): set x = x^2, which gives x(x - 1) = 0, so the switch points are x = 0 and x = 1.

• For x < 0: x^2 > x (since x is negative), so f(x) = x^2 and f'(x) = 2x.
• For 0 < x < 1: x > x^2, so f(x) = x and f'(x) = 1.
• For x > 1: x^2 > x, so f(x) = x^2 and f'(x) = 2x.

At x = 0: f'(0^-) = 2(0) = 0 and f'(0^+) = 1. They disagree — not differentiable.

At x = 1: f'(1^-) = 1 and f'(1^+) = 2(1) = 2. They disagree — not differentiable.

Implicit trig substitutions and differentiability

Some functions look complicated but simplify dramatically under a trigonometric substitution. The classic example is

Substitute x = \tan\theta with \theta \in (-\pi/2, \pi/2). Then

So f(x) = \sin^{-1}(\sin 2\theta) = 2\theta — but only when 2\theta lies in [-\pi/2, \pi/2], i.e., when |\theta| \le \pi/4, which means |x| \le 1.

The full picture:

On each piece, the derivative is \frac{-2}{1 + x^2} (for x > 1 and x < -1) or \frac{2}{1 + x^2} (for |x| < 1). At x = 1, the left derivative is \frac{2}{1 + 1} = 1 and the right derivative is \frac{-2}{1 + 1} = -1. They disagree — the function is not differentiable at x = \pm 1.

The lesson: when a function is defined through an inverse trigonometric expression containing a rational function of x, always check if the substitution changes branches. The apparent formula \frac{d}{dx}\sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \cdot u' is valid only when u stays strictly between -1 and 1. At the boundary, the function often has a corner.

Differentiation of determinants

When the entries of a determinant are functions of x, you can differentiate the determinant with respect to x. The rule is elegant: differentiate one row at a time, keeping the other rows unchanged, and add the results.

Why does this work? Expand the 2 \times 2 determinant: \Delta(x) = f_1(x) g_2(x) - f_2(x) g_1(x). Differentiate using the product rule:

Regroup: (f_1' g_2 - f_2' g_1) + (f_1 g_2' - f_2 g_1'). The first group is the determinant with row 1 differentiated. The second is the determinant with row 2 differentiated. This is exactly the formula.

The 3 \times 3 case follows the same logic — expand the determinant using cofactors along the first row, apply the product rule to each term, and regroup. Each product gives a pair: one part where the row-1 entry is differentiated and the cofactor is kept, and one where the entry is kept and the cofactor is differentiated. The cofactors themselves are 2 \times 2 determinants, and differentiating them produces the terms for rows 2 and 3. The whole thing telescopes into three determinants.

A 3 \times 3 determinant example

Take \Delta(x) = \begin{vmatrix} x & x^2 & x^3 \\ 1 & 2x & 3x^2 \\ 0 & 2 & 6x \end{vmatrix}.

The first determinant has rows 1 and 2 identical — its value is 0. The second determinant has rows 2 and 3 identical — its value is 0. Only the third survives.

Expand the third determinant along row 3. The only nonzero entry in that row is 6 in position (3, 3). Its cofactor is \begin{vmatrix} x & x^2 \\ 1 & 2x \end{vmatrix} = 2x^2 - x^2 = x^2. So \Delta'(x) = 6 \cdot x^2 = 6x^2.

Common confusions

• "The derivative of \lfloor x \rfloor is zero everywhere." Almost. It is zero wherever the derivative exists — which is everywhere except the integers. At the integers, the derivative does not exist because the function is discontinuous. Writing \frac{d}{dx}\lfloor x \rfloor = 0 without the caveat "for non-integer x" is technically wrong.

• "|f(x)| is never differentiable where f(x) = 0." False. If f has a zero of even multiplicity (like f(x) = x^2 at x = 0), then |f(x)| = f(x) near that zero (because f does not change sign), and the derivative exists. It is only when f changes sign through the zero — a simple zero — that the absolute value creates a corner.

• "You can always use the chain rule on \sin^{-1}(g(x))." You can, as long as g(x) stays strictly in (-1, 1). When g(x) = \pm 1, the derivative \frac{1}{\sqrt{1 - g^2}} has a vertical asymptote, and the formula breaks down. You need to check the original function's behaviour directly.

• "Differentiating a determinant means differentiating every entry." No — that gives a wrong answer. You differentiate one row at a time, not all at once. The determinant is a sum of products (by the Leibniz formula), and the product rule generates one term per row, not one term per entry.

Going deeper

Why piecewise differentiability is not enough for "smooth"

A function can be differentiable everywhere and still have a derivative that is discontinuous. The function f(x) = x^2 \sin(1/x) from Example 2 demonstrates this. Its derivative exists at every real number, but f' itself has a jump-like oscillation at x = 0 — the \cos(1/x) term prevents f' from having a limit at 0.

This means that even after you have checked differentiability everywhere, you cannot assume the derivative is well-behaved. The derivative of a differentiable function is guaranteed to satisfy the intermediate value property (by Darboux's theorem — the derivative hits every value between any two of its values), but it is not guaranteed to be continuous. This is one of the subtler facts in real analysis.

The n-th derivative of a determinant

The row-at-a-time rule extends to higher derivatives. For a 2 \times 2 determinant with entries that depend on x:

The pattern mirrors the binomial expansion of (D_1 + D_2)^n, where D_1 differentiates row 1 and D_2 differentiates row 2. For the n-th derivative:

This is Leibniz's rule applied to determinants — elegantly combining two ideas from different parts of calculus.

Weierstrass's monster

You know that |x| is continuous but not differentiable at x = 0 — one bad point. Can a function be continuous everywhere but differentiable nowhere?

In 1872, Weierstrass constructed exactly such a function:

where 0 < a < 1 and b is a positive odd integer with ab > 1 + 3\pi/2. This function is continuous (the series converges uniformly) but has a corner at every point — the graph is so jagged that no tangent line exists anywhere. The piecewise strategy above relies on having finitely many bad points with smooth stretches between them. Weierstrass showed that nature is not always so kind.

Where this leads next

• Differentiability — the formal theory of when derivatives exist, with left and right derivatives, and the relationship between differentiability and continuity.
• Reasons for Non-Differentiability — a systematic classification of all the ways a function can fail to be differentiable: corners, cusps, vertical tangents, and discontinuities.
• Special Functions — Part 1 — the floor function, ceiling function, fractional part, and signum function, with their graphs and properties.
• Differentiation of Determinants — a deeper look at the row-by-row differentiation rule, with applications to Wronskians and systems of differential equations.
• Functional Equations and Differentiation — using differentiability to solve functional equations, a technique that combines the ideas from this article with algebraic reasoning.