🪜Higher-Order Derivatives

Every craft in the Derivative Spire so far has stopped after one differentiation. But f'(x) is a function, not a number — it has a value at every x, it has a graph, and there is no reason at all why it should not be differentiated in its turn.

Do that and you get the second derivative, written f''(x). Differentiate again for f'''(x), and again for the fourth. After three primes the notation gets unreadable, so the order moves into a bracket up top:

f(x), f'(x), f''(x), f'''(x), f⁴(x), f⁵(x), ...

The bracket matters: f⁴(x) is the fourth derivative, while f(x)⁴ would be the function raised to the fourth power. Two completely different things.

What does it measure? f'(x) answers "how fast is f changing". So f''(x) answers "how fast is that rate changing" — a rate of a rate. The everyday example is motion. Let s(t) be where something is at time t:

• ◆s'(t) is the velocity: how fast the position is changing.
• ◆s''(t) is the acceleration: how fast the velocity is changing.
• ◆s'''(t) is called the jerk: how fast the acceleration is changing, which is what you feel when a train's braking suddenly eases off.

A car cruising at a steady 100 km/h has a large velocity and zero acceleration. A car pulling away from rest has a small velocity and a large acceleration. The first derivative and the second derivative are genuinely independent pieces of information, and knowing one tells you nothing about the other.

The other notation. In Leibniz's notation the first derivative is dy/dx, so the second is d/dx applied to dy/dx, which is written

d^2y/dx²

Read it as "d two y by d x squared", and read it as a label, not as a calculation. Nothing there is being squared. The 2 on top counts differentiations; the 2 on the bottom records that all of them were with respect to x. In the same way d^3y/dx³ is the third derivative.

What the sign says. f'(x) > 0 means f is increasing. f''(x) > 0 means f' is increasing — the slope is getting steeper as you move right — and that is exactly what "the graph bends upwards", or concave up, means. So a second derivative is the algebraic name for curvature, which is why curve sketching leans on it so heavily. And be careful about what a second derivative does *not* say: f''(3) = -4 tells you the graph is concave down at x = 3. It says nothing whatever about f(3) or about f'(3).

Polynomials are the easiest place to see a ladder of derivatives, because each rung is one more use of the power rule and nothing else. Take f(x) = 2x⁴ - 5x³ + x - 9:

f'(x) = 8x³ - 15x² + 1 f''(x) = 24x² - 30x f'''(x) = 48x - 30 f⁴(x) = 48 f⁵(x) = 0

Two things happen on every rung. The degree drops by exactly 1, because every exponent falls by 1. And the constant term disappears, because a constant has no slope.

So the ladder of a polynomial always ends. A polynomial of degree n reaches a non-zero constant at rung n, and the very next derivative, order n + 1, is 0 for every x — and so is every derivative after it. In the example the degree is 4, f⁴ is the constant 48, and the first order that is 0 everywhere is the 5th. Notice the count: it is n + 1, not n, because the constant left standing at rung n is not itself 0.

The star of the pattern is xⁿ. Differentiate x⁵ five times:

5x⁴, 20x³, 60x², 120x, 120

The multipliers 5, 4, 3, 2 and 1 have all come down and been collected, and their product is 5! = 120. In general the nth derivative of xⁿ is the constant n!, and the (n+1)th is 0. If there is a coefficient in front, it simply rides along: the 4th derivative of 3x⁴ is 3 times 4! = 72.

Two slips live here, and both are worth naming now. The first is bringing the exponent down as a multiplier but forgetting to lower it — x⁵ becomes 5x⁴, never 5x⁵. The second is stopping one rung early or going one too far. Count the primes against the order you were asked for, every time, and write each rung on its own line so the count is visible.

Fractions behave no differently. For f(x) = 3x⁵/4 the first rung is 15x⁴/4, the second is 15x³, and the third is 45x² — a denominator that survives one rung can easily be cancelled by the next multiplier that comes down.

Some functions never run out of derivatives. They fall into patterns instead, and once you see the pattern you can jump straight to the 20th derivative without writing the first nineteen.

Sine and cosine run in a cycle of four.

sin(x) -> cos(x) -> -sin(x) -> -cos(x) -> sin(x) -> ...

Four differentiations bring you back to where you started, so only the remainder of the order on division by 4 matters. For sin(x), an order that leaves remainder 1 lands on cos(x), remainder 2 on -sin(x), remainder 3 on -cos(x) and remainder 0 back on sin(x). The 5th derivative of sin(x) is therefore cos(x), because 5 leaves remainder 1. Cosine runs the same loop, one place further round: cos -> -sin -> -cos -> sin.

Now put a k inside. For sin(kx) the chain rule pays a factor of k on every rung, so after n differentiations the answer carries kⁿ:

d/dx sin(2x) = 2cos(2x) d²/dx² sin(2x) = -4sin(2x) d³/dx³ sin(2x) = -8cos(2x) d⁵/dx⁵ sin(2x) = 32cos(2x)

The single most common error in the whole craft is writing cos(2x) or 2cos(2x) for that last one. It is kⁿ, not k, and not 1.

The exponential is the simplest of all. e^kx never changes shape; every derivative is a multiple of the same e^kx, and each rung collects one more k. So the nth derivative of e^kx is kⁿ e^kx: the 4th derivative of e^3x is 81e^3x, because 3⁴ = 81. (Note that 3⁴ is not the same as 4 times 3. The chain factors multiply together, so they build a power.)

And 1/x alternates. Write it as x^-1 and let the power rule run:

f(x) = 1/x f'(x) = -1/x² f''(x) = 2/x³ f'''(x) = -6/x⁴ f⁴(x) = 24/x⁵

The numerators 1, 2, 6, 24 are the factorials, the power underneath is always one more than the order, and the sign alternates because each new exponent that comes down is negative. In symbols the nth derivative of 1/x is (-1)ⁿ n! / xⁿ⁺¹. The same reading works for square roots: sqrt(x) = x^1/2 gives exponents 1/2, -1/2, -3/2, -5/2 in turn, so f''(x) = -1/(4x^3/2) and f''(4) = -1/32.

There is no "second product rule" to learn. To find f''(x) for a product, use the product rule once, then use it again on whatever comes out. The only discipline needed is to differentiate the whole of f'(x), including the parts that are themselves products.

x e^x. With u = x and v = e^x:

f'(x) = (1)(e^x) + (x)(e^x) = (1 + x)e^x f''(x) = (e^x) + (e^x + x e^x) = (2 + x)e^x

The e^x term arrives twice on the second rung, once from each factor, which is where the 2 comes from. With e^kx inside, every rung also pays its factor of k, and x e^kx gives f''(x) = (2k + k² x)e^kx.

x sin(x). With u = x and v = sin(x):

f'(x) = sin(x) + x cos(x) f''(x) = cos(x) + (cos(x) - x sin(x)) = 2cos(x) - x sin(x)

Again a term lands on top of one already there, and the 2 is the evidence that the product rule was used properly on the second rung.

The mistake to name out loud. The second derivative of a product is not u''v + uv''. For x sin(x) that would give 0 - x sin(x) = -x sin(x), and the whole 2cos(x) would have vanished. Every term of f'(x) has to be differentiated, cross terms included.

Compositions are gentler, because the chain rule simply applies twice. For (ax + b)ⁿ:

f'(x) = na(ax + b)^n-1 f''(x) = n(n-1)a² (ax + b)^n-2

Notice a². Two differentiations means the inside derivative is collected twice, exactly as k² appeared for e^kx. So the second derivative of (3x + 5)⁴ is (4)(3)(9)(3x + 5)² = 108(3x + 5)².

The same move handles a root. For sqrt(ax + b) = (ax + b)^1/2:

f'(x) = (a/2)(ax + b)^-1/2 f''(x) = (a/2)(-1/2)(a)(ax + b)^-3/2 = -a²/(4(ax + b)^3/2)

The minus sign is not optional: it comes from the exponent -1/2 coming down, and it is why the graph of any square root is concave down everywhere it exists.

Here is a fact that looks like a curiosity and turns out to be one of the most important ideas in all of mathematics.

Take a polynomial and evaluate its derivatives at x = 0:

f(x) = 5 - 2x + 6x² + 3x³ f(0) = 5 f'(x) = -2 + 12x + 9x², so f'(0) = -2 f''(x) = 12 + 18x, so f''(0) = 12 f'''(x) = 18, so f'''(0) = 18

Look at the pairs. The constant term is f(0). The coefficient of x is f'(0). The coefficient of x² is 6, and f''(0) = 12 = 2! times 6. The coefficient of x³ is 3, and f'''(0) = 18 = 3! times 3. The rule is

f^k(0) = k! times (the coefficient of x^k)

and the reason is exactly the ladder above. Differentiate k times: every term below x^k has been flattened to 0, every term above x^k still carries an x and so vanishes when 0 is substituted, and the x^k term has collected the multipliers k, k-1, ..., 1 on its way down. Only one term can survive, and it arrives multiplied by k!.

It runs both ways. Given f(0) = 5, f'(0) = -2, f''(0) = 12 and f'''(0) = 18, you can rebuild the cubic: divide each value by the matching factorial to recover its coefficient, giving 5, -2, 12/2 = 6 and 18/6 = 3, so f(x) = 5 - 2x + 6x² + 3x³. The coefficient need not be whole — if f'''(0) = 20, the coefficient of x³ is 20/6 = 10/3.

The trap in both directions is the factorial. A coefficient of 7 in the x⁴ place is not f⁴(0); f⁴(0) is 4! times 7 = 168. And dividing by k rather than by k! is a near miss that only shows up from k = 3 onwards, where 3! = 6 and 3 are finally different.

Why does this matter so much? Because it says that a polynomial's coefficients and its derivatives at a point carry exactly the same information. Know all the derivatives at 0 and you know the polynomial. That is the seed of the Taylor series, which does the same trick for functions like sin(x) and e^x that have no last derivative at all — and it is why anyone bothers to compute a 5th or a 7th derivative in the first place.