The table of derivatives

In the first lesson we found derivatives from the definition: the increment of the argument, the increment of the function, the ratio, the limit — the same long road for every new function. Repeating that limit for sin x or √x in every problem would be slow and tiring. Mathematicians did the work once for all the basic functions and collected the answers in the table of derivatives. It works like the multiplication table: you do not count 7 · 8 on your fingers, you simply know it is 56. In this lesson you will learn the table, see where each line comes from, and practise until using it becomes automatic.

The table at a glance

Below are the derivatives of all the basic functions of the school course. The left column is the function, the middle column its derivative, and the right column says where the formula holds. Two general rules apply: a formula works only at the points where the function itself is defined (and differentiable), and in the trigonometric formulas x is always measured in radians.

1. 1
   Recognise the type

   Where is x? In the base (xⁿ): a power function; in the exponent (aˣ): an exponential function; under ln or log: a logarithmic one; under sin, cos, tan, cot: a trigonometric one. If there is no x at all, it is a constant and its derivative is 0.

2. 2
   Rewrite if needed

   Write roots and fractions as powers: √x = x^(1/2), 1/x³ = x⁻³. After that it is an ordinary power function.

3. 3
   Take the line from the table

   Write f′(x) from the table and check the conditions: x > 0 for ln x and √x, x in radians for the trigonometric functions.

4. 4
   Substitute the point

   If the value of the derivative at x₀ is asked, substitute x₀ into f′(x), not into f(x), and simplify.

Constants, linear functions and powers

A constant does not change: for f(x) = C we have Δf = C − C = 0 for every Δx, so Δf / Δx = 0 and C′ = 0. Its graph is a horizontal line, and a horizontal line has slope 0. For f(x) = kx + b the increment is Δf = k(x + Δx) + b − (kx + b) = k · Δx, so Δf / Δx = k for every Δx. The derivative of a linear function is its slope, the same at every point; in particular, x′ = 1.

Now take f(x) = xⁿ with a natural n. Write (x + Δx)ⁿ as a product of n brackets (x + Δx). Taking x from every bracket gives xⁿ. Taking Δx from exactly one bracket and x from all the others gives xⁿ⁻¹ · Δx, and that bracket can be chosen in n ways, so together we get n · xⁿ⁻¹ · Δx. Every other term contains at least (Δx)², so after dividing by Δx it still contains Δx and tends to 0.

For example, for n = 3: (x + Δx)³ = x³ + 3x²Δx + 3x(Δx)² + (Δx)³. Then Δf = 3x²Δx + 3x(Δx)² + (Δx)³ and Δf / Δx = 3x² + 3xΔx + (Δx)² → 3x². So (x³)′ = 3x²: the exponent came down as a factor and dropped by 1.

The same formula holds for negative and fractional exponents. Let us check two cases directly. For 1/x = x⁻¹: Δf = 1/(x + Δx) − 1/x = −Δx / (x(x + Δx)), so Δf / Δx = −1 / (x(x + Δx)) → −1/x². The formula gives the same: (x⁻¹)′ = −1 · x⁻² = −1/x². For √x = x^(1/2), multiply the numerator and the denominator by √(x + Δx) + √x: Δf / Δx = (√(x + Δx) − √x) / Δx = 1 / (√(x + Δx) + √x) → 1/(2√x), and the formula gives ½ · x^(−1/2) = 1/(2√x). The general proof for any real n belongs to higher mathematics; at school the formula is simply used.

Exponential and logarithmic functions

In y = aˣ the variable is in the exponent, so the power rule does not apply. Let us compute the increment: a^(x + Δx) − aˣ = aˣ · (a^Δx − 1), so Δf / Δx = aˣ · (a^Δx − 1) / Δx. The second factor does not depend on x: as Δx → 0 it tends to the slope of the graph of aˣ at the point (0, 1). So the derivative of an exponential function is the function itself times a constant.

This constant depends on the base: for 2ˣ it is about 0.69, for 3ˣ about 1.10. You can check with a calculator: (2^0.001 − 1) / 0.001 ≈ 0.693 and (3^0.001 − 1) / 0.001 ≈ 1.099. So somewhere between 2 and 3 there is a base for which the slope at (0, 1) is exactly 1. This base is the number e.

For a = e the constant is 1, so (eˣ)′ = eˣ: the function equals its own derivative. At every point the slope of the graph of eˣ equals its height: at x = 0 both are 1, at x = 2 both are e² ≈ 7.39. For any other base write a = e^(ln a); then aˣ = e^(kx) with the constant k = ln a. Multiplying x by k multiplies every slope of the graph by k (the exact reason is the chain rule of lesson 5), so (aˣ)′ = k · e^(kx) = aˣ · ln a. Check: ln 2 ≈ 0.693 and ln 3 ≈ 1.099 — exactly the numbers we measured above.

The graphs of mutually inverse functions are symmetric about the line y = x. Under this reflection Δx and Δy swap places, so a line with slope k turns into a line with slope 1/k. Take the point (t, eᵗ) on the graph of eˣ: the slope there is eᵗ. The symmetric point (eᵗ, t) lies on the graph of ln x, and the slope there is 1/eᵗ. Writing x = eᵗ, we get (ln x)′ = 1/x.

A logarithm to any other base reduces to ln by the change-of-base formula: logₐ x = ln x / ln a. Here 1/ln a is just a constant factor, so (logₐ x)′ = 1/(x · ln a). For the common logarithm log₁₀ x: (log₁₀ x)′ = 1/(x · ln 10) ≈ 0.434/x.

Trigonometric functions

Everything rests on one limit: with x in radians, sin h / h → 1 as h → 0. Geometrically: on the unit circle, a small arc of length h and the perpendicular sin h dropped from its end to the x-axis are almost equal. In numbers: sin 0.1 / 0.1 ≈ 0.9983, sin 0.01 / 0.01 ≈ 0.99998.

From trigonometry we know the formula for a difference of sines: sin α − sin β = 2 cos((α + β)/2) · sin((α − β)/2). So sin(x + h) − sin x = 2 cos(x + h/2) · sin(h/2), and dividing by h:
(sin(x + h) − sin x) / h = cos(x + h/2) · sin(h/2) / (h/2) → cos x · 1 = cos x.
In the same way the difference of cosines, cos(x + h) − cos x = −2 sin(x + h/2) · sin(h/2), gives (cos x)′ = −sin x.

The tangent and cotangent are quotients: tan x = sin x / cos x, cot x = cos x / sin x. In lesson 4 we will derive their derivatives with the quotient rule; for now we take them from the table. Note that 1/cos² x = 1 + tan² x, so (tan x)′ is always positive: the tangent increases on every interval of its domain. Likewise (cot x)′ = −1/sin² x < 0: the cotangent decreases on every interval.

Working with the table: checks and practice

The table also works backwards: if the value of the derivative is known, we can find the point. This skill is needed in tangent problems (where is the slope equal to a given number?) and in lesson 6 (where is the derivative zero?).

The table is the foundation of the next two lessons. In lesson 4 we learn to differentiate sums, products and quotients of table functions (for example x² · sin x), and in lesson 5 composite functions such as sin 3x or e^(x²). Every such computation ends with a line of this table — which is why it is worth knowing by heart.