Differentiation Rules

⚡ Key Takeaways

Differentiation rules are a toolkit of algebraic shortcuts for computing derivatives without resorting to the limit definition each time. Key result: the derivative of any combination of standard functions (sums, products, quotients, compositions, powers) can be found by applying a small set of rules — the power, product, quotient, and chain rules — in sequence. Where it applies: any differentiable function built from polynomials, trig, exponentials, logarithms, and their combinations. Who/when: developed by Newton and Leibniz in the 1670s–1680s; formalized in the 19th century.

💡 The Core Insight

Every elementary function is assembled from simple pieces using four operations: addition, multiplication, division, and composition. Differentiation rules tell you how the derivative behaves under each of these operations. If you know how to differentiate the pieces, you can differentiate the whole — no limits required.

DefinitionThe Derivative

The derivative of a function $f$ at a point $x$ is the limit of the difference quotient:

1

\quad f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Derivation hint: This is the slope of the secant line through $(x, f(x))$ and $(x+h, f(x+h))$ as the second point slides toward the first. Here $f'(x)$ (read "f prime of x") is also written $\dfrac{df}{dx}$ or $\dfrac{dy}{dx}$ , where $y = f(x)$ . The symbol $h$ is the small horizontal increment that shrinks to zero.

TheoremThe Core Differentiation Rules

Let $f$ and $g$ be differentiable functions and $c$ a constant. Then:

Constant rule:

2

\quad \frac{d}{dx}[c] = 0

Derivation hint: A constant function has zero slope everywhere; the difference quotient is $0/h = 0$ .

Power rule:

3

\quad \frac{d}{dx}[x^n] = n\,x^{\,n-1}

Derivation hint: Expand $(x+h)^n$ by the binomial theorem; all terms but $n x^{n-1}h$ carry a factor $h^2$ or higher and vanish.

Constant multiple rule:

4

\quad \frac{d}{dx}[c\,f(x)] = c\,f'(x)

Derivation hint: A constant factor pulls straight out of the limit defining $f'$ .

Sum/difference rule:

5

\quad \frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)

Derivation hint: The limit of a sum is the sum of limits; split the difference quotient into two.

Product rule:

6

\quad \frac{d}{dx}[f(x)\,g(x)] = f'(x)\,g(x) + f(x)\,g'(x)

Derivation hint: Add and subtract $f(x+h)g(x)$ in the numerator to split into two manageable quotients.

Quotient rule:

7

\quad \frac{d}{dx}\!\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{[g(x)]^2}

Derivation hint: Write the quotient as a product $f\cdot g^{-1}$ and apply the product and chain rules, or use the add-subtract trick directly.

Chain rule:

8

\quad \frac{d}{dx}\big[f(g(x))\big] = f'(g(x))\cdot g'(x)

Derivation hint: Multiply and divide the difference quotient by $\Delta g$ so it becomes (rate of $f$ in $g$ ) times (rate of $g$ in $x$ ).

DefinitionDerivatives of Standard Functions

The basic building blocks, memorized once:

9

\quad \frac{d}{dx}[\sin x] = \cos x, \qquad \frac{d}{dx}[\cos x] = -\sin x

Derivation hint: Use the limit definition with the angle-addition formulas and the limits $\lim_{h\to 0}\frac{\sin h}{h}=1$ and $\lim_{h\to 0}\frac{\cos h - 1}{h}=0$ .

10

\quad \frac{d}{dx}[e^x] = e^x, \qquad \frac{d}{dx}[\ln x] = \frac{1}{x}

Derivation hint: $e$ is defined precisely so that $e^x$ is its own derivative; differentiate $e^{\ln x}=x$ implicitly to get the log rule.

11

\quad \frac{d}{dx}[\tan x] = \sec^2 x

Derivation hint: Write $\tan x = \sin x / \cos x$ and apply the quotient rule, using $\sin^2 + \cos^2 = 1$ .

ProofThe Product Rule, Step by Step

We prove equation [6]. Start from the limit definition applied to the product $p(x) = f(x)g(x)$ :

Step 1. Write the difference quotient:

12

\quad p'(x) = \lim_{h\to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}

Derivation hint: Direct substitution of $p = fg$ into equation [1].

Step 2. Add and subtract $f(x+h)g(x)$ in the numerator (it cancels to zero net):

13

\quad p'(x) = \lim_{h\to 0} \frac{f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)}{h}

Derivation hint: This is the key algebraic trick — inserting a "bridge" term lets us factor.

Step 3. Factor and split into two limits:

14

\quad p'(x) = \lim_{h\to 0} f(x+h)\,\frac{g(x+h)-g(x)}{h} + \lim_{h\to 0} g(x)\,\frac{f(x+h)-f(x)}{h}

Derivation hint: Group the first two terms by their common factor $f(x+h)$ , the last two by $g(x)$ .

Step 4. Evaluate each limit. As $h\to 0$ , $f(x+h)\to f(x)$ (continuity), the first quotient $\to g'(x)$ , and the second $\to f'(x)$ :

15

\quad p'(x) = f(x)\,g'(x) + g(x)\,f'(x) \qquad \blacksquare

Derivation hint: Apply the product law for limits and the definition of derivative to each surviving factor.

∎

IN

Isaac Newton

1643–1727

Newton developed his "method of fluxions" around 1665–1666 during the plague years at his family farm. He thought of variables as quantities flowing in time, and the derivative — his fluxion, written $\dot{x}$ — as a velocity. His physical instinct (rates of change of position, velocity, acceleration) made the product and chain rules natural: the rate of a combined quantity comes from how each part flows. Newton's notation survives today in physics, where $\dot{x}$ and $\ddot{x}$ denote first and second time derivatives.

🧠 Intuition

Think of the product rule geometrically. Imagine a rectangle with sides $f(x)$ and $g(x)$ , so its area is the product $f\cdot g$ . When $x$ increases a little, both sides grow: $f$ by $\Delta f$ , $g$ by $\Delta g$ . The new area gains two thin strips — one of size $g\cdot \Delta f$ (bottom) and one of size $f\cdot \Delta g$ (side) — plus a tiny corner $\Delta f \cdot \Delta g$ . The corner is negligibly small (a product of two small things), so the rate of area change is $f\,g' + g\,f'$ . That is exactly equation [6]. The chain rule is about gears meshing. If gear A turns twice as fast as gear B, and gear B turns three times as fast as gear C, then A turns six times as fast as C: rates multiply. Composition $f(g(x))$ chains two gears — multiply their rates.

Figure

The product rule as growing rectangle area: the two outer strips are

g\,\Delta f

and

f\,\Delta g

; the corner

\Delta f\,\Delta g

is negligible.

ExampleProduct and Power Rules Combined

Given: Differentiate $y = (3x^2 + 1)(x^3 - 5x)$ .

Step 1. Identify the two factors: $f(x) = 3x^2 + 1$ and $g(x) = x^3 - 5x$ .

Step 2. Differentiate each using power and sum rules (equations [3], [5]): $f'(x) = 6x$ and $g'(x) = 3x^2 - 5$ .

Step 3. Apply the product rule, equation [6]:

16

\quad y' = (6x)(x^3 - 5x) + (3x^2 + 1)(3x^2 - 5)

Derivation hint: Substitute $f', g, f, g'$ directly into $f'g + fg'$ .

Step 4. Expand and collect: $y' = 6x^4 - 30x^2 + 9x^4 - 15x^2 + 3x^2 - 5$ .

\boxed{y' = 15x^4 - 42x^2 - 5}

(Check: expanding $y$ first gives $3x^5 - 14x^3 - 5x$ , whose derivative is also $15x^4 - 42x^2 - 5$ . ✓)

ExampleChain Rule with Trigonometry

Given: A particle's height is $h(t) = \sin(2t^2 + 1)$ metres, with $t$ in seconds. Find the velocity $h'(t)$ and evaluate at $t = 1\text{ s}$ .

Step 1. Identify the composition: outer $f(u) = \sin u$ , inner $u = g(t) = 2t^2 + 1$ .

Step 2. Differentiate each (equations [9], [3]): $f'(u) = \cos u$ and $g'(t) = 4t$ .

Step 3. Apply the chain rule, equation [8]:

17

\quad h'(t) = \cos(2t^2 + 1)\cdot 4t

Derivation hint: Multiply the outer derivative evaluated at the inner function by the inner derivative.

Step 4. Evaluate at $t = 1$ : the inner is $2(1)^2 + 1 = 3$ , so $h'(1) = \cos(3)\cdot 4(1) = 4\cos(3)$ . With $\cos(3\text{ rad}) \approx -0.990$ :

\boxed{h'(1) \approx -3.96\ \text{m/s}}

⚠️ Common mistake

The product rule is NOT $(fg)' = f'g'$ . The derivative of a product is not the product of derivatives. Likewise the quotient rule is not $(f/g)' = f'/g'$ . These are among the most common errors. You must use the full formulas [6] and [7]. Quick check: $\frac{d}{dx}[x\cdot x] = \frac{d}{dx}[x^2] = 2x$ , but $f'g' = (1)(1) = 1 \neq 2x$ .

Counterexample

The chain rule cannot be skipped. Suppose a student writes $\frac{d}{dx}[\sin(x^2)] = \cos(x^2)$ , forgetting the inner derivative. This is wrong. The correct answer is $\cos(x^2)\cdot 2x$ . To see the error concretely at $x = 2$ : the function is $\sin(4)$ , and its true rate of change is $\cos(4)\cdot 4 \approx (-0.654)(4) \approx -2.62$ , not the incorrect $\cos(4)\approx -0.654$ . Forgetting $\cdot\, g'(x)$ understated the rate by a factor of 4.

connection

Differentiation rules are the computational engine behind nearly all of calculus. They feed directly into Implicit Differentiation, where you differentiate both sides of an equation and apply the chain rule to terms in $y$ . They underlie Related Rates problems, where the chain rule links the rates of change of connected quantities. Reversing each rule gives techniques of Integration by Parts (reverse product rule) and Integration by Substitution (reverse chain rule). The Mean Value Theorem guarantees the derivatives these rules compute actually describe average slopes, and Taylor Series uses repeated differentiation to approximate functions.

🔬 Where this appears

⏳ History

- c. 1629 — Pierre de Fermat develops a method of finding maxima and minima ("adequality") that anticipates setting a derivative to zero.

- 1665–1666 — Isaac Newton, during the plague years, invents his "method of fluxions," treating derivatives as rates of flow in time.

- 1675 — Gottfried Wilhelm Leibniz independently develops differential calculus and introduces the elegant

\dfrac{dy}{dx}

notation and the

d

symbol still used today.

- 1684 — Leibniz publishes the first calculus paper, Nova Methodus, stating the product and quotient rules in print.

- 1696 — Guillaume de l'Hôpital publishes the first calculus textbook, spreading Leibniz's rules across Europe.

- 1734 — Bishop George Berkeley critiques the logical foundations in The Analyst, mocking infinitesimals as "ghosts of departed quantities."

- 1797 — Joseph-Louis Lagrange introduces the prime notation

f'(x)

and emphasizes a more algebraic foundation.

- 1821 — Augustin-Louis Cauchy gives rigorous limit-based definitions in Cours d'analyse, finally placing differentiation rules on solid ground.

- 1872 — Karl Weierstrass completes the rigorous

\varepsilon

–

\delta

formulation of limits, fully justifying every differentiation rule.

open-problem

The rules assume differentiability — but not every function is differentiable. A function can be continuous yet have no derivative at a corner (like $|x|$ at $x=0$ ) or a vertical tangent (like $x^{1/3}$ at $0$ ). Weierstrass shocked mathematicians in 1872 with a function that is continuous everywhere yet differentiable nowhere. The differentiation rules also assume you can identify the building blocks; for functions defined implicitly, by integrals, or as infinite series, more careful machinery is needed. A deeper subtlety: the chain rule's proof must handle the case where the inner increment $\Delta g$ equals zero, which the naive "multiply and divide" argument glosses over — a rigorous treatment requires a small extra term. These caveats motivate the careful analysis studied in real analysis.

Formal Statement

Why It Holds

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also