Derivatives — Rules, Chain Rule, Implicit Differentiation

⚡ Key Takeaways

The derivative measures instantaneous rate of change. A small set of rules (power, sum, product, quotient) plus the chain rule (for compositions) and implicit differentiation (for relations not solved for $y$ ) let you differentiate virtually any function built from elementary pieces. Used everywhere rates matter — physics, economics, engineering. Codified by Newton and Leibniz (1660s–1680s).

💡 The Core Insight

Differentiation is local and linear. Near any point, a smooth function looks like a straight line, and the derivative is that line's slope. Because the operation is linear and respects how functions are built (added, multiplied, composed), each construction rule has a matching derivative rule. The chain rule is the deepest: rates of change multiply along a chain of dependencies.

DefinitionThe Derivative

The derivative of $f$ at $x$ is

1

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

Derivation hint: this is the limit of the slope of a secant line through $(x,f(x))$ and $(x+h,f(x+h))$ as the second point slides toward the first. Here $h$ is a small increment in the input; $f(x+h)-f(x)$ is the resulting change in output; their ratio is the average rate of change, and the limit is the instantaneous rate of change. Notations: $f'(x)$ , $\frac{dy}{dx}$ , $\frac{d}{dx}f(x)$ , $Df$ .

TheoremThe Differentiation Rules

Let $u$ and $v$ be differentiable functions of $x$ , and $c$ , $n$ constants.

Constant rule:

2

\quad \frac{d}{dx}(c)=0

Derivation hint: a constant function has $f(x+h)-f(x)=0$ , so the difference quotient is $0$ for all $h$ .

Power rule:

3

\quad \frac{d}{dx}(x^n)=n\,x^{n-1}

Derivation hint: expand $(x+h)^n$ with the binomial theorem; the linear-in- $h$ term gives $nx^{n-1}$ , higher powers vanish in the limit.

Constant multiple rule:

4

\quad \frac{d}{dx}\bigl(c\,u\bigr)=c\,u'

Derivation hint: factor the constant $c$ out of the difference quotient and out of the limit.

Sum rule:

5

\quad \frac{d}{dx}(u+v)=u'+v'

Derivation hint: split the difference quotient of $u+v$ into the difference quotients of $u$ and $v$ ; the limit of a sum is the sum of limits.

Product rule:

6

\quad \frac{d}{dx}(uv)=u'v+uv'

Derivation hint: add and subtract $u(x+h)v(x)$ in the numerator to split into two difference quotients.

Quotient rule:

7

\quad \frac{d}{dx}\!\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^2}, \qquad v\neq 0

Derivation hint: write $u/v$ as $u\cdot v^{-1}$ and combine the product rule with the chain rule applied to $v^{-1}$ .

TheoremThe Chain Rule

If $y=f(g(x))$ — a composition — then

8

\quad \frac{dy}{dx}=f'\bigl(g(x)\bigr)\cdot g'(x)

Derivation hint: write the difference quotient as $\frac{\Delta f}{\Delta g}\cdot\frac{\Delta g}{\Delta x}$ and let both increments shrink; the outer rate times the inner rate. In Leibniz notation, with $u=g(x)$ :

9

\quad \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}

Derivation hint: the differentials "cancel" formally — a mnemonic justified rigorously by the limit argument above.

TheoremImplicit Differentiation

If a relation $F(x,y)=0$ defines $y$ as a function of $x$ (at least locally), differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ and applying the chain rule to every $y$ -term. Then solve algebraically for $\frac{dy}{dx}$ . For example, from $x^2+y^2=r^2$ :

10

\quad 2x+2y\frac{dy}{dx}=0 \;\Longrightarrow\; \frac{dy}{dx}=-\frac{x}{y}

Derivation hint: $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$ by the chain rule, since $y$ depends on $x$ ; isolate $\frac{dy}{dx}$ .

ProofThe Product Rule

We compute the limit directly.

Step 1. Start from the definition applied to $uv$ :

11

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

Derivation hint: just substitute $f=uv$ into equation [1].

Step 2. Add and subtract $u(x+h)v(x)$ in the numerator:

12

\quad (uv)'=\lim_{h\to 0}\frac{u(x+h)\bigl[v(x+h)-v(x)\bigr]+v(x)\bigl[u(x+h)-u(x)\bigr]}{h}

Derivation hint: this clever insertion creates two recognizable difference quotients without changing the value.

Step 3. Split the limit and recognize each piece:

13

\quad (uv)'=\lim_{h\to 0}u(x+h)\cdot\frac{v(x+h)-v(x)}{h}+v(x)\cdot\frac{u(x+h)-u(x)}{h}

Derivation hint: limit of a sum splits; constants relative to $h$ pull out.

Step 4. As $h\to 0$ : $u(x+h)\to u(x)$ (continuity, since $u$ is differentiable), the first quotient $\to v'$ , the second $\to u'$ . Hence

14

\quad (uv)'=u\,v'+v\,u'

Derivation hint: substitute each limit; this is equation [6] rearranged. $\blacksquare$

∎

IN

Isaac Newton

1643–1727

Newton developed his "method of fluxions" in the mid-1660s while sheltering from the plague at Woolsthorpe. He thought of variables as flowing quantities ( $x$ , $y$ ) whose rates of change he called fluxions (written $\dot{x}$ , $\dot{y}$ ). His motivation was physical: velocity is the fluxion of position, acceleration the fluxion of velocity. Newton's notation survives today in physics, where $\dot{x}$ means $dx/dt$ . He used these ideas to derive Kepler's laws from his law of gravitation in the Principia (1687).

🧠 Intuition

Think of the chain rule as gears. If gear A turns 3 times for each turn of gear B, and gear B turns 2 times for each turn of gear C, then A turns $3\times 2 = 6$ times per turn of C. Rates multiply through the linkage. That product $\frac{dA}{dB}\cdot\frac{dB}{dC}=\frac{dA}{dC}$ is exactly equation [9]. For implicit differentiation, picture standing on a circle. You cannot write "the" $y$ as one function of $x$ (the top and bottom semicircles disagree), but at any single point the curve still has a definite tangent slope. Implicit differentiation finds that slope without forcing you to pick a branch — it works with the relation as a whole.

Figure

Tangent line as the limit of secant lines; the derivative is the limiting slope

ExampleChain Rule with the Power Rule

Given: $y=(3x^2+1)^4$ . Find $\frac{dy}{dx}$ .

Step 1. Identify the composition. Outer function: $u^4$ . Inner function: $u=3x^2+1$ .

Step 2. Differentiate outer: $\frac{dy}{du}=4u^3=4(3x^2+1)^3$ .

Step 3. Differentiate inner: $\frac{du}{dx}=6x$ .

Step 4. Multiply (equation [9]):

15

\quad \frac{dy}{dx}=4(3x^2+1)^3\cdot 6x = 24x(3x^2+1)^3

Derivation hint: outer derivative evaluated at the inner function, times the inner derivative.

Check at $x=1$ : $\frac{dy}{dx}=24(1)(4)^3=24\cdot 64=1536$ .

\boxed{\dfrac{dy}{dx}=24x\,(3x^2+1)^3}

ExampleImplicit Differentiation — slope on a curve

Given: the ellipse $x^2+4y^2=8$ . Find the slope of the tangent at the point $(2,1)$ .

Step 1. Differentiate both sides with respect to $x$ , treating $y$ as $y(x)$ :

16

\quad 2x+8y\frac{dy}{dx}=0

Derivation hint: $\frac{d}{dx}(4y^2)=4\cdot 2y\,y'=8y\,y'$ by the chain rule.

Step 2. Solve for $\frac{dy}{dx}$ :

17

\quad \frac{dy}{dx}=-\frac{2x}{8y}=-\frac{x}{4y}

Derivation hint: subtract $2x$ , divide by $8y$ .

Step 3. Substitute the point $(2,1)$ :

\frac{dy}{dx}\Big|_{(2,1)}=-\frac{2}{4(1)}=-\frac{1}{2}.

Verify the point is on the curve: $2^2+4(1)^2=4+4=8.$ ✓

\boxed{\text{slope}=-\tfrac{1}{2}\ \text{(dimensionless)}}

⚠️ Common mistake

The product rule is NOT $u'v'$ . Many students guess $\frac{d}{dx}(uv)=u'v'$ . This is wrong — you must use equation [6], $u'v+uv'$ . Test it: $\frac{d}{dx}(x\cdot x)=\frac{d}{dx}(x^2)=2x$ , but $u'v'=1\cdot 1=1\neq 2x$ .

Don't forget the inner derivative. Writing $\frac{d}{dx}\sin(x^2)=\cos(x^2)$ is incomplete — you must multiply by $\frac{d}{dx}(x^2)=2x$ , giving $2x\cos(x^2)$ . The missing factor is the single most common chain-rule error.

In implicit differentiation, $y$ is a function, not a constant. $\frac{d}{dx}(y^3)$ is not $3y^2$ ; it is $3y^2\frac{dy}{dx}$ .

Counterexample

Consider $f(x)=x^2$ and $g(x)=x^3$ . If the false rule $(fg)'=f'g'$ held, then $(x^5)'=(2x)(3x^2)=6x^3$ . But the correct power rule gives $(x^5)'=5x^4$ . Since $6x^3\neq 5x^4$ in general (they're equal only at $x=0$ and $x=6/5$ ), the naive product "rule" fails — confirming the genuine product rule [6] is needed.

connection

These rules are the computational engine of calculus. They depend on Limits and Continuity, since every rule is ultimately a limit of difference quotients. They feed directly into Applications of Derivatives — finding maxima, minima, and rates. The chain rule reappears as the heart of Related Rates, where quantities change in time, and of Integration by Substitution, which reverses it. Implicit differentiation generalizes into Partial Derivatives and the multivariable chain rule. Together they make The Fundamental Theorem of Calculus computationally usable.

🔬 Where this appears

⏳ History

- c. 250 BCE — Archimedes computes tangents to spirals and areas by exhaustion, foreshadowing limits.

- 1629 — Pierre de Fermat introduces "adequality," a method for tangents and extrema that anticipates the difference quotient.

- 1665–1666 — Isaac Newton develops fluxions during the plague years, treating curves as traced by flowing motion.

- 1675 — Gottfried Leibniz invents the

\frac{dy}{dx}

and

\int

notations and states the product and quotient rules explicitly.

- 1684 — Leibniz publishes Nova Methodus, the first printed account of differential calculus, including the chain rule and quotient rule.

- 1687 — Newton publishes the Principia, applying fluxions to celestial mechanics.

- 1696 — L'Hôpital writes the first calculus textbook, spreading Leibniz's methods across Europe.

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

- 1821 — Augustin-Louis Cauchy rigorously grounds the derivative in the limit concept, resolving Berkeley's objections.

open-problem

The rules above assume differentiability, which is stronger than continuity. The function $f(x)=|x|$ is continuous everywhere but has no derivative at $x=0$ (the left and right slopes disagree). More dramatically, the Weierstrass function is continuous everywhere yet differentiable nowhere — a function so jagged it has no tangent at any point. Implicit differentiation also has a hidden assumption: that $F(x,y)=0$ actually defines $y$ as a function near the point of interest. The Implicit Function Theorem specifies exactly when this works — roughly, when $\frac{\partial F}{\partial y}\neq 0$ . At points where it vanishes (like the top/bottom of a circle, where the tangent is vertical), $\frac{dy}{dx}$ is undefined or infinite, and the method must be handled with care.

Derivatives — Rules, Chain Rule, Implicit Differentiation

Formal Statement

Why It Holds

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also