Chapter 03

Medium

Chain Rule, Implicit Differentiation

00 · Symbol Glossary

$f(g(x))$f of g of x — composite function

A composition of two functions: first apply $g$ to $x$ , then apply $f$ to the result. $g$ is the inner function, $f$ is the outer function. Read "f of g of x."

$u$u — substitution variable

A placeholder variable used to name the inner function during a chain-rule computation: $u=g(x)$ . Once substituted, the outer function is differentiated as a function of $u$ alone.

$\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}$chain rule in Leibniz form

The chain rule written with Leibniz notation (Chapter 02, Section 00): differentiate the outer layer with respect to the inner variable $u$ , then multiply by the derivative of the inner layer with respect to $x$ . The $du$ terms visually "cancel" as a mnemonic, though this is not literal fraction cancellation.

$y'$y prime — shorthand for dy/dx

A compact alternative to $\dfrac{dy}{dx}$ when the dependent variable $y$ is understood from context, used heavily once implicit differentiation is introduced.

01 · Composite Functions

Many functions are built by feeding the output of one function into another. $y = (x^2+1)^5$ is not a bare power of $x$ — it is the fifth power applied to the result of $x^2+1$ . Recognizing this layered structure is the first step before differentiating it.

Definition — Composite Function

Given two functions $f$ and $g$ , the composition $f(g(x))$ is formed by using $g(x)$ as the input to $f$ .

(f\circ g)(x) = f(g(x))

$g$ — the inner function, applied first.

$f$ — the outer function, applied second, to the output of $g$ .

Example — Decomposing $y=(x^2+1)^5$

Identify the layers: the innermost operation is squaring and adding $1$ , so $g(x)=x^2+1$ . The outermost operation is raising to the fifth power, so $f(u)=u^5$ . Then $y = f(g(x)) = (g(x))^5 = (x^2+1)^5$ , matching the original expression.

02 · The Chain Rule

Neither the power rule nor the product rule (Chapter 02) can differentiate $(x^2+1)^5$ directly — it is a single power of an expression, not of $x$ alone. The chain rule handles composite functions by differentiating each layer separately and multiplying the results.

Definition — Chain Rule

If $y = f(g(x))$ , then

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

"Derivative of the outer function, evaluated at the inner function (leaving the inner function untouched inside), times the derivative of the inner function."

Step-by-step — Differentiate $y=(x^2+1)^5$

Identify inner and outer:

g(x)=x^2+1

(inner),

f(u)=u^5

(outer), so

y=f(g(x))

Differentiate the outer function, keeping the inner intact:

f'(u)=5u^4

, so

f'(g(x)) = 5(x^2+1)^4

— the exponent

5

came down as a multiplier and the inner expression

x^2+1

is substituted back in unexpanded.

Differentiate the inner function:

g'(x) = 2x

(power rule).

Multiply the two results:

\dfrac{dy}{dx} = 5(x^2+1)^4 \cdot 2x

Simplify:

\dfrac{dy}{dx} = 10x(x^2+1)^4

. The coefficient

10

came from

5\times2

❌ Forgetting to multiply by the inner derivative

A common shortcut error on the same function: writing $\dfrac{dy}{dx} = 5(x^2+1)^4$ and stopping there.

Why it breaks: this only accounts for the outer layer's rate of change, treating $x^2+1$ as if it were the variable itself changing at rate $1$ . But $x^2+1$ changes at rate $2x$ , not $1$ — omitting the factor $g'(x)=2x$ silently assumes the inner function's slope is always $1$ , which is false everywhere except where $2x=1$ .

Consequence: the resulting derivative is wrong at every $x$ except $x=\frac12$ — a coincidental match, not a valid method. The multiplication by $g'(x)$ is not optional.

03 · Chain Rule with Leibniz Notation

Writing $u=g(x)$ explicitly and using $\dfrac{dy}{du}$ , $\dfrac{du}{dx}$ makes multi-layer compositions easier to track, especially once three or more layers are nested.

Step-by-step — Differentiate $y = \sqrt{3x^2-x}$ using Leibniz notation

Name the inner expression: let

u = 3x^2-x

, so

y = \sqrt{u} = u^{1/2}

Differentiate $y$ with respect to $u$ :

\dfrac{dy}{du} = \frac12 u^{-1/2} = \dfrac{1}{2\sqrt{u}}

(power rule with

n=\frac12

Differentiate $u$ with respect to $x$ :

\dfrac{du}{dx} = 6x-1

(power rule term by term on

3x^2-x

Multiply and substitute $u$ back:

\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx} = \dfrac{1}{2\sqrt{u}}\cdot(6x-1) = \dfrac{6x-1}{2\sqrt{3x^2-x}}

Chaining through three or more layers

For $y=\sin\big((x^3+1)^2\big)$ -style triple compositions (outer trig-like layer, middle power layer, inner polynomial), apply the chain rule repeatedly: differentiate the outermost layer, multiply by the derivative of the next layer in, multiply by the derivative of the layer after that, and so on until reaching a bare polynomial in $x$ .

04 · Implicit Differentiation

Every derivative computed so far assumed $y$ was written explicitly as a function of $x$ , like $y=x^2+1$ . Some equations mix $x$ and $y$ together with no way to isolate $y$ cleanly — for example, $x^2+y^2=25$ . Implicit differentiation finds $\dfrac{dy}{dx}$ directly from such an equation, without first solving for $y$ .

Definition — Implicit Differentiation

To differentiate an equation containing both $x$ and $y$ :

Differentiate both sides of the equation with respect to $x$ , term by term.
Every time a term contains $y$ , apply the chain rule: treat $y$ as an inner function of $x$ , so $\dfrac{d}{dx}[y^n] = n y^{n-1}\cdot\dfrac{dy}{dx}$ .
Solve the resulting equation algebraically for $\dfrac{dy}{dx}$ .

Step-by-step — Find $\dfrac{dy}{dx}$ for $x^2+y^2=25$

Differentiate both sides with respect to $x$ :

\dfrac{d}{dx}[x^2] + \dfrac{d}{dx}[y^2] = \dfrac{d}{dx}[25]

Differentiate $x^2$ normally: power rule gives

2x

Differentiate $y^2$ using the chain rule: treat

y

as a function of

x

;

\dfrac{d}{dx}[y^2] = 2y\cdot\dfrac{dy}{dx}

— the extra factor

\dfrac{dy}{dx}

appears precisely because

y

is itself changing with

x

, unlike the plain variable

x

Differentiate the right side:

25

is a constant, so its derivative is

0

Assemble the differentiated equation:

2x + 2y\dfrac{dy}{dx} = 0

Solve algebraically for $\dfrac{dy}{dx}$ :

2y\dfrac{dy}{dx} = -2x \implies \dfrac{dy}{dx} = \dfrac{-2x}{2y} = -\dfrac{x}{y}

Example — Slope on the circle at a specific point

For $x^2+y^2=25$ (a circle of radius $5$ ), find the slope at the point $(3,4)$ — check first that this point satisfies the equation: $3^2+4^2=9+16=25$ . ✓

Using $\dfrac{dy}{dx}=-\dfrac{x}{y}$ from the step-block above: $\dfrac{dy}{dx}\Big|_{(3,4)} = -\dfrac{3}{4}$ .

Geometrically, the radius to $(3,4)$ has slope $\dfrac{4}{3}$ ; the tangent line's slope $-\dfrac34$ is its negative reciprocal — the tangent to a circle is always perpendicular to the radius at the point of tangency.

❌ Treating y as a constant during implicit differentiation

On the same equation $x^2+y^2=25$ , a common mistake is writing $\dfrac{d}{dx}[y^2]=2y$ , omitting the chain-rule factor $\dfrac{dy}{dx}$ entirely — treating $y$ as if it does not depend on $x$ .

Why it breaks: $y$ is defined implicitly in terms of $x$ through the equation itself; wherever $x$ moves, $y$ generally moves too, at the (unknown, to-be-solved-for) rate $\dfrac{dy}{dx}$ . Omitting that factor is the same category of error as omitting $g'(x)$ in the chain rule (Section 02) — because differentiating $y^2$ with respect to $x$ is an application of the chain rule with inner function $y(x)$ .

Consequence: the resulting equation $2x+2y=0$ gives $y=-x$ , a straight line — nothing like the correct slope formula $\dfrac{dy}{dx}=-x/y$ , and wrong at every point on the circle except where it happens to coincide by accident.

Implicit differentiation extends naturally to situations where every quantity in an equation is itself changing with respect to time $t$ — differentiate both sides with respect to $t$ instead of $x$ , applying the chain rule to every variable.

Example — Related rates, compressed

A trading desk's exposure follows $V = P\cdot n$ , where $P$ is price per share and $n$ is number of shares held, both changing with time $t$ . Differentiating both sides with respect to $t$ using the product rule (Chapter 02) combined with the chain rule: $\dfrac{dV}{dt} = \dfrac{dP}{dt}\cdot n + P\cdot\dfrac{dn}{dt}$ .

If $P=50$ dollars, $n=200$ shares, $\dfrac{dP}{dt}=2$ dollars/hour, and the desk is unwinding the position at $\dfrac{dn}{dt}=-10$ shares/hour, then $\dfrac{dV}{dt} = 2(200) + 50(-10) = 400-500=-100$ : total exposure is falling by $100$ dollars/hour, even though the price itself is rising, because the position is being reduced faster than the price gain compensates.

06 · Exercises

EXERCISE 3.1

Identify the inner and outer functions first: the inner is the expression being cubed, the outer is the cubing operation itself.

$y=(4x-3)^3$ . Inner $g(x)=4x-3$ , outer $f(u)=u^3$ . $f'(u)=3u^2 \implies f'(g(x))=3(4x-3)^2$ . $g'(x)=4$ . Chain rule: $y' = 3(4x-3)^2\cdot4 = 12(4x-3)^2$ .

Differentiate $y=(4x-3)^3$ using the chain rule. Identify the inner and outer functions explicitly.

EXERCISE 3.2

Rewrite the square root as a $\frac12$ power before applying the chain rule.

$y=\sqrt{x^3+2x}=(x^3+2x)^{1/2}$ . Outer derivative: $\frac12(x^3+2x)^{-1/2}$ . Inner derivative: $3x^2+2$ . Chain rule: $y' = \frac12(x^3+2x)^{-1/2}(3x^2+2) = \dfrac{3x^2+2}{2\sqrt{x^3+2x}}$ .

Differentiate $y=\sqrt{x^3+2x}$ .

EXERCISE 3.3

Differentiate both sides with respect to $x$ , remembering to apply the chain rule ( $n y^{n-1}\,dy/dx$ ) to every $y$ term, then isolate $dy/dx$ algebraically.

$x^3+y^3=9$ . Differentiate: $3x^2 + 3y^2\dfrac{dy}{dx} = 0$ . Solve: $3y^2\dfrac{dy}{dx} = -3x^2 \implies \dfrac{dy}{dx} = \dfrac{-3x^2}{3y^2} = -\dfrac{x^2}{y^2}$ .

Find $\dfrac{dy}{dx}$ for $x^3+y^3=9$ using implicit differentiation.

EXERCISE 3.4

This equation has a product of $x$ and $y$ — differentiating that term requires the product rule combined with the chain rule, since $y$ depends on $x$ .

$xy + y^2 = 6$ . Differentiate $xy$ with the product rule: $\dfrac{d}{dx}[xy] = 1\cdot y + x\cdot\dfrac{dy}{dx} = y + x\dfrac{dy}{dx}$ . Differentiate $y^2$ with the chain rule: $2y\dfrac{dy}{dx}$ . Right side derivative: $0$ . Assemble: $y + x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0$ . Group $\dfrac{dy}{dx}$ terms: $\dfrac{dy}{dx}(x+2y) = -y$ . Solve: $\dfrac{dy}{dx} = \dfrac{-y}{x+2y}$ .

Find $\dfrac{dy}{dx}$ for $xy+y^2=6$ . Note that one term requires both the product rule and the chain rule together.

EXERCISE 3.5

Check the point satisfies the equation first, then use the implicit-differentiation formula for slope, then plug in the specific coordinates.

$x^2+y^2=25$ at $(-4,3)$ : check $(-4)^2+3^2=16+9=25$ ✓. From Section 04, $\dfrac{dy}{dx}=-\dfrac{x}{y}$ . At $(-4,3)$ : $\dfrac{dy}{dx} = -\dfrac{-4}{3} = \dfrac43$ .

Find the slope of the tangent line to $x^2+y^2=25$ at the point $(-4,3)$ .

EXERCISE 3.6

Differentiate the given relation with respect to time $t$ using the product rule, since both quantities in the product are changing with $t$ .

$R = q\cdot p$ (revenue = quantity sold times price), both functions of $t$ . $\dfrac{dR}{dt} = \dfrac{dq}{dt}\cdot p + q\cdot\dfrac{dp}{dt}$ . With $q=1000$ units, $p=20$ dollars, $\dfrac{dq}{dt}=50$ units/day, $\dfrac{dp}{dt}=-0.50$ dollars/day: $\dfrac{dR}{dt} = 50(20) + 1000(-0.50) = 1000-500=500$ . Revenue is increasing at $500$ dollars/day — the volume gain outweighs the price decline.

Revenue is $R=qp$ where quantity sold $q$ and price $p$ both change with time $t$ . Given $q=1000$ units, $p=20$ dollars, $\dfrac{dq}{dt}=50$ units/day, and $\dfrac{dp}{dt}=-0.50$ dollars/day, find $\dfrac{dR}{dt}$ using implicit (related-rate) differentiation.

07 · Chapter Summary

Concept	Formula / Rule
Composite function	$f(g(x))$ : apply inner $g$ , then outer $f$
Chain rule	$\dfrac{dy}{dx} = f'(g(x))\cdot g'(x)$
Chain rule (Leibniz)	$\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}$
Implicit differentiation	Differentiate both sides w.r.t. $x$ ; chain-rule every $y$ term as $ny^{n-1}\dfrac{dy}{dx}$
Product + chain together	Terms like $xy$ need the product rule with a $\dfrac{dy}{dx}$ factor on the $y$ half
Related rates	Differentiate a governing equation with respect to $t$ to link multiple changing quantities

Next: Chapter 04 — Applications of Derivatives: Core, where the derivative rules built here are used to analyze how functions rise, fall, and curve.