The Timpson Lyceum · Calculus 1 (2024)

Chain Rule and Implicit Differentiation - Lecture 10 (Calculus 1)

Overview

This lecture introduces the chain rule, the differentiation rule for composite functions (a function nested inside another function), and then applies the same idea to implicit differentiation. The chain rule is the last of the core differentiation rules — after the power, sum, product, and quotient rules — and it is the one students forget to apply most often. It lets you differentiate expressions like $(3x^2+1)^2$ or $(5x^3-x^4)^7$ without expanding them, and it underpins implicit differentiation, where $y$ is defined implicitly by an equation rather than solved for explicitly. By the end you can find slopes, tangent lines, and normal lines for curves like circles and other implicit relations.

Key Concepts

Review of the basic rules

• Constant: $\frac{d}{dx}(c) = 0$.
• Power rule: $\frac{d}{dx}(x^n) = n\,x^{n-1}$.
• Sum rule: the derivative of a sum is the sum of the derivatives.
• Product rule: $\frac{d}{dx}(uv) = u\,v' + v\,u'$ ("first times derivative of the second plus second times derivative of the first").
• Quotient rule: $\frac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{v\,u' - u\,v'}{v^2}$ ("low d-high minus high d-low, over the bottom squared").

The chain rule

• For a composition $f(g(x))$, the derivative is the derivative of the outside (evaluated at the inside) times the derivative of the inside: $\frac{d}{dx}\,f\!\big(g(x)\big) = f'\!\big(g(x)\big)\cdot g'(x).$
• Leibniz form: if $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}.$
• In practice you do not relabel everything as $f$ and $g$; you just read it as "derivative of the outer shell $\times$ derivative of what's inside," working from the outermost operation inward.
• Note that the outer function $f$ itself does not appear in the answer — only $f'$ evaluated at $g(x)$ and $g'(x)$ do.

Why the proof works (sketch)

• Since $g$ is differentiable at $x_0$, it is continuous there, so as $x \to x_0$, $g(x)\to g(x_0)$.
• Starting from the limit definition and multiplying top and bottom by $g(x)-g(x_0)$, the difference quotient splits into two factors whose limits are $f'(g(x_0))$ and $g'(x_0)$ respectively. In Leibniz notation this is the intuitive (though not literally a fraction) cancellation $\frac{dy}{du}\cdot\frac{du}{dx} = \frac{dy}{dx}$.

Chained compositions

• For three nested functions $y=f(u)$, $u=g(w)$, $w=h(x)$: $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dw}\cdot\frac{dw}{dx}.$
• The chain extends to any depth; you can even view a "plain" derivative as a chain rule whose inner derivative is $1$ (e.g. $\frac{d}{dx}\sin x = \cos x \cdot 1$).

Common chain-rule trick for quotients/roots

• Rewrite $\dfrac{1}{3x-2}$ as $(3x-2)^{-1}$ and $\sqrt{x}$ as $x^{1/2}$, then apply the chain rule to avoid the quotient rule. For example $\frac{d}{dx}(3x-2)^{-1} = -1(3x-2)^{-2}\cdot 3 = \dfrac{-3}{(3x-2)^2}$.

Implicit functions and implicit differentiation

• An implicit function is one defined by an equation that implies $y$ in terms of $x$ without giving $y$ explicitly. Example: $y^2 = x$ implies $y = \pm\sqrt{x}$ (two outputs per input).
• Implicit differentiation: differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$. Every time you differentiate a $y$-term you must apply the chain rule, attaching a factor of $\frac{dy}{dx}$ (written $y'$).
• Key contrast: $\frac{dx}{dx}=1$, but $\frac{dy}{dx}$ is not $1$ — it is the unknown you solve for. After differentiating, collect all $y'$ terms on one side, factor out $y'$, and divide.
• Tangent and normal lines: the tangent line at a point has slope $m = \frac{dy}{dx}$ there; the normal line is perpendicular, with slope $-\frac{1}{m}$ (negative reciprocal). Use point-slope form $y - y_1 = m(x - x_1)$ for each.

Worked Example

Chain rule. Using $\frac{d}{dx}f(g(x)) = f'(g(x))\,g'(x)$, differentiate $y=(3x^2+1)^2$ with outer $f(u)=u^2$ and inner $g(x)=3x^2+1$:

which matches expanding first and differentiating directly.

Implicit differentiation. Find the slope of the circle $x^2 + y^2 = 25$ at the point $(3,4)$. Differentiate both sides with respect to $x$, applying the chain rule to the $y^2$ term, then solve for $y'$:

So the curve has slope $-\tfrac{3}{4}$ at $(3,4)$. The tangent line is $y - 4 = -\tfrac{3}{4}(x-3)$, and the normal line (negative reciprocal slope $\tfrac{4}{3}$) is $y - 4 = \tfrac{4}{3}(x-3)$.

Summary

• The chain rule differentiates a composite function as the derivative of the outside (at the inside) times the derivative of the inside: $\frac{d}{dx}f(g(x)) = f'(g(x))\,g'(x)$, or $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$.
• Apply it from the outermost operation inward; it lets you differentiate powers, roots, and trig of complicated expressions without expanding (e.g. $(5x^3-x^4)^7$).
• The chain rule extends to any depth of nesting: $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dw}\cdot\frac{dw}{dx}$.
• Rewriting reciprocals and roots as powers (e.g. $(3x-2)^{-1}$, $x^{1/2}$) often replaces a messy quotient rule with a cleaner chain rule.
• Implicit differentiation differentiates both sides with respect to $x$, attaching a $y'$ factor (via the chain rule) to every $y$-term, then solves algebraically for $y'$.
• Use the resulting slope $m=\frac{dy}{dx}$ for the tangent line and its negative reciprocal $-\frac1m$ for the normal line, each through the given point in point-slope form.