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Lesson 53 — The Chain Rule

Derivative of a composite function: if y = f(g(x)), then dy/dx = f'(g(x))·g'(x). The most used rule in all of applied calculus.

Used in: 2nd Year High School (16 years old) · Equivalent Japanese Math II/III §微分 · Equivalent German Klasse 11 Abitur

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

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Rigorous notation, full derivation, hypotheses

Definition and theory

Formal Statement

"The chain rule states that the derivative of $f(g(x))$ is $f'(g(x)) \cdot g'(x)$ . That is, we differentiate the outer function $f$ , evaluate it at the inner function $g(x)$ , and multiply by the derivative of the inner function." — OpenStax Calculus Volume 1, §3.6

"Think of the process from outside in: identify the outer function, differentiate it while keeping the inner function unchanged, then multiply by the derivative of the inner function." — Boelkins, Active Calculus §2.5

Rigorous Proof

The difficulty lies in the fact that $g(x+h) - g(x)$ can be zero for $h \neq 0$ , invalidating the naive argument of canceling $\Delta g$ . The solution uses the auxiliary function:

$Q(y) = \begin{cases} \dfrac{f(y) - f(g(a))}{y - g(a)} & y \neq g(a) \\ f'(g(a)) & y = g(a) \end{cases}$

$Q$ is continuous at $g(a)$ (by differentiability of $f$ ). Since $f(g(a+h)) - f(g(a)) = Q(g(a+h)) \cdot [g(a+h) - g(a)]$ , dividing by $h$ and taking the limit $h \to 0$ yields $(f \circ g)'(a) = f'(g(a)) \cdot g'(a)$ .

Fundamental Special Cases

Composite Function	Derivative
$[g(x)]^n$	$n\,[g(x)]^{n-1} \cdot g'(x)$
$\sin(g(x))$	$\cos(g(x)) \cdot g'(x)$
$\cos(g(x))$	$-\sin(g(x)) \cdot g'(x)$
$e^{g(x)}$	$e^{g(x)} \cdot g'(x)$
$\ln(g(x))$	$g'(x)/g(x)$
$\sqrt{g(x)}$	$g'(x) / (2\sqrt{g(x)})$
$a^{g(x)}$	$a^{g(x)} \ln a \cdot g'(x)$

Triple Composition

For $h(x) = f(g(k(x)))$ :

(f \circ g \circ k)'(x) = f'(g(k(x))) \cdot g'(k(x)) \cdot k'(x)

(tripla)

what this means · The chain rule generalizes: multiply the derivative of each layer, always from outside in.

Composition Diagram

Flow of composition: input x, processed by g to yield u, then by f to yield y. The total rate dy/dx is the product of the individual rates.

Solved Examples

Example— 1· Generalized Power Rule — Composition with a Polynomial

Problem: Calculate the derivative of $h(x) = (x^2 + 1)^5$ .

Strategy: The function $h$ is a composition: the outer function is $u^5$ and the inner function is $u = x^2 + 1$ . We apply the chain rule by explicitly identifying each layer.

Solution:

Identify the layers:
- Outer function: $f(u) = u^5$
- Inner function: $g(x) = x^2 + 1$
Derivative of the outer layer: $f'(u) = 5u^4$ . Evaluated at $g(x)$ : $f'(g(x)) = 5(x^2+1)^4$ .
Derivative of the inner layer: $g'(x) = 2x$ .
Apply the chain rule: $h'(x) = f'(g(x)) \cdot g'(x) = 5(x^2+1)^4 \cdot 2x = 10x(x^2 + 1)^4$

Verification: For $x = 0$ : $h(0) = 1^5 = 1$ , $h'(0) = 10 \cdot 0 \cdot 1 = 0$ . The function has a minimum at $x = 0$ (a shifted parabola raised to the 5th power), which is consistent with $h'(0) = 0$ .

Source. Active Calculus §2.5 — Boelkins · 2024 · CC-BY-NC-SA.

Example— 2· Chain Rule with Exponential and Trigonometry

Problem: Calculate the derivative of $f(x) = e^{\sin x}$ .

Strategy: Composition with the exponential function. The formula $\frac{d}{dx}[e^{g(x)}] = e^{g(x)} \cdot g'(x)$ is a direct special case of the chain rule.

Solution:

Outer layer: $f(u) = e^u$ , derivative $e^u$ .
Inner layer: $g(x) = \sin x$ , derivative $\cos x$ .
Result: $f'(x) = e^{\sin x} \cdot \cos x$

Interpretation: The derivative changes sign whenever $\cos x = 0$ , i.e., at $x = \pi/2 + k\pi$ . At the maxima of $\sin x$ , the derivative becomes zero — because $\cos(\pi/2) = 0$ .

Source. OpenStax Calculus Volume 1 §3.6 — OpenStax · 2016 · CC-BY-NC-SA.

Example— 3· Chain Rule with Natural Logarithm

Problem: Calculate the derivative of $g(x) = \ln(3x^2 + 5)$ .

Strategy: Composition with the natural logarithm. The formula $\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}$ is the direct special case.

Solution:

Outer layer: $f(u) = \ln u$ , derivative $1/u$ . Evaluated at $g(x)$ : $\dfrac{1}{3x^2 + 5}$ .
Inner layer: $g(x) = 3x^2 + 5$ , derivative $g'(x) = 6x$ .
Result: $g'(x) = \frac{1}{3x^2 + 5} \cdot 6x = \frac{6x}{3x^2 + 5}$

Verification: For large $x$ , $g'(x) \approx \frac{6x}{3x^2} = \frac{2}{x} \to 0$ . This makes sense: the logarithm grows slower and slower.

Source. Active Calculus §2.5 — Boelkins · 2024 · CC-BY-NC-SA.

Example— 4· Triple Composition — Nested Chain

Problem: Calculate the derivative of $h(x) = \sin(e^{x^2})$ .

Strategy: There are three nested layers. We apply the chain rule from outside to inside, one layer at a time.

Solution:

Write the layers explicitly:

Layer 1 (outermost): $f_1(u) = \sin u$ , derivative $\cos u$
Layer 2 (intermediate): $f_2(v) = e^v$ , derivative $e^v$
Layer 3 (innermost): $f_3(x) = x^2$ , derivative $2x$

Applying the chain rule from outside in:

$h'(x) = \cos(e^{x^2}) \cdot e^{x^2} \cdot 2x$

Verification: For $x = 0$ : $h'(0) = \cos(1) \cdot 1 \cdot 0 = 0$ . Numerically verifiable: $h(0+0.001) - h(0) \approx 0$ .

Source. OpenStax Calculus Volume 1 §3.6, Example 3.47 — CC-BY-NC-SA.

Example— 5· Chain Rule Combined with Product Rule

Problem: Calculate the derivative of $f(x) = x^2 \cdot e^{-x^2}$ .

Strategy: This is a product of two functions, one of which is composite. We apply the product rule first, then the chain rule to the second factor.

Solution:

Identify the two factors of the product:

$u(x) = x^2$ , derivative $u'(x) = 2x$
$v(x) = e^{-x^2}$ , derivative via chain rule: $v'(x) = e^{-x^2} \cdot (-2x) = -2x e^{-x^2}$

Applying the product rule $[u \cdot v]' = u' v + u v'$ :

$f'(x) = 2x \cdot e^{-x^2} + x^2 \cdot (-2x e^{-x^2}) = 2x e^{-x^2} - 2x^3 e^{-x^2}$

Factoring out $2x e^{-x^2}$ :

$f'(x) = 2x e^{-x^2}(1 - x^2)$

Interpretation: The critical points are $x = 0$ and $x = \pm 1$ . This function appears in the normal distribution: the $(1-x^2)$ factor determines the extrema at $x = \pm 1$ .

Source. Active Calculus §2.5, Activity 2.5.3 — Boelkins · 2024 · CC-BY-NC-SA.

Exercise list

40 exercises · 10 with worked solution (25%)

26 4 6 3 1

Ex. 53.1Answer key
Calculate $\dfrac{d}{dx}[(2x+3)^5]$ .
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Ex. 53.2
Calculate $(\sin(4x))'$ .
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Ex. 53.3
Calculate $(e^{x^2})'$ .
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Ex. 53.4Answer key
Calculate $(\ln(x^2 + 1))'$ .
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Ex. 53.5Answer key
Calculate $(\sqrt{x^2+1})'$ .
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Ex. 53.6
Calculate $(\cos^2 x)'$ .
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Ex. 53.7
Calculate $(\tan(2x))'$ .
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Ex. 53.8
Calculate $(\sin(\cos x))'$ .
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Ex. 53.9
Calculate $(e^{\sin x})'$ .
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Ex. 53.10
Calculate $((\ln x)^3)'$ .
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Ex. 53.11
Calculate $(\sqrt{1-x^2})'$ .
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Ex. 53.12
Calculate $\left(\dfrac{1}{x^2+4}\right)'$ .
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Ex. 53.13Answer key
Calculate $(2^{x^2})'$ .
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Ex. 53.14
Calculate $(\sin^3(2x))'$ .
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Ex. 53.15
Calculate $(\arctan(x^2))'$ . (Ans: $2x/(1+x^4)$ .)
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Ex. 53.16
Calculate $(\ln(\sin x))'$ .
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Ex. 53.17Answer key
Calculate $(e^{\cos(2x)})'$ .
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Ex. 53.18
Calculate $(\sqrt{\tan x})'$ .
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Ex. 53.19Answer key
Calculate $(\sin(\sqrt{x}))'$ .
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Ex. 53.20
Calculate $((3x+5)^{10})'$ .
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Ex. 53.21Answer key
Calculate $(\cos(3x^2+2))'$ .
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Ex. 53.22
Calculate $(\ln(\ln x))'$ .
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Ex. 53.23
Calculate $(x \cdot \sin(x^2))'$ .
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Ex. 53.24
Calculate $(x \cdot e^{x^2})'$ .
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Ex. 53.25
Calculate $(\sin(\sqrt{x^2+1}))'$ .
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Ex. 53.26Answer key
Calculate $(\tan^2(3x))'$ .
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Ex. 53.27
Find the tangent line to the curve $y = (x^2+1)^3$ at the point $x = 1$ . (Ans: $y = 24x - 16$ .)
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Ex. 53.28
Error Analysis. A student writes $(e^{x^2})' = xe^{x^2}$ . What specific error was made?
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Ex. 53.29
Conceptual. To differentiate $\sin(x^2)$ , which rule applies? Why isn't it the product rule?
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Ex. 53.30
Calculate $(e^{e^x})'$ .
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Ex. 53.31
Physics. The position of a particle in simple harmonic motion is $s(t) = \sin(\omega t)$ . Calculate the acceleration $s''(t)$ and show that $s''(t) = -\omega^2 s(t)$ .
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Ex. 53.32Answer key
Nuclear Physics. Radioactive decay follows $N(t) = N_0 e^{-\lambda t}$ . Calculate $N'(t)$ and show that $N'(t) = -\lambda N(t)$ .
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Ex. 53.33Answer key
Biology. Logistic growth is $P(t) = \dfrac{K}{1 + e^{-rt}}$ . Calculate $P'(0)$ .
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Ex. 53.34
Statistics. Calculate $f'(x)$ for the standard normal density $f(x) = \dfrac{1}{\sqrt{2\pi}} e^{-x^2/2}$ .
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Ex. 53.35
Finance. The present value of a cash flow $S$ discounted at rate $r$ is $V(t) = S\,e^{-rt}$ . Calculate $dV/dt$ and interpret the result.
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Ex. 53.36
Physics. Kinetic energy is $E(v) = \tfrac{1}{2}mv^2$ and velocity is $v(t) = at$ . Calculate $dE/dt$ using the chain rule and verify with direct differentiation.
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Ex. 53.37
Calculate $(\sin(\cos(\sin x)))'$ .
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Ex. 53.38
Calculate $\dfrac{d}{dx}\left[(x + \sqrt{x^2+1})^n\right]$ .
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Ex. 53.39
Calculate $(\sin(e^{x^2}))'$ .
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Ex. 53.40
Proof. Explain why the naive argument $\frac{\Delta f}{\Delta g} \cdot \frac{\Delta g}{h}$ fails as a rigorous proof of the chain rule. How does the auxiliary function $Q(y)$ resolve the issue?
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Sources

Active Calculus 2.0 — Boelkins, Austin, Schlicker · 2024 · §2.5. Primary source. CC-BY-NC-SA.
Calculus Volume 1 — OpenStax (Herman et al.) · 2016 · §3.6. CC-BY-NC-SA.
APEX Calculus — Hartman et al. · 2024 · v5 · §2.5. CC-BY-NC.