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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)

The Chain Rule: the derivative of a composite function $f(g(x))$ is the derivative of the outer function, evaluated at the inner function, multiplied by the derivative of the inner function. In Leibniz notation: $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ , where $u = g(x)$ .

Choose your door

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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Inner: $g(x) = 2x+3$ , derivative $g'(x) = 2$ . Outer: $f(u) = u^5$ , derivative $5u^4$ evaluated at $g(x)$ : $5(2x+3)^4$ . Result: $5(2x+3)^4 \cdot 2 = 10(2x+3)^4$ .
Show step-by-step (with the why)
1. Identify the layers. Write $h(x) = f(g(x))$ with $f(u) = u^5$ (outer) and $g(x) = 2x+3$ (inner). Always name the layers explicitly before differentiating.
2. Differentiate the outer layer. $f'(u) = 5u^4$ . Evaluate at $g(x)$ : $f'(g(x)) = 5(2x+3)^4$ . Note: the argument remains $2x+3$ , not $x$ .
3. Differentiate the inner layer. $g'(x) = 2$ . The derivative of a linear function is its slope.
4. Multiply. $h'(x) = 5(2x+3)^4 \cdot 2 = 10(2x+3)^4$ .
Mnemonic: for any $(\text{linear})^n$ , the derivative is $n \cdot (\text{linear})^{n-1} \cdot (\text{slope})$ . Here the slope is 2, hence the factor of 2.
Ex. 53.2
Calculate $(\sin(4x))'$ .
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Inner: $g(x) = 4x$ , derivative $4$ . Outer: $\sin u$ , derivative $\cos u$ . Result: $\cos(4x) \cdot 4 = 4\cos(4x)$ .
Ex. 53.3
Calculate $(e^{x^2})'$ .
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Inner: $g(x) = x^2$ , derivative $2x$ . Outer: $e^u$ , derivative $e^u$ . Result: $e^{x^2} \cdot 2x = 2xe^{x^2}$ . Common mistake: writing only $xe^{x^2}$ — missing the factor of 2 from the derivative of $x^2$ .
Show step-by-step (with the why)
1. Identify the layers. $f(u) = e^u$ (outer), $g(x) = x^2$ (inner).
2. Differentiate the outer layer. $f'(u) = e^u$ . Evaluated at $g(x)$ : $e^{x^2}$ . The exponential function is its own derivative.
3. Differentiate the inner layer. $g'(x) = 2x$ . Power rule on $x^2$ .
4. Multiply. $(e^{x^2})' = e^{x^2} \cdot 2x = 2xe^{x^2}$ .
Mnemonic: the exponential function "copies itself" upon differentiation — the only new factor is the derivative of the exponent. Never forget this factor.
Ex. 53.4Answer key
Calculate $(\ln(x^2 + 1))'$ .
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Formula: $(\ln(g(x)))' = g'(x)/g(x)$ . Here $g(x) = x^2 + 1$ , $g'(x) = 2x$ . Result: $\dfrac{2x}{x^2+1}$ .
Ex. 53.5Answer key
Calculate $(\sqrt{x^2+1})'$ .
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Inner: $g(x) = x^2+1$ , derivative $2x$ . Outer: $\sqrt{u}$ , derivative $1/(2\sqrt{u})$ . Result: $\frac{1}{2\sqrt{x^2+1}} \cdot 2x = \frac{x}{\sqrt{x^2+1}}$ .
Ex. 53.6
Calculate $(\cos^2 x)'$ .
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Inner: $g(x) = \cos x$ , derivative $-\sin x$ . Outer: $u^2$ , derivative $2u$ evaluated at $\cos x$ : $2\cos x$ . Result: $2\cos x \cdot (-\sin x) = -2\sin x\cos x = -\sin 2x$ (double angle identity).
Show step-by-step (with the why)
1. Rewrite. $\cos^2 x = (\cos x)^2$ . Outer layer: $f(u) = u^2$ . Inner: $g(x) = \cos x$ .
2. Differentiate the outer. $f'(u) = 2u$ evaluated at $\cos x$ : $2\cos x$ .
3. Differentiate the inner. $g'(x) = -\sin x$ .
4. Multiply. $(\cos^2 x)' = 2\cos x \cdot (-\sin x) = -2\sin x\cos x$ .
5. Simplify. Using the identity $2\sin x\cos x = \sin 2x$ : result $-\sin 2x$ .
Curiosity: differentiating $\cos^2 x$ and differentiating $1 - \sin^2 x$ yield the same result — confirm this using the chain rule on the latter for practice.
Ex. 53.7
Calculate $(\tan(2x))'$ .
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Inner: $g(x) = 2x$ , derivative $2$ . Outer: $\tan u$ , derivative $\sec^2 u$ . Result: $\sec^2(2x) \cdot 2 = 2\sec^2(2x)$ .
Ex. 53.8
Calculate $(\sin(\cos x))'$ .
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Inner: $g(x) = \cos x$ , derivative $-\sin x$ . Outer: $\sin u$ , derivative $\cos u$ . Result: $\cos(\cos x) \cdot (-\sin x) = -\sin x \cos(\cos x)$ .
Ex. 53.9
Calculate $(e^{\sin x})'$ .
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Inner: $\sin x$ , derivative $\cos x$ . Outer: $e^u$ , derivative $e^u$ . Result: $e^{\sin x} \cdot \cos x$ .
Ex. 53.10
Calculate $((\ln x)^3)'$ .
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Inner: $g(x) = \ln x$ , derivative $1/x$ . Outer: $u^3$ , derivative $3u^2$ . Result: $3(\ln x)^2 \cdot \frac{1}{x} = \frac{3(\ln x)^2}{x}$ .
Show step-by-step (with the why)
1. Identify. $(\ln x)^3$ is a composition: $f(u) = u^3$ (outer), $g(x) = \ln x$ (inner). It is not the same as $\ln(x^3)$ !
2. Differentiate the outer. $f'(u) = 3u^2$ evaluated at $\ln x$ : $3(\ln x)^2$ .
3. Differentiate the inner. $g'(x) = 1/x$ .
4. Result. $3(\ln x)^2 \cdot \frac{1}{x} = \frac{3(\ln x)^2}{x}$ .
Mnemonic: $(\ln x)^3 \neq \ln(x^3) = 3\ln x$ . The position of the exponent changes everything. Be careful with notation.
Ex. 53.11
Calculate $(\sqrt{1-x^2})'$ .
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Inner: $g(x) = 1-x^2$ , derivative $-2x$ . Outer: $\sqrt{u}$ , derivative $1/(2\sqrt{u})$ . Result: $\frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1-x^2}}$ . (Domain: $-1 < x < 1$ .)
Ex. 53.12
Calculate $\left(\dfrac{1}{x^2+4}\right)'$ .
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Rewrite as $(x^2+4)^{-1}$ . Inner: $x^2+4$ , derivative $2x$ . Outer: $u^{-1}$ , derivative $-u^{-2}$ . Result: $-(x^2+4)^{-2} \cdot 2x = -\dfrac{2x}{(x^2+4)^2}$ .
Ex. 53.13Answer key
Calculate $(2^{x^2})'$ .
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Formula: $(a^{g(x)})' = a^{g(x)} \ln a \cdot g'(x)$ . Here $a=2$ , $g(x)=x^2$ , $g'(x)=2x$ . Result: $2^{x^2} \cdot \ln 2 \cdot 2x = 2x\ln 2 \cdot 2^{x^2}$ .
Ex. 53.14
Calculate $(\sin^3(2x))'$ .
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Three layers: outermost $u^3$ , intermediate $\sin v$ , innermost $2x$ . Chain rule: $3\sin^2(2x) \cdot \cos(2x) \cdot 2 = 6\sin^2(2x)\cos(2x)$ .
Show step-by-step (with the why)
1. Identify 3 layers. $\sin^3(2x) = (\sin(2x))^3$ . Outer: $f_3(u) = u^3$ . Middle: $f_2(v) = \sin v$ . Inner: $f_1(x) = 2x$ .
2. Differentiate each layer. $f_3'(u) = 3u^2$ ; $f_2'(v) = \cos v$ ; $f_1'(x) = 2$ .
3. Assemble the product from outside in. $3[\sin(2x)]^2 \cdot \cos(2x) \cdot 2$ .
4. Simplify. $6\sin^2(2x)\cos(2x)$ .
Mnemonic: in triple composition, always assemble the product from the outermost to the innermost layer. Never skip a layer.
Ex. 53.15
Calculate $(\arctan(x^2))'$ . (Ans: $2x/(1+x^4)$ .)
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Formula: $(\arctan(g(x)))' = g'(x)/(1+[g(x)]^2)$ . Here $g(x)=x^2$ , $g'(x)=2x$ . Result: $\dfrac{2x}{1+x^4}$ .
Ex. 53.16
Calculate $(\ln(\sin x))'$ .
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Formula: $(\ln(g(x)))' = g'(x)/g(x)$ . Here $g(x)=\sin x$ , $g'(x)=\cos x$ . Result: $\cos x / \sin x = \cot x$ .
Ex. 53.17Answer key
Calculate $(e^{\cos(2x)})'$ .
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Three layers: $e^u$ , $\cos v$ , $2x$ . Chain rule: $e^{\cos(2x)} \cdot (-\sin(2x)) \cdot 2 = -2\sin(2x)\,e^{\cos(2x)}$ .
Ex. 53.18
Calculate $(\sqrt{\tan x})'$ .
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Inner: $\tan x$ , derivative $\sec^2 x$ . Outer: $\sqrt{u}$ , derivative $1/(2\sqrt{u})$ . Result: $\dfrac{\sec^2 x}{2\sqrt{\tan x}}$ . (Domain: $\tan x > 0$ .)
Ex. 53.19Answer key
Calculate $(\sin(\sqrt{x}))'$ .
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Inner: $\sqrt{x}$ , derivative $1/(2\sqrt{x})$ . Outer: $\sin u$ , derivative $\cos u$ . Result: $\cos(\sqrt{x}) \cdot \dfrac{1}{2\sqrt{x}} = \dfrac{\cos(\sqrt{x})}{2\sqrt{x}}$ .
Ex. 53.20
Calculate $((3x+5)^{10})'$ .
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Inner: $3x+5$ , derivative $3$ . Outer: $u^{10}$ , derivative $10u^9$ . Result: $10(3x+5)^9 \cdot 3 = 30(3x+5)^9$ .
Ex. 53.21Answer key
Calculate $(\cos(3x^2+2))'$ .
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Inner: $3x^2+2$ , derivative $6x$ . Outer: $\cos u$ , derivative $-\sin u$ . Result: $-\sin(3x^2+2) \cdot 6x = -6x\sin(3x^2+2)$ .
Ex. 53.22
Calculate $(\ln(\ln x))'$ .
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Inner: $\ln x$ , derivative $1/x$ . Outer: $\ln u$ , derivative $1/u$ . Result: $\frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x}$ . (Domain: $x > 1$ .)
Show step-by-step (with the why)
1. Identify. $\ln(\ln x)$ is a composition of the logarithm with itself: $f(u) = \ln u$ (outer), $g(x) = \ln x$ (inner).
2. Differentiate the outer. $f'(u) = 1/u$ evaluated at $\ln x$ : $1/\ln x$ .
3. Differentiate the inner. $g'(x) = 1/x$ .
4. Result. $(1/\ln x) \cdot (1/x) = 1/(x\ln x)$ .
Note: the natural domain requires $x > 1$ so that the inner logarithm $\ln x > 0$ (the argument of the outer logarithm must be positive).
Ex. 53.23
Calculate $(x \cdot \sin(x^2))'$ .
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Product: $u = x$ , $v = \sin(x^2)$ . Product rule: $1 \cdot \sin(x^2) + x \cdot \cos(x^2) \cdot 2x = \sin(x^2) + 2x^2\cos(x^2)$ .
Show step-by-step (with the why)
1. Recognize the product. Two factors: $u = x$ and $v = \sin(x^2)$ . Use the product rule, not the chain rule directly.
2. Differentiate each factor. $u' = 1$ . For $v' = (\sin(x^2))'$ : chain rule with inner $x^2$ , derivative $2x$ . Thus $v' = \cos(x^2) \cdot 2x$ .
3. Product rule. $(uv)' = u'v + uv' = 1 \cdot \sin(x^2) + x \cdot 2x\cos(x^2)$ .
4. Result. $\sin(x^2) + 2x^2\cos(x^2)$ .
Mnemonic: in this type of exercise, first identify IF it is a product or a composition. $x \cdot \sin(x^2)$ is a product. $\sin(x \cdot x^2) = \sin(x^3)$ would be pure composition.
Ex. 53.24
Calculate $(x \cdot e^{x^2})'$ .
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Product: $u=x$ , $v=e^{x^2}$ . Derivative: $1 \cdot e^{x^2} + x \cdot 2xe^{x^2} = e^{x^2}(1+2x^2)$ .
Ex. 53.25
Calculate $(\sin(\sqrt{x^2+1}))'$ .
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Three layers: $\sin u$ , $\sqrt{v}$ , $x^2+1$ . Chain rule: $\cos(\sqrt{x^2+1}) \cdot \frac{1}{2\sqrt{x^2+1}} \cdot 2x = \frac{x\cos(\sqrt{x^2+1})}{\sqrt{x^2+1}}$ .
Ex. 53.26Answer key
Calculate $(\tan^2(3x))'$ .
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Three layers: $u^2$ , $\tan v$ , $3x$ . Chain rule: $2\tan(3x) \cdot \sec^2(3x) \cdot 3 = 6\tan(3x)\sec^2(3x)$ .
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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At $x=1$ : $y(1) = (1^2+1)^3 = 8$ . Derivative: $y' = 3(x^2+1)^2 \cdot 2x$ . At $x=1$ : $y'(1) = 3 \cdot 4 \cdot 2 = 24$ . Tangent line: $y - 8 = 24(x - 1)$ , or $y = 24x - 16$ .
Ex. 53.28
Error Analysis. A student writes $(e^{x^2})' = xe^{x^2}$ . What specific error was made?
Select the correct option
$A)\$ The chain rule does not apply to trigonometric compositions $B)\$ The inner derivative is $\cos x$ , yielding the correct result $C)\$ Used $x$ as the derivative of $x^2$ , missing the factor of $2$ $D)\$ Confused composition with product — category error
Select an option first
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In $(e^{x^2})'$ , the inner function is $x^2$, whose derivative is $2x$, not $x$. The student multiplied by $x$ instead of $2x$. The correct result is $2xe^{x^2}$ . This is the most common error with the chain rule for exponentials.
Ex. 53.29
Conceptual. To differentiate $\sin(x^2)$ , which rule applies? Why isn't it the product rule?
Select the correct option
$A)\$ Product rule, because $\sin$ and $x^2$ are two multiplied factors $B)\$ Chain rule, because $x^2$ is inside $\sin$ as its argument $C)\$ No special rule — direct derivative $D)\$ Quotient rule
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See the referenced source for the detailed solution.
Ex. 53.30
Calculate $(e^{e^x})'$ .
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Inner: $e^x$ , derivative $e^x$ . Outer: $e^u$ , derivative $e^u$ . Result: $e^{e^x} \cdot e^x$ . The function $e^{e^x}$ grows immensely fast: at $x=0$ , $(e^{e^x})'|_{x=0} = e^1 \cdot 1 = e \approx 2.72$ .
Show step-by-step (with the why)
1. Identify layers. Outermost: $f(u) = e^u$ , derivative $e^u$ . Inner: $g(x) = e^x$ , derivative $e^x$ .
2. Evaluate derivatives at composed arguments. Outer evaluated: $e^{e^x}$ . Inner: $e^x$ .
3. Multiply. $e^{e^x} \cdot e^x$ .
Observation: at $x=0$ the derivative is $e$, consistent with rapid growth.
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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Velocity: $s'(t) = \omega\cos(\omega t)$ (chain rule). Acceleration: $s''(t) = \omega \cdot (-\sin(\omega t)) \cdot \omega = -\omega^2\sin(\omega t) = -\omega^2 s(t)$ . Acceleration is proportional to the negative of displacement — simple harmonic oscillator law.
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See the referenced source for the step-by-step walkthrough.
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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$N(t) = N_0 e^{-\lambda t}$ . Chain rule: inner $-\lambda t$ , derivative $-\lambda$ . Outer $e^u$ . Result: $N'(t) = N_0 e^{-\lambda t} \cdot (-\lambda) = -\lambda N(t)$ . The decay rate is proportional to the amount present — law of radioactive decay.
Ex. 53.33Answer key
Biology. Logistic growth is $P(t) = \dfrac{K}{1 + e^{-rt}}$ . Calculate $P'(0)$ .
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At $t=0$ : $P(0) = K/(1+1) = K/2$ . Rewrite $P = K(1+e^{-rt})^{-1}$ . Chain rule: $P'(t) = K \cdot (-1)(1+e^{-rt})^{-2} \cdot (-re^{-rt}) = \dfrac{rKe^{-rt}}{(1+e^{-rt})^2}$ . At $t=0$ : $P'(0) = \dfrac{rK \cdot 1}{4} = \dfrac{rK}{4}$ .
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1. P(0). $P(0) = K/(1+e^0) = K/2$ . The population starts at half the carrying capacity.
2. Rewrite to apply chain rule. $P(t) = K(1+e^{-rt})^{-1}$ . Outer: $u^{-1}$ . Inner: $1 + e^{-rt}$ .
3. Differentiate. Outer: $-u^{-2}$ . Inner: the derivative of $e^{-rt}$ is $-re^{-rt}$ (inner chain rule). Thus $P'(t) = K \cdot (-(1+e^{-rt})^{-2}) \cdot (-re^{-rt}) = rKe^{-rt}/(1+e^{-rt})^2$ .
4. At t=0. $P'(0) = rK/(1+1)^2 = rK/4$ . The maximum growth rate occurs at $P = K/2$ .
Curiosity: the inflection point of logistic growth occurs exactly when the population reaches half the carrying capacity — this is when growth is fastest.
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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$f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$ . Chain rule on $e^{-x^2/2}$ : inner $-x^2/2$ , derivative $-x$ . Result: $f'(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \cdot (-x) = -x \cdot f(x)$ . The derivative of the Gaussian is itself multiplied by $-x$ .
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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Product: $u = e^{-rt}$ (derivative $-re^{-rt}$ via chain rule), $v = S$ (constant, derivative zero). Product rule: $V'(t) = -re^{-rt} \cdot S + e^{-rt} \cdot 0 = -rSe^{-rt} = -rV(t)$ . The present value decays at a rate $r$ multiplied by its own value.
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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$E(t) = \frac{1}{2}m v(t)^2 = \frac{1}{2}m(at)^2 = \frac{1}{2}ma^2 t^2$ . Direct derivative: $E'(t) = ma^2 t$ . Via chain rule: $\frac{dE}{dt} = \frac{dE}{dv} \cdot \frac{dv}{dt} = mv \cdot a = m(at) \cdot a = ma^2 t$ . Both methods agree.
Ex. 53.37
Calculate $(\sin(\cos(\sin x)))'$ .
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Three layers: $f_3 = \sin u$ , $f_2 = \cos v$ , $f_1 = \sin x$ . Chain rule from outside in: $\cos(\cos(\sin x)) \cdot (-\sin(\sin x)) \cdot \cos x = -\cos x \cdot \sin(\sin x) \cdot \cos(\cos(\sin x))$ .
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1. Identify the 3 layers. Outermost: $f_3(u) = \sin u$ , derivative $\cos u$ . Intermediate: $f_2(v) = \cos v$ , derivative $-\sin v$ . Innermost: $f_1(x) = \sin x$ , derivative $\cos x$ .
2. Evaluate each derivative at its composed argument. Outer: $\cos(\cos(\sin x))$ . Intermediate: $-\sin(\sin x)$ . Inner: $\cos x$ .
3. Product. $\cos(\cos(\sin x)) \cdot (-\sin(\sin x)) \cdot \cos x$ .
Mnemonic: in triple composition, set up a table of layers and derivatives before multiplying — this prevents errors in order or missing factors.
Ex. 53.38
Calculate $\dfrac{d}{dx}\left[(x + \sqrt{x^2+1})^n\right]$ .
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$f(x) = (x+\sqrt{x^2+1})^n$ . Outer: $u^n$ , derivative $nu^{n-1}$ . Inner: $x + \sqrt{x^2+1}$ , derivative $1 + x/\sqrt{x^2+1}$ (via chain rule on the square root). Result: $n(x+\sqrt{x^2+1})^{n-1} \cdot \left(1 + \dfrac{x}{\sqrt{x^2+1}}\right)$ .
Ex. 53.39
Calculate $(\sin(e^{x^2}))'$ .
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Three layers: $\sin u$ , $e^v$ , $x^2$ . Chain rule: $\cos(e^{x^2}) \cdot e^{x^2} \cdot 2x$ .
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1. Layers. Outermost: $\sin u$ , derivative $\cos u$ . Intermediate: $e^v$ , derivative $e^v$ . Innermost: $x^2$ , derivative $2x$ .
2. Evaluate each derivative at its composed argument. Outer: $\cos(e^{x^2})$ . Intermediate: $e^{x^2}$ . Inner: $2x$ .
3. Product. $\cos(e^{x^2}) \cdot e^{x^2} \cdot 2x$ .
Note: at $x=0$ the derivative is zero (factor $2x = 0$ ), consistent with a possible extremum. Check numerically: $h(0.001) - h(-0.001) \approx 0$ .
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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The naive argument writes $\frac{f(g(a+h))-f(g(a))}{h} = \frac{\Delta f}{\Delta g} \cdot \frac{\Delta g}{h}$ . The problem: $\Delta g = g(a+h) - g(a)$ can be zero for $h \neq 0$ (e.g., oscillating $g$ ), leading to division by zero. The solution is to define the auxiliary function $Q(y)$ which equals the difference quotient when $y \neq g(a)$ and $f'(g(a))$ when $y = g(a)$ . The continuity of $Q$ at $g(a)$ (guaranteed by differentiability of $f$ ) allows completing the argument without division by zero.

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.