Chain rule — proof, composite function derivatives: Exercises

The one-line tool we lean on the whole way down:

Level 1 — Recognition

Can you identify the outer function $f$ and inner function $g$ , and apply the rule once?

Problem 1.1

Differentiate $h (x) = (2 x + 7)^{4}$ .

Recall Solution 1.1

Identify the two machines. Outer $f (u) = u^{4}$ , inner $g (x) = 2 x + 7$ . Differentiate each. $f^{'} (u) = 4 u^{3}$ (power rule on the shell); $g^{'} (x) = 2$ (derivative of a line). Multiply, plug the inside back in. $h^{'} (x) = 4 (2 x + 7)^{3} \cdot 2 = 8 (2 x + 7)^{3} .$

Problem 1.2

Differentiate $h (x) = cos (5 x)$ .

Recall Solution 1.2

Outer $f (u) = cos u \Rightarrow f^{'} (u) = - sin u$ . Inner $g (x) = 5 x \Rightarrow g^{'} (x) = 5$ . $h^{'} (x) = - sin (5 x) \cdot 5 = - 5 sin (5 x) .$ Notice the inside stays $5 x$ inside the $sin$ — only the outer rate changes.

Problem 1.3

Differentiate $h (x) = e^{3 x}$ .

Recall Solution 1.3

Outer $f (u) = e^{u} \Rightarrow f^{'} (u) = e^{u}$ (the exponential is its own derivative). Inner $g (x) = 3 x \Rightarrow g^{'} (x) = 3$ . $h^{'} (x) = e^{3 x} \cdot 3 = 3 e^{3 x} .$

Level 2 — Application

Combine the chain rule with power, product-of-constants, and known derivatives.

Problem 2.1

Differentiate $h (x) = 9 - x^{2}$ .

Recall Solution 2.1

Rewrite as $(9 - x^{2})^{1/2}$ . Outer $f (u) = u^{1/2} \Rightarrow f^{'} (u) = \frac{1}{2} u^{- 1/2}$ . Inner $g (x) = 9 - x^{2} \Rightarrow g^{'} (x) = - 2 x$ . $h^{'} (x) = \frac{1}{2} (9 - x^{2})^{- 1/2} \cdot (- 2 x) = \frac{- x}{9 - x ^{2}} .$ The $\frac{1}{2}$ and the $- 2$ combine to $- 1$ — clean cancellation.

Problem 2.2

Differentiate $h (x) = tan (x^{3})$ .

Recall Solution 2.2

Outer $f (u) = tan u \Rightarrow f^{'} (u) = sec^{2} u$ . Inner $g (x) = x^{3} \Rightarrow g^{'} (x) = 3 x^{2}$ . $h^{'} (x) = sec^{2} (x^{3}) \cdot 3 x^{2} = 3 x^{2} sec^{2} (x^{3}) .$

Problem 2.3

Differentiate $h (x) = ln (x^{2} + 1)$ .

Recall Solution 2.3

Outer $f (u) = ln u \Rightarrow f^{'} (u) = \frac{1}{u}$ . Inner $g (x) = x^{2} + 1 \Rightarrow g^{'} (x) = 2 x$ . $h^{'} (x) = \frac{1}{x ^{2} + 1} \cdot 2 x = \frac{2 x}{x ^{2} + 1} .$ Note $x^{2} + 1 > 0$ always, so $ln$ is defined for all $x$ and no domain worry arises.

Level 3 — Analysis

Multiple layers, and combining the chain rule with the Product rule / Quotient rule.

Problem 3.1

Differentiate $h (x) = sin^{3} (2 x)$ (meaning $[sin (2 x)]^{3}$ ).

Recall Solution 3.1

Three layers: outermost cube, then $sin$ , then $2 x$ . Peel outside-in.

Cube: $\frac{d}{d ( \cdot )} (\cdot)^{3} = 3 (\cdot)^{2}$ evaluated at $sin (2 x)$ ⇒ $3 sin^{2} (2 x)$ .
$sin$ : derivative $cos (2 x)$ .
$2 x$ : derivative $2$ . Multiply all three: $h^{'} (x) = 3 sin^{2} (2 x) \cdot cos (2 x) \cdot 2 = 6 sin^{2} (2 x) cos (2 x) .$

Problem 3.2

Differentiate $h (x) = x^{2} e^{- x^{2}}$ .

Recall Solution 3.2

This is a product $x^{2} \cdot e^{- x^{2}}$ , and the second factor needs the chain rule. Product rule: $(uv)^{'} = u^{'} v + u v^{'}$ with $u = x^{2}$ , $v = e^{- x^{2}}$ .

$u^{'} = 2 x$ .
$v^{'} = e^{- x^{2}} \cdot (- 2 x)$ (chain rule: outer $e^{(\cdot)}$ , inner $- x^{2} \Rightarrow - 2 x$ ). Assemble: $h^{'} (x) = 2 x e^{- x^{2}} + x^{2} (- 2 x e^{- x^{2}}) = 2 x e^{- x^{2}} - 2 x^{3} e^{- x^{2}} = 2 x (1 - x^{2}) e^{- x^{2}} .$

Problem 3.3

Differentiate $h (x) = \frac{cos x}{1 + x ^{2}}$ .

Recall Solution 3.3

Quotient rule: $(\frac{p}{q})^{'} = \frac{p ^{'} q - p q ^{'}}{q ^{2}}$ with $p = cos x$ , $q = (1 + x^{2})^{1/2}$ .

$p^{'} = - sin x$ .
$q^{'} = \frac{1}{2} (1 + x^{2})^{- 1/2} \cdot 2 x = \frac{x}{1 + x ^{2}}$ (chain rule).
$q^{2} = 1 + x^{2}$ . $h^{'} (x) = \frac{- s i n x 1 + x ^{2} - c o s x \cdot \frac{x}{1 + x ^{2}}}{1 + x ^{2}} .$ Multiply top and bottom by $1 + x^{2}$ to clear the inner fraction: $h^{'} (x) = \frac{- s i n x ( 1 + x ^{2} ) - x c o s x}{( 1 + x ^{2} ) ^{3/2}} .$

Level 4 — Synthesis

Build the composite yourself: implicit differentiation and inverse-function derivatives are the chain rule in disguise.

Problem 4.1

The curve $x^{2} + y^{2} = 25$ passes through $(3, 4)$ . Using Implicit differentiation, find $\frac{d y}{d x}$ there.

Recall Solution 4.1

Treat $y$ as a function of $x$ : then $y^{2}$ is a composite $[y (x)]^{2}$ , so its derivative needs the chain rule: $\frac{d}{d x} (y^{2}) = 2 y \cdot \frac{d y}{d x} .$ Differentiate both sides of $x^{2} + y^{2} = 25$ : $2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{x}{y} .$ At $(3, 4)$ : $\frac{d y}{d x} = - \frac{3}{4}$ . Geometry check: the radius to $(3, 4)$ has slope $\frac{4}{3}$ ; the tangent is perpendicular, slope $- \frac{3}{4}$ . ✓ (see figure).

Figure — Chain rule — proof, composite function derivatives

Problem 4.2

Let $f (x) = x^{3} + x$ (strictly increasing, so invertible). Find $(f^{- 1})^{'} (2)$ using the Inverse function derivative.

Recall Solution 4.2

The identity $f (f^{- 1} (x)) = x$ is a composite. Chain-rule both sides: $f^{'} (f^{- 1} (x)) \cdot (f^{- 1})^{'} (x) = 1 \Rightarrow (f^{- 1})^{'} (x) = \frac{1}{f ^{'} ( f ^{- 1} ( x ))} .$ Find $f^{- 1} (2)$ : solve $x^{3} + x = 2$ . Try $x = 1$ : $1 + 1 = 2$ . ✓ So $f^{- 1} (2) = 1$ . $f^{'} (x) = 3 x^{2} + 1$ , so $f^{'} (1) = 4$ . $(f^{- 1})^{'} (2) = \frac{1}{f ^{'} ( 1 )} = \frac{1}{4} .$

Problem 4.3

A balloon is a sphere with volume $V = \frac{4}{3} π r^{3}$ . Air is pumped so that $\frac{d r}{d t} = 0.2 cm/s$ . Using Related rates, find $\frac{d V}{d t}$ when $r = 5 cm$ .

Recall Solution 4.3

$V$ depends on $r$ , and $r$ depends on $t$ — a composite $V (r (t))$ . Chain rule in Leibniz form: $\frac{d V}{d t} = \frac{d V}{d r} \cdot \frac{d r}{d t} .$ $\frac{d V}{d r} = 4 π r^{2}$ . At $r = 5$ : $4 π (25) = 100 π$ . $\frac{d V}{d t} = 100 π \cdot 0.2 = 20 π \approx 62.83 cm^{3} / s .$

Level 5 — Mastery

Subtle cases: repeated nesting, the proof's own machinery, and where naïve cancellation would break.

Problem 5.1

Differentiate $h (x) = e^{s i n (c o s x)}$ (a genuine triple nest).

Recall Solution 5.1

Layers, outside-in: $e^{(\cdot)}$ , then $sin (\cdot)$ , then $cos x$ .

$e^{(\cdot)}$ derivative at $sin (cos x)$ : $e^{s i n (c o s x)}$ .
$sin (\cdot)$ derivative at $cos x$ : $cos (cos x)$ .
$cos x$ derivative: $- sin x$ . Multiply all three: $h^{'} (x) = e^{s i n (c o s x)} \cdot cos (cos x) \cdot (- sin x) = - sin x cos (cos x) e^{s i n (c o s x)} .$

Problem 5.2

Let $g (x) = x^{2} sin (1/ x)$ for $x \neq = 0$ and $g (0) = 0$ . It is known that $g^{'} (0) = 0$ . This $g$ hits $g (x) = 0$ infinitely often as $x \to 0$ , so $Δ u = g (x + Δ x) - g (0)$ is zero for infinitely many nonzero $Δ x$ . Explain in one paragraph why the chain rule for $h (x) = sin (g (x))$ still gives $h^{'} (0) = cos (g (0)) \cdot g^{'} (0) = 0$ , even though the naïve "divide by $Δ u$ " derivation is illegal here.

Recall Solution 5.2

The error-function proof never divides by $Δ u$ . Recall from the parent note: with $u = g (0) = 0$ and $f = sin$ , write $f (u + k) - f (u) = (f^{'} (u) + E (k)) k, E (k) \to 0 as k \to 0, E (0) = 0.$ Set $k = Δ u$ . This equation is a multiplication, valid even when $Δ u = 0$ (both sides are then $0 = 0$ ). Dividing by $Δ x \neq = 0$ : $\frac{h ( Δ x ) - h ( 0 )}{Δ x} = (f^{'} (0) + E (Δ u)) \frac{Δ u}{Δ x} .$ As $Δ x \to 0$ : $g$ continuous ⇒ $Δ u \to 0$ ⇒ $E (Δ u) \to 0$ , and $\frac{Δ u}{Δ x} \to g^{'} (0) = 0$ . So the whole product $\to (cos 0 + 0) \cdot 0 = 0$ . The infinitely-many zero $Δ u$ 's do no harm because they never sit in a denominator. This is exactly the pathology the [!mistake] "steel-man" in the parent note warned about.

Problem 5.3

In the rigorous proof the term $E (Δ u) \to 0$ relies on $Δ u \to 0$ . Which single property of $g$ guarantees $Δ u \to 0$ as $Δ x \to 0$ , and what theorem supplies it?

Recall Solution 5.3

Continuity of $g$ at $x$ : $lim_{Δ x \to 0} (g (x + Δ x) - g (x)) = 0$ , i.e. $Δ u \to 0$ . It is guaranteed because $g$ is differentiable, and differentiable ⇒ continuous — see Continuity implies differentiability fails converse. (The converse fails: continuity alone would not give us $g^{'} (x)$ , which we also need for the $\frac{Δ u}{Δ x}$ factor.)

Connections

Chain rule — proof, composite function derivatives — the parent; every problem is one application of its rule.
Derivative — limit definition — the source of the difference quotient behind the proof (5.2, 5.3).
Product rule · Quotient rule — combined with the chain rule in L3.
Implicit differentiation — the chain rule on $y (x)$ (4.1).
Inverse function derivative — the chain rule on $f (f^{- 1} (x)) = x$ (4.2).
Related rates — the chain rule in $\frac{d V}{d t} = \frac{d V}{d r} \frac{d r}{d t}$ (4.3).
Continuity implies differentiability fails converse — supplies $Δ u \to 0$ (5.3).