Calculus I · 2A · lesson

The Basic Chain Rule

Learn the basic chain rule with clear exposition, guided derivations, worked examples, visuals, checks, and interpretation.

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Section overview

The chain rule and compositions

What this section is building

Learn the basic chain rule with clear exposition, guided derivations, worked examples, visuals, checks, and interpretation.

Notice

A small input change passes through a sequence of machines; the total response multiplies the response at each stage.

Decide

List the layers, differentiate one layer at a time, and stop only when every input-dependent layer contributes.

Avoid

Differentiating the outside and leaving the inside unchanged without its derivative factor.

Concept

Learning objectives

Apply the chain rule to powers of linear and polynomial inner functions.

Differentiate the Outside, Then the Inside

Explanation

Before the formulas

In The Basic Chain Rule, the phrase "outside to inside" is useful only when paired with structure. Keep the inner expression unchanged while differentiating the outer layer, then multiply by the derivative of that inner expression. If the inner expression is itself composite, continue inward.

A missing inner derivative is the signature chain-rule error. Check units or scaling to catch it. If an inner quantity changes three times as fast, the final output rate should reflect that factor of three. The chain rule is a rate-conversion law, not punctuation attached to parentheses.

Explanation

Differentiate the outside, keep the inside, then multiply by the inside rate

The phrase "outside times inside" is incomplete. The precise move is: differentiate the outer function while leaving its input placeholder unchanged, then multiply by the derivative of the inner function. For $(g(x))^n$ , this gives $n(g(x))^{n-1}g'(x)$ .

Check the result by asking whether every changing layer contributed a factor. If the inner expression is not simply $x$ , an inner derivative should usually appear somewhere in the answer.

The phrase "differentiate the outside, keep the inside, multiply by the derivative of the inside" is a useful first algorithm, but it should not become empty choreography. The inner derivative is present because a one-unit change in $x$ need not produce a one-unit change in the inner expression.

Before differentiating, name the inner function mentally. If $y=(3x^2+1)^5$ , then the outer power reacts to changes in $u=3x^2+1$ , while $u$ reacts to changes in $x$ . The chain rule connects the two reactions.

Theorem

Chain rule

If $y=F(u)$ and $u=g(x)$ , then

\boxed{\frac{dy}{dx}=\frac{dF}{du}\frac{du}{dx}}.

Equivalently,

\boxed{(F\circ g)'(x)=F'(g(x))g'(x)}.

Guided walkthrough

A power of a linear function

Differentiate

y=(3x-2)^5.

Answer reveal

Worked solution

Your setup or answer

Write a real attempt before opening the supplied answer.

Worked example

A polynomial inside a reciprocal power

y=(x^2+4)^{-3}.

Then

\begin{aligned} y'&=-3(x^2+4)^{-4}(2x)\\ &=\boxed{-\frac{6x}{(x^2+4)^4}}. \end{aligned}

Proof idea

Why rates multiply

For small changes,

\frac{\Delta y}{\Delta x} =\frac{\Delta y}{\Delta u}\frac{\Delta u}{\Delta x}.

As the changes shrink, the two ratios approach $dy/du$ and $du/dx$ . The chain rule is the exact limit version of multiplying conversion rates through an intermediate variable.

Interactive checkchain-basic-01

Differentiate $(4x+1)^3$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Differentiate the outer cube and multiply by the derivative of 4x+1.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Modeling lab

A calibrated temperature sensor

A sensor produces voltage

V(T)=0.02T^2+0.5T

when temperature is $T$ degrees Celsius. A display converts voltage to a reading

R(V)=100\sqrt{V}.

The display sensitivity to temperature is

\frac{dR}{dT}=\frac{dR}{dV}\frac{dV}{dT} =\frac{50}{\sqrt V}(0.04T+0.5).

At a chosen temperature, first compute $V(T)$ , then evaluate the product. The chain rule separates sensor physics from display calibration.

Optional advanced note

Analysis preview: composition of linear maps

Near $x=a$ , suppose $g$ behaves like

g(a+h)\approx g(a)+g'(a)h.

Near $g(a)$ , suppose $f$ behaves like

f(g(a)+k)\approx f(g(a))+f'(g(a))k.

Substituting the first local model into the second makes the total linear coefficient

f'(g(a))g'(a).

In multivariable calculus, the same idea becomes matrix multiplication of derivative maps.

Interactive checkchain-extra-01

Differentiate $(x^2+1)^5$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Outer power derivative times inner derivative.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Source & rights

Original instruction with traceable references.

BetterGrades-original composition declared by source handoff; owner provenance review required before public release

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

Differentiate the Outside, Then the Inside

Explanation

Before the formulas

Explanation

Differentiate the outside, keep the inside, then multiply by the inside rate

Check the result by asking whether every changing layer contributed a factor. If the inner expression is not simply $x$ , an inner derivative should usually appear somewhere in the answer.

x

need not produce a one-unit change in the inner expression.

Before differentiating, name the inner function mentally. If

y=(3x^2+1)^5

, then the outer power reacts to changes in

u=3x^2+1

, while

u

reacts to changes in

x

. The chain rule connects the two reactions.

Theorem

Chain rule

If $y=F(u)$ and $u=g(x)$ , then

\boxed{\frac{dy}{dx}=\frac{dF}{du}\frac{du}{dx}}.

Equivalently,

\boxed{(F\circ g)'(x)=F'(g(x))g'(x)}.

Guided walkthrough

A power of a linear function

Differentiate

y=(3x-2)^5.

Answer reveal

Worked solution

Your setup or answer

Write a real attempt before opening the supplied answer.

Worked example

A polynomial inside a reciprocal power

y=(x^2+4)^{-3}.

Then

\begin{aligned} y'&=-3(x^2+4)^{-4}(2x)\\ &=\boxed{-\frac{6x}{(x^2+4)^4}}. \end{aligned}

Proof idea

Why rates multiply

For small changes,

\frac{\Delta y}{\Delta x} =\frac{\Delta y}{\Delta u}\frac{\Delta u}{\Delta x}.

As the changes shrink, the two ratios approach $dy/du$ and $du/dx$ . The chain rule is the exact limit version of multiplying conversion rates through an intermediate variable.

Interactive checkchain-basic-01

Differentiate $(4x+1)^3$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Differentiate the outer cube and multiply by the derivative of 4x+1.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Modeling lab

A calibrated temperature sensor

A sensor produces voltage

V(T)=0.02T^2+0.5T

when temperature is $T$ degrees Celsius. A display converts voltage to a reading

R(V)=100\sqrt{V}.

The display sensitivity to temperature is

\frac{dR}{dT}=\frac{dR}{dV}\frac{dV}{dT} =\frac{50}{\sqrt V}(0.04T+0.5).

At a chosen temperature, first compute $V(T)$ , then evaluate the product. The chain rule separates sensor physics from display calibration.

Optional advanced note

Analysis preview: composition of linear maps

Near $x=a$ , suppose $g$ behaves like

g(a+h)\approx g(a)+g'(a)h.

Near $g(a)$ , suppose $f$ behaves like

f(g(a)+k)\approx f(g(a))+f'(g(a))k.

Substituting the first local model into the second makes the total linear coefficient

f'(g(a))g'(a).

In multivariable calculus, the same idea becomes matrix multiplication of derivative maps.

Interactive checkchain-extra-01

Differentiate $(x^2+1)^5$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Outer power derivative times inner derivative.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Source & rights

Original instruction with traceable references.

BetterGrades-original composition declared by source handoff; owner provenance review required before public release

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

Differentiate the Outside, Then the Inside

Related resources

Original instruction with traceable references.

Differentiate the Outside, Then the Inside

Related resources

Original instruction with traceable references.