MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: the chain rule (MATH 203)

This sheet is not a summary of section 3.6 of Thomas' Calculus: you already have the course notes. It answers one question, what makes students lose marks on the chain rule in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The rule fits on one line; the marks are lost in the algebra around it. Rewriting before, factoring at the lowest power after, and reading every derivative at the right point are the three habits this sheet trains, with a number attached to every claim so that you can check it yourself.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next and tells you which chapter to tackle next. Create your space, an email is enough.

The thread of the chapter

Name the inside function $u$ , differentiate the outside function at $u$ left alone and multiply by $u'$ , one factor per layer; then finish the algebra. In MATH 203 the marks go in the rewriting before (a root or a reciprocal as a power, the exponent $n - 1$ ) and in the factoring after (each bracket at its LOWEST power).

This chapter is part of MATH 203, Calculus I (Concordia)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Chain rule
Outside-Inside Rule
Power Chain Rule
Composite function
Negative and fractional exponents
Laws of exponents
Exponential function with base a
Factoring the lowest power
Second derivative of a composite
Derivatives from a table
Horizontal and vertical tangent lines
Rates of change with units

The essentials

One factor per layer: name u, differentiate the outside at u, multiply by u'

• Thomas 3.6: $(f \circ g)'(x) = f'(g(x))\,g'(x)$ , or $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$ with $u = g(x)$ . Hypotheses: $g$ differentiable at $x$ AND $f$ differentiable at $g(x)$ .
• The Outside-Inside Rule: differentiate the outside function, evaluate it at the inside LEFT ALONE, multiply by the derivative of the inside.
• Power Chain Rule: $\frac{d}{dx}u^n = nu^{n-1}u'$ for every real $n$ , including $n = -3$ , $n = \frac{2}{3}$ , $n = 0.16$ .
• Several layers, one factor each: $\frac{d}{dx}(3 + \sin 3x)^{-2} = -2(3 + \sin 3x)^{-3}\cdot\cos 3x\cdot 3$ . Count the factors against the layers.
• $y' = f'(g(x))\,g'(x)$ is a PRODUCT: it is zero when the outside factor is zero OR when $u' = 0$ . Both kinds of zeros give horizontal tangents.

y = (x^2 - 4x + 3)^2

has

y' = 2u\,u'

: the orange tangents at

x = 1

and

x = 3

come from

u = 0

, the red one at

x = 2

from

u' = 0

The first line of the write-up names the inside function: “let $u = 2x^2 + 1$ , so $y = 6u^{-3}$ ”. That sentence earns the method mark and forces the question “and $u'$ ?”.

Rewrite before, factor after: the algebra where MATH 203 marks are lost

• BEFORE: a constant over a power is a negative power, $\frac{2}{(x^2 + 3)^4} = 2(x^2 + 3)^{-4}$ ; a root is a fractional power, $\sqrt[3]{u^2} = u^{2/3}$ ; $\frac{1}{a^x} = a^{-x}$ and $\sqrt{a^x} = a^{x/2}$ .
• DURING: the new exponent is $n - 1$ , computed carefully: $-4 - 1 = -5$ , $\frac{2}{3} - 1 = -\frac{1}{3}$ , $-\frac{1}{2} - 1 = -\frac{3}{2}$ . It always moves AWAY from the old one by $1$ , downward.
• AFTER: factor each bracket at its LOWEST power, the most negative exponent: $-\frac{1}{2}$ before $\frac{1}{2}$ , $-3$ before $-2$ . A term in $u^n$ keeps $u^{n - m}$ after $u^m$ is taken out.
• The bracket left over must contain only non-negative powers. A fraction inside the bracket means the wrong power was taken out.
• A negative exponent in the final answer goes to the denominator: $(x^2 - 1)^{-2/3}$ becomes $\frac{1}{(x^2 - 1)^{2/3}}$ , and the zeros of that denominator are where $y'$ does not exist.

Evaluate the unfactored sum and your factored form at one convenient $x$ : equal values in ten seconds prove the factorisation.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Which power to factor out of a derivative

Read a line as: when the derivative contains these two powers of the same bracket $u$ , take out the one of the middle column; the last column is the finished form. The red line is a factorisation copied from real papers.

Powers present	Take out	Factored derivative
$u^{1/2}$ and $u^{-1/2}$	$u^{-1/2}$	$\frac{\text{polynomial}}{\sqrt u}$
Example: $\left(x^2\sqrt{2x + 1}\right)' = 2x(2x + 1)^{1/2} + x^2(2x + 1)^{-1/2} = \frac{x(5x + 2)}{\sqrt{2x + 1}}$ , which is $\frac{88}{3}$ at $x = 4$ .
$u^{-2}$ and $u^{-3}$	$u^{-3}$	$\frac{\text{polynomial}}{u^3}$
Example: $\left(\frac{(x + 1)^3}{(x - 2)^2}\right)' = 3(x + 1)^2(x - 2)^{-2} - 2(x + 1)^3(x - 2)^{-3} = \frac{(x + 1)^2(x - 8)}{(x - 2)^3}$ , which is $-80$ at $x = 3$ .
$u^{2/3}$ and $u^{-1/3}$	$u^{-1/3}$	$\frac{\text{polynomial}}{u^{1/3}}$
Example: $\left(x(x - 1)^{2/3}\right)' = (x - 1)^{2/3} + \frac{2x}{3}(x - 1)^{-1/3} = \frac{5x - 3}{3(x - 1)^{1/3}}$ , which is $7$ at $x = 9$ .
$u^{1/2}$ and $u^{-1/2}$	$u^{1/2}$	$x\sqrt u\,(x + 2)$ wrong power taken out
Example: For $2x(2x + 1)^{1/2} + x^2(2x + 1)^{-1/2}$ , taking out $(2x + 1)^{1/2}$ and writing $x(2x + 1)^{1/2}(x + 2)$ gives $72$ at $x = 4$ , instead of $\frac{88}{3}$ . What to do: Take out the LOWER exponent, $-\frac{1}{2}$ : taking out $u^{1/2}$ leaves $x^2u^{-1}$ , a fraction, inside the bracket.

The rule is the same for $x$ itself: $x^2$ and $x^3$ give $x^2$ , and $x^{-1}$ and $x$ give $x^{-1}$ .

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Forgetting the inside derivative of a root

half the marks of the question, and every later part built on it

What not to write

“ $\frac{d}{dx}\sqrt{1 - 4x^2} = \frac{1}{2\sqrt{1 - 4x^2}}$ .”

What to write

“Let $u = 1 - 4x^2$ , $u' = -8x$ : $\frac{d}{dx}\sqrt u = \frac{u'}{2\sqrt u} = -\frac{4x}{\sqrt{1 - 4x^2}}$ .”

Why: The outside layer alone is correct, which is why the error survives rereading. At $x = \frac{1}{4}$ the true slope is $-\frac{2}{\sqrt 3} \approx -1.1547$ , negative, while the false answer is positive.

2. Computing the new exponent in the wrong direction

1 to 2 marks, and a wrong value everywhere

What not to write

“ $\frac{d}{dx}(5x + 1)^{-1/2} = -\frac{5}{2}(5x + 1)^{1/2}$ .”

What to write

“ $-\frac{1}{2} - 1 = -\frac{3}{2}$ , so $\frac{d}{dx}(5x + 1)^{-1/2} = -\frac{5}{2}(5x + 1)^{-3/2}$ , which is $-\frac{5}{128}$ at $x = 3$ .”

Why: Subtracting $1$ from a negative fraction moves it further from zero. The false answer at $x = 3$ is $-10$ , and it grows with $x$ while the function flattens.

3. Differentiating the denominator of a constant over a power

the whole question

What not to write

“ $\frac{d}{dx}\frac{2}{(x^2 + 3)^4} = \frac{2}{4(x^2 + 3)^3 \cdot 2x}$ .”

What to write

“Rewrite $2(x^2 + 3)^{-4}$ ; then $y' = -8(x^2 + 3)^{-5}\cdot 2x = -\frac{16x}{(x^2 + 3)^5}$ , which is $-\frac{1}{64}$ at $x = 1$ .”

Why: The derivative of $\frac{1}{v}$ is not $\frac{1}{v'}$ . Rewriting as a negative power removes the temptation and the quotient rule at once.

4. Factoring out the higher power

2 to 3 marks, and every horizontal tangent found from it

What not to write

“ $2x(2x + 1)^{1/2} + x^2(2x + 1)^{-1/2} = x(2x + 1)^{1/2}(x + 2)$ .”

What to write

“Lowest power $-\frac{1}{2}$ : $x(2x + 1)^{-1/2}\left[2(2x + 1) + x\right] = \frac{x(5x + 2)}{\sqrt{2x + 1}}$ .”

Why: Taking out $(2x + 1)^{1/2}$ leaves $x^2(2x + 1)^{-1}$ in the second term, not $x^2$ . At $x = 4$ the false form gives $72$ , the true one $\frac{88}{3}$ .

5. Reading the table in the row of x instead of g(x)

the whole question, on every table item of the exam

What not to write

“ $g(3) = 5$ , $g'(3) = 2$ , $f'(3) = -1$ , $f'(5) = 4$ , so $(f \circ g)'(3) = f'(3)\,g'(3) = -2$ .”

What to write

“ $(f \circ g)'(3) = f'(g(3))\,g'(3)$ . Since $g(3) = 5$ , it is $f'(5) \cdot 2 = 8$ .”

Why: The outside function never sees $x$ , only the value that $g$ hands it. Write $g(3)$ as a number first, then go to that row.

6. Applying the power rule to an exponential in base a

the whole question

What not to write

“ $\frac{1}{2^x} = 2^{-x}$ , so its derivative is $-x\,2^{-x-1}$ .”

What to write

“ $2^{-x} = e^{-x\ln 2}$ with $\ln 2$ a constant, so its derivative is $-\ln 2 \cdot 2^{-x}$ .”

Why: The rewriting was right, the rule was not: the power rule needs a variable BASE and a constant exponent. At $x = 0$ the false formula gives $0$ , the true slope is $-\ln 2 \approx -0.6931$ .

7. Missing the points where the derivative does not exist

1 to 2 marks on a graph or tangent question

What not to write

“ $y = \sqrt[3]{x^2 - 1}$ has $y' = 0$ only at $x = 0$ , so its tangent turns smoothly everywhere.”

What to write

“ $y' = \frac{2x}{3(x^2 - 1)^{2/3}}$ is $0$ at $x = 0$ and UNDEFINED at $x = \pm 1$ , where the tangent is vertical.”

y = \sqrt[3]{x^2 - 1}

: a horizontal tangent at

(0, -1)

, where

y' = 0

, and vertical tangents at

x = \pm 1

, where the denominator of

y'

vanishes.

Why: The negative exponent $-\frac{2}{3}$ puts $x^2 - 1$ in the denominator. Its zeros are points of the curve where $y'$ does not exist.

8. Skipping the product rule in a second derivative

the whole question

What not to write

“ $g(x) = f(x^2)$ , so $g''(x) = 4x^2 f''(x^2)$ .”

What to write

“ $g'(x) = 2x\,f'(x^2)$ is a product, so $g''(x) = 2f'(x^2) + 4x^2f''(x^2)$ .”

Why: The first derivative of a composite is a product. With $f'(4) = -1$ and $f''(4) = 3$ , $g''(2) = 46$ ; the false formula gives $48$ .

Which method to choose

Before differentiating: rewrite according to the FORM of the expression

Look at how the expression is written, rewrite it as powers or a single exponential, then differentiate

If a constant over a power of a bracket, $\frac{c}{u^k}$ → rewrite as $c\,u^{-k}$ , then the Power Chain Rule, exponent $-k - 1$

Example: $\frac{3}{(1 - x^2)^2} = 3(1 - x^2)^{-2}$ , derivative $\frac{12x}{(1 - x^2)^3}$
If a root of a power, $\sqrt[n]{u^m}$ → rewrite as $u^{m/n}$ , the index in the denominator

Example: $\sqrt[3]{(x^2 + 4)^2} = (x^2 + 4)^{2/3}$ , derivative $\frac{4x}{3\sqrt[3]{x^2 + 4}}$
If one over a root, $\frac{1}{\sqrt u}$ → rewrite as $u^{-1/2}$ , exponent of the derivative $-\frac{3}{2}$

Example: $\frac{1}{\sqrt{2x + 5}}$ has derivative $-(2x + 5)^{-3/2}$
If a constant base $a$ with the variable in the exponent, possibly under a root or a fraction → laws of exponents until one $a^u$ remains, then $a^u\ln a \cdot u'$

Example: $\frac{1}{3^{x^2}} = 3^{-x^2}$ , derivative $-2x\ln 3 \cdot 3^{-x^2}$
If a product of powers of brackets → product rule, chain rule inside each factor, then factor each bracket at its lowest power

Example: $x(x - 1)^{2/3}$ gives $\frac{5x - 3}{3(x - 1)^{1/3}}$
If a quotient whose denominator is a power of a bracket → rewrite as a product with a negative exponent, then as above

Example: $\frac{(x + 1)^3}{(x - 2)^2}$ gives $\frac{(x + 1)^2(x - 8)}{(x - 2)^3}$

the quotient rule is correct too, but it ends in a complex fraction that needs the same factorisation

Several branches can apply in turn: $\frac{4}{\sqrt{\sin x}} = 4(\sin x)^{-1/2}$ takes the root branch, then the trigonometric derivative as the inside factor.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Differentiate and simplify completely

When to use it: The question says “differentiate and simplify”, or asks next for the horizontal tangents or the points where $y'$ does not exist

1 Rewrite roots and reciprocals as powers; name each inside function.
2 Apply the product or quotient rule if the LAST operation calls for it, then the Power Chain Rule on each factor, one factor per layer.
3 Factor each bracket at its lowest power and simplify the bracket that remains to a polynomial.
4 Write one fraction with positive exponents, and state where the denominator is zero.

Concluding sentence

“Let $u = x^2 - 1$ . Then $y' = (x^2 - 1)^{1/3} + \frac{2x(x + 2)}{3}(x^2 - 1)^{-2/3} = \frac{5x^2 + 4x - 3}{3(x^2 - 1)^{2/3}}$ , undefined at $x = \pm 1$ .”

The trap: Stopping at the unfactored sum: it is correct, earns part of the marks, and makes the next question impossible.

Marking: Typically 1 mark for the rewriting, 2 for the derivative, 1 for the factorisation, 1 for the final form.

A chain of rates with units

When to use it: A quantity depends on a second one, which depends on time, each link given by a formula or a graph

1 Write the chain in Leibniz form, $\frac{dT}{dt} = \frac{dT}{dh}\cdot\frac{dh}{dt}$ , and check that the units cancel to those of the answer.
2 Compute the CURRENT value of each intermediate variable at the given time.
3 Read each derivative at the value of its own variable, and check that the outside function is differentiable there.
4 Multiply, and interpret the sign and the unit in a sentence.

Concluding sentence

“At $t = 20$ min, $h = 6$ km, so $\frac{dT}{dt} = T'(6)\cdot h'(20) = -6.5 \cdot 0.3 = -1.95$ °C per minute: the reading falls by about two degrees each minute.”

The trap: Reading $T'$ at $20$ , a number of minutes, instead of at the altitude $h = 6$ km.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

As many factors as layers

Count the layers of the function, then the factors of your derivative. The innermost factor, often a plain number, is the one that disappears.

$(3 + \sin 3x)^{-2}$ has three layers, so its derivative has three factors: $-2(3 + \sin 3x)^{-3}$ , $\cos 3x$ and $3$ .
The new exponent went down by exactly 1

Compare the exponent of the derivative with the exponent you started from. It must be one less, including for negative and fractional exponents.

From $-\frac{1}{2}$ you must land on $-\frac{3}{2}$ ; landing on $\frac{1}{2}$ means you added.
Factored form against unfactored sum, at one point

Evaluate both expressions at a convenient x. If they differ, the factorisation is wrong, usually taken at the higher power.

At $x = 4$ , $2x(2x + 1)^{1/2} + x^2(2x + 1)^{-1/2} = 24 + \frac{16}{3} = \frac{88}{3}$ , and $\frac{x(5x + 2)}{\sqrt{2x + 1}} = \frac{88}{3}$ too.
The sign matches the graph

Decide from the function whether it rises or falls at your point before trusting the sign of your derivative. Two minus signs that should cancel are the usual culprit.

$\frac{1}{(3x - 1)^2}$ decreases for $x > \frac{1}{3}$ , so its derivative there must be negative: $-\frac{6}{(3x - 1)^3}$ is, $\frac{1}{6(3x - 1)}$ is not.

The typical problem, taken apart

Differentiate, factor, and read the tangents

Let $y = x\sqrt{9 - x^2}$ on $[-3, 3]$ . Find $y'$ in factored form, the points where the tangent is horizontal, and the points where $y'$ does not exist.

Give exact values, and justify every step as on a MATH 203 final.

The curve starts and ends on the

x

-axis at

x = \pm 3

and has one hump on each side of the origin: expect two horizontal tangents and something special at the endpoints.

Step 1

Rewrite: $y = x(9 - x^2)^{1/2}$ . The last operation is a product; the second factor is a composite with inside $u = 9 - x^2$ , $u' = -2x$ .

Why

Naming the last operation picks the first rule; naming $u$ secures the inside factor.

Step 2

$y' = (9 - x^2)^{1/2} + x \cdot \frac{1}{2}(9 - x^2)^{-1/2}(-2x) = (9 - x^2)^{1/2} - x^2(9 - x^2)^{-1/2}$ .

Why

Product rule outside, Power Chain Rule inside, with $\frac{1}{2} - 1 = -\frac{1}{2}$ . This sum is correct but cannot yet be solved.

Step 3

Lowest power $-\frac{1}{2}$ : $y' = (9 - x^2)^{-1/2}\left[(9 - x^2) - x^2\right] = \frac{9 - 2x^2}{\sqrt{9 - x^2}}$ .

Why

Taking out $u^{-1/2}$ leaves $u^{1/2 - (-1/2)} = u$ in the first term: the bracket becomes a polynomial.

Step 4

Horizontal tangents: $9 - 2x^2 = 0$ , $x = \pm\frac{3}{\sqrt 2}$ , and $y = \pm\frac{3}{\sqrt 2}\cdot\frac{3}{\sqrt 2} = \pm\frac{9}{2}$ .

Why

A fraction is zero when its numerator is, provided the denominator is not: here $9 - x^2 = \frac{9}{2} \ne 0$ .

Step 5

$y'$ does not exist at $x = \pm 3$ , where $\sqrt{9 - x^2} = 0$ while the numerator is $-9$ : the tangents at the two endpoints are vertical.

Why

The negative exponent put $9 - x^2$ in the denominator; its zeros are the points to report.

The conclusion, written out

“ $y' = \frac{9 - 2x^2}{\sqrt{9 - x^2}}$ ; horizontal tangents at $\left(\frac{3}{\sqrt 2}, \frac{9}{2}\right)$ and $\left(-\frac{3}{\sqrt 2}, -\frac{9}{2}\right)$ ; $y'$ undefined at $x = \pm 3$ , where the tangent is vertical.”

The classic mistake on this problem: Factoring out $(9 - x^2)^{1/2}$ and writing $y' = (9 - x^2)^{1/2}(1 - x^2)$ , which puts the horizontal tangents at $x = \pm 1$ .

Learn by heart

• $(f \circ g)'(x) = f'(g(x))\,g'(x)$ : outside derivative AT the inside, times the inside derivative.
• One factor per layer; count them.
• Rewrite first: $\frac{c}{u^k} = cu^{-k}$ , $\sqrt[n]{u^m} = u^{m/n}$ , $\frac{1}{a^x} = a^{-x}$ .
• $\frac{d}{dx}u^n = nu^{n-1}u'$ ; the exponent goes DOWN by $1$ : $-\frac{1}{2} \to -\frac{3}{2}$ .
• $\frac{d}{dx}a^u = a^u\ln a\cdot u'$ , from $a^u = e^{u\ln a}$ .
• Factor each bracket at its LOWEST power; the leftover bracket is a polynomial.
• Table: compute $g(a)$ first, then read $f'$ at $g(a)$ .

Frequently asked questions

How do I know which is the inside function in the chain rule?

Read the expression the way a calculator would evaluate it. The operation performed last is the outside function, and whatever it is applied to is the inside function. In the square root of one minus four x squared, the root is done last, so the inside function is one minus four x squared. Name it u on the first line of your answer.

How do I simplify a derivative with fractional or negative exponents?

Factor each bracket at its lowest power, which is the most negative exponent. Between one half and minus one half, take out minus one half; between minus two and minus three, take out minus three. What remains in the square bracket then has only whole positive powers and simplifies to a polynomial. Finish by moving negative exponents to the denominator.

What is the derivative of a to the power x?

It is a to the x times the natural logarithm of a. Write a to the x as e to the power x times ln a, where ln a is just a constant, and apply the chain rule. With an expression u in the exponent instead of x, multiply also by the derivative of u. The power rule never applies, because the variable is in the exponent.

Why is my derivative undefined at some points where the function is defined?

After the chain rule, a fractional power often comes out with a negative exponent, which puts its bracket in the denominator. Where that bracket is zero, the derivative does not exist even though the function does: the cube root of x squared minus one is defined at x equals one, but its tangent there is vertical. Report those points with the zeros of the numerator.

Practise it

Corrected exercises: The chain rule, MATH 203 at Concordia

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Derivatives of trigonometric functions Next sheet Implicit differentiation

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math203-chain-rule. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

Looking for a MATH 203 tutor in Montreal?

Get in touch for a first session. The chain rule is the rule every later chapter of MATH 203 relies on, and the algebra around it is where most marks are won back.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: the chain rule (MATH 203)

The essentials

One factor per layer: name u, differentiate the outside at u, multiply by u'

Rewrite before, factor after: the algebra where MATH 203 marks are lost

The rules in table form

Which power to factor out of a derivative

The mistakes that cost marks

1. Forgetting the inside derivative of a root

2. Computing the new exponent in the wrong direction

3. Differentiating the denominator of a constant over a power

4. Factoring out the higher power

5. Reading the table in the row of x instead of g(x)

6. Applying the power rule to an exponential in base a

7. Missing the points where the derivative does not exist

8. Skipping the product rule in a second derivative

Which method to choose

Before differentiating: rewrite according to the FORM of the expression

How the answer is expected to be written

Differentiate and simplify completely

A chain of rates with units

Check before you hand in

The typical problem, taken apart

Differentiate, factor, and read the tangents

Learn by heart

Frequently asked questions

How do I know which is the inside function in the chain rule?

How do I simplify a derivative with fractional or negative exponents?

What is the derivative of a to the power x?

Why is my derivative undefined at some points where the function is defined?

Corrected exercises: The chain rule, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?