MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: Differentiation rules (MATH 203)

This sheet is not a summary of section 3.3 of Thomas' Calculus: you have the textbook. It answers one question, what makes students lose marks on the differentiation rules in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The rules themselves are rarely the problem. The marks go in the algebra that surrounds them: an exponent rewritten wrongly before differentiating, a sign or a factor mishandled after. Every trap below comes with the line that earns the mark, and every value is checked; where a scientific calculator helps, the sheet says how to use it as a ten-second check.

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

The rule takes one line and the marks are lost in the algebra around it: rewrite so that only the power of $x$ crosses the fraction bar, distribute the minus sign and factor the smallest power after differentiating, and substitute the point last.

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

Power rule for real exponents
Laws of exponents
Derivative of the exponential function
Product rule
Quotient rule
Reciprocal rule
Higher-order derivatives
Tangent line
Normal line
Velocity and acceleration
Marginal cost and marginal revenue

The essentials

Before differentiating: rewrite, and move only the power of x

• Power rule: $(x^n)' = nx^{n-1}$ for EVERY real $n$ , on the form $cx^n$ only.
• A power of $x$ in the denominator crosses the bar with its sign changed; a numerical coefficient does NOT move: $\frac{5}{2x^3} = \frac{5}{2}x^{-3}$ .
• Roots: $\sqrt[q]{x^p} = x^{p/q}$ ; products of powers add exponents: $x^2\sqrt x = x^{5/2}$ .
• Subtract $1$ from the exponent, even when it is negative: $-\frac{2}{3} - 1 = -\frac{5}{3}$ , further from zero.
• Letters that are numbers stay coefficients: $\frac{e^3}{x} = e^3x^{-1}$ , derivative $-\frac{e^3}{x^2}$ ; and $(e^x)' = e^x$ .

For

y = \frac{1}{2x^2} = \frac{1}{2}x^{-2}

the tangent at

x = 1

has slope

-1

; the false rewrite

2x^{-2}

gives slope

-4

, a line that cuts through the curve.

The rewritten line is the one a marker reads first: when it shows $\frac{3}{2}x^{-4}$ and not $(2x)^{-4}$ , the method mark is secured.

After differentiating: the algebra that finishes the answer

• Product: $(fg)' = f'g + fg'$ . Quotient: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ , the minus sign on the WHOLE product $fg'$ .
• Factor the SMALLEST power: $-x^{-3/2} + \frac{9}{2}x^{-5/2} = \frac{9 - 2x}{2x^{5/2}}$ . Factor $e^x$ out of every term.
• Cancel a common factor, but keep the values the original function excludes: $\frac{1 - x^2}{(x + 1)^4} = \frac{1 - x}{(x + 1)^3}$ with $x \ne -1$ .
• Tangent at $a$ : $y = f(a) + f'(a)(x - a)$ , slope a NUMBER, substituted last. Normal: slope $-\frac{1}{f'(a)}$ .
• Motion: $v = s'$ , $a = s''$ . Marginal cost $C'(x)$ and marginal revenue $R'(x)$ , in dollars per unit.

No chain rule in this chapter: a squared binomial such as $(x + 1)^2$ is expanded before it is differentiated.

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.

Rewriting before the power rule: the correct forms and the false ones

Read a line as: the expression of the first column is rewritten as in the second, then differentiated as in the third. The red lines are rewritings students write that are simply not equal to the expression.

Expression	Rewritten as	Derivative
$\frac{5}{x^3}$	$5x^{-3}$	$-15x^{-4}$
Example: At $x = 1$ the slope is $-15$ ; at $x = 2$ it is $-\frac{15}{16}$ .
$\frac{5}{2x^3}$	$\frac{5}{2}x^{-3}$	$-\frac{15}{2}x^{-4}$
Example: At $x = 1$ the slope is $-\frac{15}{2} = -7.5$ .
$\frac{5}{2x^3}$	$10x^{-3}$	not equal false rewriting
Example: At $x = 1$ : $\frac{5}{2 \cdot 1} = 2.5$ , but $10 \cdot 1 = 10$ . What to do: Only $x^3$ crosses the bar: $\frac{5}{2x^3} = \frac{5}{2}x^{-3}$ .
$x\sqrt[3]{x}$	$x^{4/3}$	$\frac{4}{3}x^{1/3}$
Example: At $x = 8$ : $\frac{4}{3} \cdot 2 = \frac{8}{3}$ .
$\frac{1}{\sqrt x}$	$x^{1/2}$	not equal false rewriting
Example: At $x = 4$ : $\frac{1}{\sqrt 4} = 0.5$ , but $4^{1/2} = 2$ . What to do: Leaving the denominator changes the sign of the exponent: $x^{-1/2}$ , derivative $-\frac{1}{2}x^{-3/2}$ .
$x^{-2/3}$	$-\frac{2}{3}x^{1/3}$ as derivative	not a rule rule that does not exist
Example: At $x = 1$ the calculator quotient gives $-0.6667$ both ways, but at $x = 8$ the true slope is $-\frac{1}{48}$ , the false one $-\frac{4}{3}$ . What to do: Subtract $1$ : $-\frac{2}{3} - 1 = -\frac{5}{3}$ , so the derivative is $-\frac{2}{3}x^{-5/3}$ .

Every red line is refuted by one number substituted on both sides. That test takes five seconds and should be run on every rewriting you are unsure of.

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. Moving the coefficient with the power

1 to 2 marks, and every number that follows

What not to write

“ $\frac{1}{2x^3} = 2x^{-3}$ , so the derivative is $-6x^{-4}$ .”

What to write

“ $\frac{1}{2x^3} = \frac{1}{2}x^{-3}$ , so the derivative is $-\frac{3}{2}x^{-4} = -\frac{3}{2x^4}$ .”

Why: Only $x^3$ is raised to a power; the $2$ is a plain number that stays in the denominator. The false answer is four times too large: $-6$ against $-1.5$ at $x = 1$ .

2. Adding 1 to a negative exponent

1 mark per term

What not to write

“ $\frac{d}{dx}x^{-3} = -3x^{-2}$ .”

What to write

“ $\frac{d}{dx}x^{-3} = -3x^{-4} = -\frac{3}{x^4}$ .”

Why: The rule subtracts $1$ , and $-3 - 1 = -4$ . At $x = 2$ the true slope is $-\frac{3}{16} = -0.1875$ ; the false formula gives $-0.75$ .

3. Stopping the minus sign of the quotient rule after one term

2 marks: the value at

0

becomes

-8

instead of

8

What not to write

“ $\left(\frac{x^2 - 4}{2x + 1}\right)' = \frac{2x(2x + 1) - 2x^2 - 8}{(2x + 1)^2}$ .”

What to write

“ $\frac{2x(2x + 1) - (x^2 - 4) \cdot 2}{(2x + 1)^2} = \frac{2x^2 + 2x + 8}{(2x + 1)^2}$ .”

Why: The minus sign multiplies the whole product $fg'$ . Write it with parentheses, $-(x^2 - 4) \cdot 2$ , then expand: $-2x^2 + 8$ .

4. Keeping an excluded value after cancelling

1 mark, and the credibility of the whole answer

What not to write

“ $q(x) = \frac{x}{(x + 1)^2}$ has $q'(x) = \frac{1 - x^2}{(x + 1)^4}$ , zero at $x = 1$ and $x = -1$ : two horizontal tangents.”

What to write

“ $q'(x) = \frac{(1 - x)(1 + x)}{(x + 1)^4} = \frac{1 - x}{(x + 1)^3}$ , for $x \ne -1$ : one horizontal tangent, at $x = 1$ .”

Why: A fraction is zero when its numerator is zero AND its denominator is not. At $x = -1$ the denominator is zero: $q$ is not even defined there.

5. Writing the tangent line without the shift by a

the whole question: the line misses the point

What not to write

“Tangent to $y = \sqrt x$ at $x = 4$ : $f(4) = 2$ , $f'(4) = \frac{1}{4}$ , so $y = \frac{1}{4}x + 2$ .”

What to write

“ $y = f(4) + f'(4)(x - 4) = 2 + \frac{1}{4}(x - 4) = \frac{1}{4}x + 1$ .”

The true tangent

y = \frac{x}{4} + 1

touches

y = \sqrt x

(4, 2)

; the false line

y = \frac{x}{4} + 2

is parallel to it and never touches the point.

Why: $y = f'(a)x + f(a)$ passes through $(0, f(a))$ , not through $(a, f(a))$ . Plug $x = 4$ : the false line gives $3$ , not $2$ .

6. Differentiating a complex fraction as it stands

2 to 3 marks, lost in the simplification

What not to write

“ $\left(\frac{1 - \frac{2}{x}}{x + \frac{3}{x}}\right)' = \frac{\frac{2}{x^2}\left(x + \frac{3}{x}\right) - \left(1 - \frac{2}{x}\right)\left(1 - \frac{3}{x^2}\right)}{\left(x + \frac{3}{x}\right)^2}$ , then lost in the algebra.”

What to write

“Multiply top and bottom by $x$ : $\frac{x - 2}{x^2 + 3}$ for $x \ne 0$ , whose derivative is $\frac{-x^2 + 4x + 3}{(x^2 + 3)^2}$ .”

Why: The direct route is legal but carries fractions of fractions in every term. Clearing first gives a quotient of two polynomials; the condition $x \ne 0$ travels with it.

7. Reading the slope of a line not solved for y

the whole question

What not to write

“The tangent is parallel to $3x - 4y = 7$ , so its slope is $3$ .”

What to write

“ $3x - 4y = 7$ gives $y = \frac{3}{4}x - \frac{7}{4}$ : slope $\frac{3}{4}$ .”

Why: The coefficient of $x$ is the slope only in the form $y = mx + b$ . Solve for $y$ first, dividing EVERY term by $-4$ .

8. Taking a zero of the velocity for a change of direction

2 marks, and a wrong distance travelled

What not to write

“ $v(t) = 4t(t - 3)^2 = 0$ at $t = 3$ , so the dolly turns back at $t = 3$ .”

What to write

“ $v(t) \ge 0$ on both sides of $t = 3$ , since $(t - 3)^2 \ge 0$ : the dolly stops for an instant and continues forward.”

Why: A change of direction needs a change of SIGN of $v$ . A squared factor vanishes without changing sign.

9. Taking the price for the marginal revenue

the whole question

What not to write

“With $p(x) = 60 - 2\sqrt x$ , the marginal revenue at $x = 100$ is the price, $40$ dollars.”

What to write

“ $R = x\,p(x)$ , so $R'(x) = p(x) + x\,p'(x) = 60 - 3\sqrt x$ , and $R'(100) = 30$ dollars per unit.”

Why: Selling one more unit lowers the price on all the units sold: the product rule counts that loss in $x\,p'(x) = -10$ . The exact extra revenue is about $29.93$ dollars.

Which method to choose

After the rule: which simplification, by the FORM of what you got

Look at the raw derivative before substituting anything: its form tells you the next algebraic step

If a quotient-rule numerator with a minus sign in front of a product → put the product in parentheses, then expand and collect

Example: $-(x^2 - 4) \cdot 2 = -2x^2 + 8$
If a sum of powers of $x$ with different negative or fractional exponents → factor out the SMALLEST power, the most negative exponent

Example: $-x^{-3/2} + \frac{9}{2}x^{-5/2} = \frac{9 - 2x}{2x^{5/2}}$
If $e^x$ in every term → factor $e^x$ , then use $e^x > 0$

Example: $e^x(x^2 + 1) - 2xe^x = e^x(x - 1)^2$
If a numerator and a denominator that share a factor → factor both, cancel, and write the excluded values

Example: $\frac{1 - x^2}{(x + 1)^4} = \frac{1 - x}{(x + 1)^3}$ , $x \ne -1$
If the question asks for zeros or horizontal tangents → keep the fully factored form; numerator zero, denominator not

Example: $\frac{x^2 - 4}{x^2} = 0$ at $x = \pm 2$
If the question asks for a slope at one point → substitute into the simplified derivative, then check with the calculator quotient

Example: $h'(1) = \frac{9 - 2}{2} = 3.5$

Simplifying is not decoration: the factored form is the one that gives zeros, signs and the next derivative in a line.

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.

Points where the tangent has a prescribed direction

When to use it: The statement asks where the tangent (or the normal) is horizontal, or parallel, or perpendicular to a given line

1 Compute $f'(x)$ and simplify it to a single factored fraction.
2 Translate the direction into a slope: horizontal means $0$ ; parallel to $Ax + By = C$ means $-\frac{A}{B}$ after solving for $y$ ; a NORMAL parallel to a line of slope $m$ means a tangent of slope $-\frac{1}{m}$ .
3 Solve $f'(x) = m$ , and reject any solution outside the domain.
4 Compute $f$ at each solution and give the POINTS, both coordinates.

Concluding sentence

“The line $3x - 4y = 7$ has slope $\frac{3}{4}$ . Solving $f'(x) = 1 - \frac{4}{x^2} = \frac{3}{4}$ gives $x^2 = 16$ , $x = \pm 4$ , both in the domain. The points are $(4, 5)$ and $(-4, -5)$ .”

The trap: Stopping at the values of $x$ : the question asks for points, and half the marks are in the second coordinate.

Marking: Typically 2 marks for the derivative, 2 for the slope, 3 for the equation, 3 for the points.

A derivative from a table of values

When to use it: The functions are known only through a table, and the question asks for the derivative of a product or a quotient at a point

1 Write the rule with letters at the point: $P'(1) = f'(1)g(1) + f(1)g'(1)$ .
2 Substitute one number per letter, negative values in parentheses.
3 Compute, watching the sign of each product of two negatives.
4 If the question gives the result and hides one entry, solve the same equation for that entry.

Concluding sentence

“ $P'(1) = f'(1)g(1) + f(1)g'(1) = 4 \cdot 2 + (-3)(-5) = 23$ .”

The trap: Reading a value from the wrong row, or writing $(-3)(-5) = -15$ .

Marking: Typically 3 marks for the rule with letters, 5 for the substitution, 2 for the value.

Check before you hand in

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

The calculator quotient

With a scientific calculator, compute f(a + 0.001) minus f(a - 0.001), divided by 0.002. It must agree with your f'(a) to about four digits. It catches every rewriting error and every lost sign.

For $h(x) = \frac{2x - 3}{x\sqrt x}$ at $a = 1$ : $\frac{h(1.001) - h(0.999)}{0.002} \approx 3.5000$ , matching $h'(1) = \frac{7}{2}$ .
Plug the point into the tangent

Replace x by a in your tangent line: you must get f(a). A line of the right slope that fails this test was written without the shift by a.

$y = -3x + 8$ at $x = 1$ gives $5 = f(1)$ for $f(x) = x + \frac{4}{x}$ : consistent.
The sign matches the graph

Where the function decreases, the derivative must be negative. Test one value.

$\frac{1}{2x^3}$ decreases for $x > 0$ ; the answer $-\frac{3}{2x^4} < 0$ is consistent.
Check a product by expanding

When both factors are simple, expand the product and differentiate term by term: both answers must agree.

$(x - 1)(x + 2)(x + 3) = x^3 + 4x^2 + x - 6$ , derivative $12$ at $x = 1$ , as the triple product rule gives.

The typical problem, taken apart

Horizontal tangent, tangent and normal after rewriting

Let $f(x) = \frac{(x + 1)^2}{\sqrt x}$ for $x > 0$ . Find the point where the tangent is horizontal, then the tangent and the normal lines at the point of abscissa $1$ .

No chain rule. Every rule must be named, as on a MATH 203 final.

The curve comes down from very high values near

0

, is flat once, then rises: the flat point is the one to find, and

(1, 4)

is already on the rising part.

Step 1

Expand and rewrite: $f(x) = (x^2 + 2x + 1)x^{-1/2} = x^{3/2} + 2x^{1/2} + x^{-1/2}$ .

Why

No chain rule for $(x + 1)^2$ , and no quotient rule needed once the denominator is a single power of $x$ : dividing term by term turns the problem into three power rules.

Step 2

Power rule: $f'(x) = \frac{3}{2}x^{1/2} + x^{-1/2} - \frac{1}{2}x^{-3/2}$ .

Why

Each exponent loses $1$ : $\frac{1}{2} - 1 = -\frac{1}{2}$ and $-\frac{1}{2} - 1 = -\frac{3}{2}$ . The sign of the last term comes from the exponent $-\frac{1}{2}$ .

Step 3

Factor the smallest power $\frac{1}{2}x^{-3/2}$ : $f'(x) = \frac{3x^2 + 2x - 1}{2x^{3/2}} = \frac{(3x - 1)(x + 1)}{2x\sqrt x}$ .

Why

Inside the bracket, $x^{1/2} = x^{-3/2} \cdot x^2$ and $x^{-1/2} = x^{-3/2} \cdot x$ . The factored form gives the zeros at once.

Step 4

$f'(x) = 0$ for $x > 0$ only at $x = \frac{1}{3}$ ( $x = -1$ is outside the domain). $f\left(\frac{1}{3}\right) = \frac{16}{9} \cdot \sqrt 3 = \frac{16\sqrt 3}{9} \approx 3.0792$ .

Why

Rejecting $x = -1$ is part of the answer. The height uses $\frac{1}{\sqrt{1/3}} = \sqrt 3$ .

Step 5

At $x = 1$ : $f(1) = 4$ and $f'(1) = \frac{2 \cdot 2}{2} = 2$ . Tangent $y = 4 + 2(x - 1) = 2x + 2$ . Normal: slope $-\frac{1}{2}$ , $y = 4 - \frac{1}{2}(x - 1) = -\frac{1}{2}x + \frac{9}{2}$ .

Why

The slope comes from the factored derivative, substituted last. The check $2 \cdot \left(-\frac{1}{2}\right) = -1$ confirms the normal, and both lines give $4$ at $x = 1$ .

The conclusion, written out

“The tangent is horizontal at $\left(\frac{1}{3}, \frac{16\sqrt 3}{9}\right)$ . At $(1, 4)$ the tangent is $y = 2x + 2$ and the normal is $y = -\frac{1}{2}x + \frac{9}{2}$ .”

The classic mistake on this problem: Writing $(x + 1)^2x^{-1/2}$ and differentiating it as $2(x + 1) \cdot \left(-\frac{1}{2}\right)x^{-3/2}$ , factor by factor: at $x = 1$ this gives $-2$ instead of $2$ .

Learn by heart

• Rewrite first: only the power of $x$ crosses the bar, $\frac{5}{2x^3} = \frac{5}{2}x^{-3}$ .
• $(x^n)' = nx^{n-1}$ : subtract $1$ , even from a negative exponent. $(e^x)' = e^x$ , $(c)' = 0$ .
• $(fg)' = f'g + fg'$ ; $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ , minus sign on the whole product.
• After the rule: factor the smallest power or $e^x$ , cancel common factors, keep the excluded values.
• Tangent: $y = f(a) + f'(a)(x - a)$ ; normal slope $-\frac{1}{f'(a)}$ .
• A zero of $v$ is a turn only if $v$ changes sign. Marginal revenue $R' = p + xp'$ , not $p$ .
• Check any derivative with $\frac{f(a + 0.001) - f(a - 0.001)}{0.002}$ on the calculator.

Frequently asked questions

How do I rewrite a fraction like 5 over 2x cubed before using the power rule?

Only the power of x moves out of the denominator, and its exponent changes sign. The number 2 is a plain coefficient and stays where it is, so 5 over 2x cubed becomes five halves times x to the minus three. Its derivative is then minus fifteen halves times x to the minus four. Moving the 2 up as well gives an answer four times too large.

Why is my quotient rule answer off by a sign?

Two usual causes. Either the terms of the numerator were swapped, and the rule is the derivative of the top times the bottom minus the top times the derivative of the bottom. Or the minus sign stopped after the first term of the product it multiplies. Put that product in parentheses before expanding, and check one value with the calculator.

Can I use my calculator to check a derivative in MATH 203?

Yes, with any scientific calculator. Compute the function a little to the right of the point, a little to the left, subtract, and divide by the gap, for instance with steps of one thousandth. The result must match your derivative to about four digits. It does not replace the rules on the exam, but it catches a lost sign or a wrong exponent in ten seconds.

Does a zero velocity always mean the object changes direction?

No. The object changes direction only if the velocity changes sign at that instant. If the velocity has a squared factor, it can touch zero and stay positive on both sides: the object stops for an instant and continues the same way. Always check the sign of the velocity just before and just after.

Is marginal revenue the same as the price?

No. Revenue is the price times the quantity, so its derivative, by the product rule, is the price plus the quantity times the derivative of the price. When selling more forces the price down, that second term is negative and the marginal revenue is smaller than the price.

Practise it

Corrected exercises: Differentiation rules, 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 The derivative at a point and as a function Next sheet Derivatives of trigonometric functions

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

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Revision sheet: Differentiation rules (MATH 203)

The essentials

Before differentiating: rewrite, and move only the power of x

After differentiating: the algebra that finishes the answer

The rules in table form

Rewriting before the power rule: the correct forms and the false ones

The mistakes that cost marks

1. Moving the coefficient with the power

2. Adding 1 to a negative exponent

3. Stopping the minus sign of the quotient rule after one term

4. Keeping an excluded value after cancelling

5. Writing the tangent line without the shift by a

6. Differentiating a complex fraction as it stands

7. Reading the slope of a line not solved for y

8. Taking a zero of the velocity for a change of direction

9. Taking the price for the marginal revenue

Which method to choose

After the rule: which simplification, by the FORM of what you got

How the answer is expected to be written

Points where the tangent has a prescribed direction

A derivative from a table of values

Check before you hand in

The typical problem, taken apart

Horizontal tangent, tangent and normal after rewriting

Learn by heart

Frequently asked questions

How do I rewrite a fraction like 5 over 2x cubed before using the power rule?

Why is my quotient rule answer off by a sign?

Can I use my calculator to check a derivative in MATH 203?

Does a zero velocity always mean the object changes direction?

Is marginal revenue the same as the price?

Corrected exercises: Differentiation rules, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?