Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: the derivative and its rules (201-SN2-RE)

This sheet does not copy out the table of derivatives again: you have it in your notes, and copying it costs nobody a single mark. It deals with what really decides the grade on the second test of Differential Calculus 201-SN2-RE (formerly 201-NYA): recognizing the structure of an expression before choosing a rule, not losing the inside derivative, and writing the equation of a tangent line the way the marker expects it.

It is written for CEGEP students in the Science program (Sciences de la nature). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Before differentiating, read the STRUCTURE of the expression, never its symbols: the rule to apply is decided by the LAST operation performed. Almost every mark lost in this chapter comes from seeing a $\sin$ or a square instead of seeing a product, a quotient or a composition.

This chapter is part of 201-SN2-RE, Differential Calculus

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

Average rate of change
Derivative as a limit
Derivative at a point
Product rule
Quotient rule
Chain rule
Implicit differentiation
Logarithmic differentiation
Derivatives of transcendental functions
Higher-order derivatives
Differentiability and continuity
Equation of the tangent line

The essentials

The derivative is a limit, and which one

• $f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ , or equivalently $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ .
• The quotient ALWAYS has the form $\frac{0}{0}$ : that is why you must transform the expression, never substitute $h=0$ .
• Interpretation: $\frac{f(a+h)-f(a)}{h}$ is the slope of the SECANT line; its limit is the slope of the TANGENT line.
• $f'$ is a FUNCTION, $f'(a)$ is a NUMBER. This distinction decides half of the tangent line questions.

The secant line through

(2, 4)

and

(3, 9)

has slope

5

; as the second point slides towards the first, it tips towards the tangent line at

(2, 4)

, with slope

4

. The limit computes that tipping, and only that.

When the question says “from the definition” or “using the limit”, any answer obtained with the rules is worth zero, even if it is correct.

The rules, in the order you look for them

• Sum and constant multiple: $(u\pm v)'=u'\pm v'$ and $(ku)'=ku'$ . These are the ONLY rules that distribute.
• Product: $(uv)'=u'v+uv'$ . Quotient: $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}$ , numerator in THAT order.
• Chain: $\left(f(g(x))\right)'=f'(g(x))\cdot g'(x)$ . The second factor is the inside derivative.
• Transcendental: $(\sin x)'=\cos x$ , $(\cos x)'=-\sin x$ , $(\tan x)'=\sec^{2}x$ , $\left(e^{x}\right)'=e^{x}$ , $(\ln x)'=\frac{1}{x}$ , all in RADIANS.
• Common compositions: $\left(e^{u}\right)'=u'e^{u}$ , $(\ln u)'=\frac{u'}{u}$ , $\left(u^{n}\right)'=nu^{n-1}u'$ .

The factor $u'$ that shows up everywhere is one and the same thing: the inside derivative. Forgetting it is the most frequent error in the chapter.

What is NOT a rule

• $(uv)'\neq u'v'$ : a product is not differentiated factor by factor.
• $\left(\frac{u}{v}\right)'\neq\frac{u'}{v'}$ : neither is a quotient.
• $\left(x^{x}\right)'\neq x\cdot x^{x-1}$ : the power rule assumes a CONSTANT exponent.
• $\left(a^{x}\right)'=a^{x}\ln a$ , not $x\,a^{x-1}$ : constant base, variable exponent, it is the exponential.
• Continuity does NOT imply differentiability; differentiability implies continuity. The implication goes one way only.

Reading non-differentiability on an exam graph

• The question does not ask for a definition, it asks for a sweep: run along the curve from left to right and stop wherever the pencil abruptly changes direction, lifts off, or shoots up vertically.
• A DISCONTINUITY is enough to conclude, with no computation: not continuous, so not differentiable. Corners and vertical tangents, on the other hand, occur on perfectly continuous pieces, which is what makes them easy to miss.
• A CONSTANT piece is not a point where the derivative is zero, it is a whole INTERVAL: answering with a single $x$ -value where $f'=0$ on $(a,b)$ is expected costs the question.
• The sign of $f'$ is discussed only at points where $f'$ EXISTS. Claiming that the derivative is positive “at” a corner is an error of logic, not a slip.

An exam graph does not sort the accidents for you: here a corner at

x=-3

, a jump at

x=0

and a vertical tangent at

x=3

, plus a constant piece on

(-6,-3)

where

f'

is zero everywhere and not at a single point.

“Where is $f$ not differentiable?” expects a LIST of $x$ -values AND one reason per value. Naming the points without naming the cause earns half the marks.

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.

The basic derivatives

The table the exam assumes you know. Each row comes with the number it produces at a point, because a wrong derivative is spotted faster on a slope than on a formula.

$f(x)$	Where it holds	$f'(x)$
$k$ , a constant	everywhere	$0$
Example: $\left(\pi^{2}\right)'=0$ , and $\left(\sqrt{2}\,x^{3}\right)'=3\sqrt{2}\,x^{2}$ : anything that does not contain $x$ is a constant.
$x^{n}$ , $n$ constant	any real $n$	$nx^{n-1}$
Example: $\left(x^{5}\right)'=5x^{4}$ , so the slope at $x=2$ is $80$ .
$\sqrt{x}$	$x>0$ strictly	$\frac{1}{2\sqrt{x}}$
Example: At $x=9$ , the slope is $\frac{1}{6}\approx 0.167$ . At $x=0$ it does not exist: the tangent line there is vertical.
$\frac{1}{x}$	$x\neq 0$	$-\frac{1}{x^{2}}$
Example: At $x=2$ , the slope is $-0.25$ : always negative, which you can see on both branches of the hyperbola.
$\mathrm{e}^{x}$	everywhere	$\mathrm{e}^{x}$
Example: The only function equal to its own derivative. At $x=0$ , the slope is $1$ ; at $x=1$ it is $\mathrm{e}\approx 2.718$ .
$a^{x}$ , $a>0$	everywhere	$a^{x}\ln a$
Example: $\left(2^{x}\right)'=2^{x}\ln 2$ , so the slope at $x=3$ is $8\ln 2\approx 5.55$ , not $3\times 2^{2}=12$ .
$\ln x$	$x>0$	$\frac{1}{x}$
Example: At $x=4$ , the slope is $0.25$ . With $\ln\|x\|$ , the same derivative $\frac{1}{x}$ holds for every $x\neq 0$ .
$\sin x$	in radians	$\cos x$
Example: At $x=0$ , the slope is $1$ : that is exactly the limit $\frac{\sin x}{x}\to 1$ from the previous chapter.
$\cos x$	in radians	$-\sin x$
Example: At $x=\frac{\pi}{2}$ , the slope is $-1$ . The minus sign is the only thing to remember from this row.
$\tan x$	$x\neq\frac{\pi}{2}+k\pi$	$\sec^{2}x=1+\tan^{2}x$
Example: At $x=0$ , the slope is $1$ ; at $x=\frac{\pi}{4}$ it is already $2$ . A derivative that never vanishes, so a function that is strictly increasing on each branch.

Every row can be read with a $u$ in place of $x$ , and a factor $u'$ behind it: $\left(\mathrm{e}^{u}\right)'=u'\mathrm{e}^{u}$ , $(\ln u)'=\frac{u'}{u}$ , $\left(u^{n}\right)'=nu'u^{n-1}$ . The table gives the outside; you write the $u'$ .

The five operations, and what they produce

Read the LAST operation in the expression, the one you would perform last on a calculator, and take the corresponding row.

Form of $f$	What you set up on scrap paper	$f'$
$u+v$	nothing, differentiate each piece	$u'+v'$
Example: $\left(x^{3}+\sin x\right)'=3x^{2}+\cos x$ , which equals $3+\cos 1\approx 3.540$ at $x=1$ .
$ku$ , $k$ constant	take $k$ out	$ku'$
Example: $\left(5\ln x\right)'=\frac{5}{x}$ , so the slope at $x=5$ is exactly $1$ .
$uv$	$u$ , $v$ , $u'$ , $v'$	$u'v+uv'$
Example: $\left(x^{2}\sin x\right)'=2x\sin x+x^{2}\cos x$ , which equals $-\pi^{2}\approx -9.87$ at $x=\pi$ . With $u'v'$ you would get $2\pi\cos\pi\approx -6.28$ , a different slope.
$\frac{u}{v}$	$u$ , $v$ , $u'$ , $v'$	$\frac{u'v-uv'}{v^{2}}$
Example: $\left(\frac{x}{x+1}\right)'=\frac{1}{(x+1)^{2}}$ , so $0.25$ at $x=1$ . The numerator is read in THAT order, otherwise the slope changes sign.
$v\bigl(u(x)\bigr)$	the inside $u$ and $u'$	$u'\times v'\bigl(u(x)\bigr)$
Example: $\left(\sqrt{x^{2}+1}\right)'=\frac{x}{\sqrt{x^{2}+1}}$ , which equals $\frac{2}{\sqrt{5}}\approx 0.894$ at $x=2$ .
$u^{n}$	$u$ and $u'$	$nu'u^{n-1}$
Example: $\left((3x+1)^{4}\right)'=12(3x+1)^{3}$ , so $768$ at $x=1$ . Expanding first would give the same number after many more lines.

None of these rows distributes except the first two: $(uv)'\neq u'v'$ and $\left(\frac{u}{v}\right)'\neq\frac{u'}{v'}$ . It is the most frequent error on the second test, the one on the differentiation rules, and it is detected by comparing two numerical slopes.

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. Differentiating a product factor by factor

2 marks, and the whole question if what follows depends on it

What not to write

$\left(x^{2}\sin x\right)'=2x\cos x$

What to write

$\left(x^{2}\sin x\right)'=2x\sin x+x^{2}\cos x$

With

u=v=x

: the tangent line to

y=x^{2}

1

has slope

2

(green), not

u'v'=1

(red). The product of the derivatives gives a line that cuts the curve instead of touching it.

Why: One counterexample settles the matter for good: with $u=v=x$ , the product is $x^{2}$ , whose derivative is $2x$ , whereas $u'v'=1$ . The product rule does not have two terms for elegance, it has two because each of the two factors varies.

2. Losing the inside derivative in the chain rule

1 mark per occurrence, and there are many

What not to write

$\left(\sin(3x)\right)'=\cos(3x)$

What to write

$\left(\sin(3x)\right)'=3\cos(3x)$

At the origin,

\sin 3x

(orange) rises three times as fast as

\sin x

(blue): its tangent line has slope

3\cos 0=3

(green). Forgetting the inside derivative gives the slope

1

\sin x

(red).

Why: The forgotten factor here is $3$ , so the answer is three times too small. The ten-second check: the derivative of $\sin(3x)$ at $0$ must equal the slope of the curve at the origin, visibly steeper than that of $\sin x$ . An answer equal to $1$ instead of $3$ is spotted without any computation.

3. Reversing the order in the numerator of the quotient rule

the whole question, because the sign of the result is reversed

What not to write

$\left(\frac{u}{v}\right)'=\frac{uv'-u'v}{v^{2}}$

What to write

$\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}$ , with the TOP differentiated first

\frac{1}{x}

goes down: its derivative at

1

-1

(green). The reversed numerator gives

+1

(red), a line that rises where the curve falls.

Why: The result is not approximate, it is the opposite of the correct one: a sign chart built on it is entirely wrong. Instant check on $\frac{1}{x}$ , whose derivative must be $-\frac{1}{x^{2}}$ , hence negative: the correct formula gives $\frac{0\cdot x-1\cdot 1}{x^{2}}=-\frac{1}{x^{2}}$ ✓, the wrong one gives $+\frac{1}{x^{2}}$ .

4. Writing the tangent line with the derivative function instead of the derivative at the point

2 marks, even though the derivative was correct

What not to write

“ $f'(x)=3x^{2}$ , so the tangent line at $x=2$ is $y=3x^{2}(x-2)+8$ .”

What to write

“ $f'(2)=12$ , so the tangent line is $y=12(x-2)+8$ , that is $y=12x-16$ .”

The tangent line to

y=x^{3}

(2, 8)

is a LINE,

y=12x-16

(green). Keeping

f'(x)=3x^{2}

gives a curve (red): it is no longer a tangent line.

Why: A tangent line is a LINE: its equation cannot contain any $x^{2}$ . Substitute $a$ into $f'$ before writing anything. The check is visual: if the final equation is not of the form $y=mx+p$ , the substitution was not done.

5. Believing that a continuous function is differentiable

the whole differentiability question, often 3 marks

What not to write

“ $f$ is continuous at $0$ , so $f'(0)$ exists.”

What to write

“ $f$ is continuous at $0$ , but the two limits of the difference quotient differ, so $f$ is not differentiable at $0$ .”

Three continuous curves, none differentiable at the marked point: the corner of

|x|

has two opposite slopes, the cusp of

x^{2/3}

two infinite slopes,

x^{1/3}

a single slope, vertical.

Why: $|x|$ , $x^{2/3}$ and $\sqrt[3]{x}$ are continuous everywhere and yet not differentiable at $0$ , for three different reasons: a corner, a cusp, a vertical tangent. The implication only goes the other way: differentiable implies continuous, which is how you show that a discontinuous function is certainly not differentiable.

6. Applying the power rule to a variable exponent

the whole question

What not to write

$\left(x^{x}\right)'=x\cdot x^{x-1}=x^{x}$

What to write

$y=x^{x}$ gives $\ln y=x\ln x$ , then $\frac{y'}{y}=\ln x+1$ , so $\left(x^{x}\right)'=x^{x}\left(\ln x+1\right)$

x=2

, the tangent line to

y=x^{x}

has slope

4(\ln 2+1)\approx 6.77

(green). The power rule wrongly applied gives

4

(red): a line that is too flat.

Why: The rule $\left(x^{n}\right)'=nx^{n-1}$ requires a CONSTANT exponent, and $\left(a^{x}\right)'=a^{x}\ln a$ requires a constant base. When both vary, neither applies and logarithmic differentiation is the only way. Numerical check at $x=1$ : the correct formula gives $1\cdot(0+1)=1$ , which the difference quotient confirms.

7. Forgetting the factor $y'$ in implicit differentiation

the whole question, since the equation obtained no longer contains the unknown

What not to write

$\frac{d}{dx}\left(y^{2}\right)=2y$

What to write

$\frac{d}{dx}\left(y^{2}\right)=2y\,y'$ , because $y$ is a function of $x$

Why: In implicit differentiation, every $y$ is an unknown function of $x$ : differentiating a $y$ always produces a $y'$ , exactly as the chain rule says. The alarm bell is immediate: if the equation you obtain contains no $y'$ , there is nothing left to solve for and the solution stops dead.

8. Differentiating a sine with an argument in degrees

1 mark, plus all those that depend on the numerical value

What not to write

“The calculator is in degrees, so $(\sin x)'=\cos x$ still works.”

What to write

“All the trigonometric differentiation formulas assume RADIANS; in degrees, $(\sin x)'=\frac{\pi}{180}\cos x$ .”

Why: The formula comes from $\lim_{h\to 0}\frac{\sin h}{h}=1$ , which equals $\frac{\pi}{180}\approx 0.01745$ in degrees. A slope that comes out fifty-seven times too small, or the number $0.01745$ appearing on the screen, are the two telltale signs of this setting.

9. Differentiating a constant because it contains a symbol

1 to 2 marks, on the question that is supposed to be the easy one

What not to write

“In $f(x)=5x^{4}+\sqrt{2}\,x^{3}-\pi^{2}$ , the term $-\pi^{2}$ differentiates to $-2\pi$ .”

What to write

“ $\pi$ is a NUMBER, so $\pi^{2}$ is a constant and its derivative is $0$ ; $\sqrt{2}$ is a coefficient, so $\left(\sqrt{2}\,x^{3}\right)'=3\sqrt{2}\,x^{2}$ . In total $f'(x)=20x^{3}+3\sqrt{2}\,x^{2}$ .”

Why: The rule has no exceptions: anything that does not contain the variable is a constant, however complicated it looks. The same reflex makes students stumble on $\sqrt[3]{8x^{4}}$ , which only needs to be rewritten as $2x^{4/3}$ to fall back on the power rule and get $\frac{8}{3}x^{1/3}$ . Before differentiating, circle what contains $x$ : the rest is a number.

10. Believing that a tangent line touches the curve at only one point

the whole question when you are asked where the tangent line meets the curve again

What not to write

“This line crosses the curve again further on, so it is not the tangent line.”

What to write

“Being tangent is a LOCAL property, which describes the contact at one point. The tangent line to $f(x)=x^{3}-\frac{9}{2}x^{2}-12x+2$ at $(0,2)$ is $y=-12x+2$ , and it crosses the curve again at $(4.5, -52)$ .”

The tangent line at

(0, 2)

y=-12x+2

, touches the curve at that point and then CROSSES it again at

(4.5, -52)

. Being tangent is a local property.

Why: Solving $f(x)=-12x+2$ gives $x^{2}\left(x-\frac{9}{2}\right)=0$ : the DOUBLE root at $0$ is the signature of tangential contact, the simple root at $\frac{9}{2}$ is an ordinary intersection. It is the multiplicity of the root that distinguishes touching from crossing, never the number of intersection points. The reasoning works both ways and is used to find a point of tangency when the question gives the line instead of the point.

Which method to choose

Which rule, according to the last operation in the expression

Do not look at which symbols are present; look for the LAST operation you would perform to evaluate the expression at a number. That operation names the rule.

If the last operation is an addition or a subtraction → differentiate term by term; it is the only rule that distributes

Example: $\left(x^{3}+\sin x\right)'=3x^{2}+\cos x$
If the last operation is a multiplication of two factors that both depend on $x$ → product rule $u'v+uv'$

Example: $\left(x^{2}e^{x}\right)'=2xe^{x}+x^{2}e^{x}=xe^{x}(x+2)$

If one of the factors is a CONSTANT, it is not a product in the sense of the rule: the constant simply comes out.
If the last operation is a division → quotient rule $\frac{u'v-uv'}{v^{2}}$ , or rewriting as a negative power when the denominator is a monomial

Example: $\frac{3}{x^{2}}=3x^{-2}$ differentiates to $-6x^{-3}$ , without the quotient rule
If the last operation is applying a function to an expression, in other words a composition → chain rule: differentiate the outside while keeping the inside, then multiply by the inside derivative

Example: $\left(\sqrt{x^{2}+1}\right)'=\frac{2x}{2\sqrt{x^{2}+1}}=\frac{x}{\sqrt{x^{2}+1}}$
If the exponent contains the variable → logarithmic differentiation: take the logarithm of both sides, differentiate, then multiply by $y$

Example: $y=\left(\sin x\right)^{x}$ gives $\ln y=x\ln\sin x$
If the equation is not solved for $y$ → implicit differentiation: differentiate both sides with respect to $x$ , each $y$ producing a $y'$ , then solve for $y'$

Example: $x^{2}+y^{2}=25$ gives $2x+2yy'=0$ , so $y'=-\frac{x}{y}$

A composition with three layers is differentiated from the outside in, one layer at a time, and the result is a PRODUCT of three factors. Counting the layers before you start keeps you from forgetting one.

Should you simplify before differentiating

One minute of rewriting often saves five minutes of computation and removes half of the sign errors.

If the denominator is a monomial → rewrite with negative powers and differentiate term by term

Example: $\frac{x^{3}+2x}{x^{2}}=x+2x^{-1}$ , whose derivative is $1-2x^{-2}$
If the expression is a product of roots or of powers of the same base → combine the exponents before differentiating

Example: $\sqrt{x}\cdot x^{2}=x^{5/2}$ , whose derivative is $\frac{5}{2}x^{3/2}$
If the expression contains the logarithm of a product, a quotient or a power → expand with the properties of logarithms before differentiating

Example: $\ln\frac{x^{2}\sqrt{x+1}}{x-3}=2\ln x+\frac{1}{2}\ln(x+1)-\ln(x-3)$
If none of these forms → differentiate directly, and simplify only at the end if the question asks for it

Be careful, though: simplifying can change the DOMAIN. $\frac{x^{2}-4}{x-2}$ equals $x+2$ except at $2$ , where the original function is not defined and therefore has no derivative.

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.

The equation of a tangent line, written up for the marker

When to use it: As soon as the question asks for the tangent line, the normal line, or the linear approximation at a point.

1 Compute $f(a)$ and write it down: “ $f(2)=8$ , so the point of tangency is $A(2, 8)$ .”
2 Compute the derivative FUNCTION $f'(x)$ , naming the rule used.
3 Substitute: “ $f'(2)=12$ ”, a NUMBER, which is the slope.
4 Write the equation in point-slope form $y-f(a)=f'(a)(x-a)$ , then expand.
5 Check that the point of tangency satisfies the equation you found.

Concluding sentence

The tangent line to the graph of $f$ at the point with $x$ -coordinate $2$ has equation $y=12(x-2)+8$ , that is $y=12x-16$ .

The trap: Leaving an $x$ in the slope. A tangent line is a line: if the final equation contains an $x^{2}$ or a $\sin x$ , the substitution of $a$ into $f'$ was forgotten.

Marking: Usually 1 mark for $f(a)$ , 1 mark for $f'(x)$ , 1 mark for $f'(a)$ , 1 mark for the final equation in slope-intercept form.

The derivative from the definition, when the question requires it

When to use it: The question says “from the definition”, “using the limit” or “without using the rules”. Any answer obtained otherwise is worth zero.

1 Write the complete limit, with the $\lim_{h\to 0}$ , before any computation.
2 Expand $f(a+h)$ without simplifying too fast, and set up the difference $f(a+h)-f(a)$ .
3 Note the form $\frac{0}{0}$ and transform: factor out $h$ , or multiply by the conjugate if a square root appears.
4 Cancel the $h$ , saying explicitly that $h\neq 0$ , since $h$ TENDS to zero without reaching it.
5 Take the limit and conclude with a sentence.

Concluding sentence

The difference quotient equals $2a+h$ for every $h\neq 0$ , so $f'(a)=\lim_{h\to 0}(2a+h)=2a$ .

The trap: Substituting $h=0$ before simplifying, which gives $\frac{0}{0}$ and stalls the solution. The indeterminate form is not an obstacle, it is the signal to transform the expression.

Marking: The limit written at the start is often worth as much as the computation itself: it proves you understand what you are computing.

Check before you hand in

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

Check the derivative with a numerical difference quotient

Choose a simple $a$ , then compute $\frac{f(a+0.001)-f(a-0.001)}{0.002}$ on the calculator. The result must agree with $f'(a)$ to two or three decimal places.

For $f(x)=x^{2}\sin x$ at $a=1$ : the formula gives $2\sin 1+\cos 1\approx 2.223$ , and the numerical quotient shows $2.223$ ✓. The wrong derivative $2\cos 1\approx 1.081$ is ruled out in ten seconds.
Count the expected terms

A product of two factors gives TWO terms, a quotient gives two terms in the numerator, a chain with three layers gives a product of THREE factors. A result that does not have the right shape signals a skipped rule.

A derivative of $x^{2}e^{x}$ that fits in a single term is necessarily wrong; a derivative of $\sin\left(\sqrt{x^{2}+1}\right)$ with only two factors has lost a layer.
Check the sign at a point where the direction of change is obvious

Take a point where the function is visibly rising and check that the derivative is positive there, and conversely. This is what catches the reversed numerator in the quotient rule.

$\frac{1}{x}$ decreases on $\left(0,+\infty\right)$ , so its derivative must be negative: a $+\frac{1}{x^{2}}$ gives away the reversed formula.
Test the formula on a known special case

Substitute a value that reduces the expression to a case you already know, and compare.

The formula $\left(u^{n}\right)'=nu^{n-1}u'$ with $u=x$ must give back $nx^{n-1}$ , since $u'=1$ . If it does not, the formula you memorized is wrong.
Reread the verb in the question

“Differentiate”, “compute $f'(2)$ ”, “give the equation of the tangent line” and “show that $f$ is differentiable” ask for four answers of different kinds: a function, a number, the equation of a line, a comparison of two limits.

A flawless derivative in answer to “show that $f$ is differentiable at $0$ ” earns nothing: the question was about whether the limit of the difference quotient exists.

The typical problem, taken apart

Tangent line to an implicitly defined curve, as on the exam

The curve with equation $x^{2}+xy+y^{2}=7$ passes through the point $A(1, 2)$ .

Find the equation of the tangent line to this curve at the point $A$ .

For a single

x

-value, the ellipse

x^{2}+xy+y^{2}=7

gives TWO

y

-values: no function

y=f(x)

describes all of it, hence implicit differentiation.

Step 1

Preliminary check: $1^{2}+1\times 2+2^{2}=1+2+4=7$ ✓, the point $A$ does lie on the curve.

Why

Ten seconds that keep you from solving the wrong problem. A point off the curve makes everything that follows pointless, and the marker always starts there.

Step 2

Differentiate both sides with respect to $x$ , treating $y$ as an unknown function of $x$ : $\frac{d}{dx}\left(x^{2}\right)=2x$ , $\frac{d}{dx}\left(y^{2}\right)=2y\,y'$ , and $\frac{d}{dx}(xy)=1\cdot y+x\cdot y'$ by the product rule. The right-hand side is constant, so its derivative is zero. Hence $2x+y+xy'+2yy'=0$ .

Why

This is the step that carries the whole chapter: each differentiated $y$ produces a $y'$ through the chain rule, and the cross term $xy$ is a genuine product of two functions of $x$ . A solution that writes $2x+y+2yy'=0$ has treated $xy$ as a constant times $y$ .

Step 3

Collect the terms in $y'$ : $y'(x+2y)=-(2x+y)$ , hence $y'=-\frac{2x+y}{x+2y}$ .

Why

Solving for $y'$ BEFORE substituting gives a formula valid at every point of the curve, reusable if the question asks for a second tangent line. Substituting first would mean starting all over again.

Step 4

Substitute the coordinates of $A$ : $y'(1, 2)=-\frac{2(1)+2}{1+2(2)}=-\frac{4}{5}$ .

Why

In implicit differentiation, the slope depends on BOTH coordinates, not only on $x$ . That is the algebraic translation of the fact that a single $x$ -value carries two points of the curve, with two different tangent lines.

Step 5

Point-slope form: $y-2=-\frac{4}{5}(x-1)$ , that is $y=-\frac{4}{5}x+\frac{4}{5}+2=-\frac{4}{5}x+\frac{14}{5}$ .

Why

Expand to deliver the equation in slope-intercept form, which the marking scheme almost always asks for, but write the point-slope form first because it is the least risky: it contains the point and the slope, visible at a glance.

Step 6

Check: at $x=1$ , the tangent line gives $-\frac{4}{5}+\frac{14}{5}=\frac{10}{5}=2$ ✓, the $y$ -coordinate of $A$ . And the slope is negative, which the graph confirms: at the point $A$ , the curve goes down.

Why

Two independent checks, one algebraic and one graphical. They catch a sign error just as well as an arithmetic error in the expansion.

The conclusion, written out

The tangent line to the curve $x^{2}+xy+y^{2}=7$ at the point $A(1, 2)$ has equation $y=-\frac{4}{5}x+\frac{14}{5}$ .

The classic mistake on this problem: Writing $\frac{d}{dx}(xy)=y'$ , forgetting the product rule, which gives $2x+xy'+2yy'=0$ and then a slope of $-\frac{2}{5}$ . The cross term $xy$ is the only place in the computation where two rules combine, product AND chain, and that is where the question is won or lost.

Learn by heart

• $f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ , always of the form $\frac{0}{0}$ : transform, do not substitute.
• $(uv)'=u'v+uv'$ and $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}$ , the numerator of the quotient in THAT order.
• Chain: $\left(f(g(x))\right)'=f'(g(x))\,g'(x)$ . The second factor, the inside derivative, is the one people forget.
• $\left(e^{u}\right)'=u'e^{u}$ , $(\ln u)'=\frac{u'}{u}$ , $\left(u^{n}\right)'=nu^{n-1}u'$ : the same $u'$ everywhere.
• Variable exponent: logarithmic differentiation. $\left(a^{x}\right)'=a^{x}\ln a$ , never $x\,a^{x-1}$ .
• Implicit differentiation: each differentiated $y$ gives a $y'$ ; with no $y'$ in the equation, there is nothing to solve for.
• Tangent line: $y=f'(a)(x-a)+f(a)$ , where $f'(a)$ is a NUMBER. No tangent line contains an $x^{2}$ .
• Differentiable implies continuous; the converse is false, as $|x|$ at $0$ shows.
• Three ways to lose the derivative while staying continuous: a corner (two different finite slopes), a cusp (two infinite slopes of opposite signs), a vertical tangent (the same infinite slope on both sides). A discontinuity, for its part, always rules out the derivative.
• Anything that does not contain the variable is a constant: $\left(\pi^{2}\right)'=0$ and $\left(\sqrt{2}\,x^{3}\right)'=3\sqrt{2}\,x^{2}$ .

Frequently asked questions

How do you know which differentiation rule to use?

Find the last operation you would perform to evaluate the expression. An addition is differentiated term by term, a multiplication calls for the product rule, a division for the quotient rule, and a function applied to an expression calls for the chain rule. It is the structure of the expression that decides, never the symbols that appear in it.

Why do you have to multiply by the inside derivative?

Because the change in the inside expression is amplified or reduced before it enters the outside function. If the inside changes three times as fast, the composite function changes three times as fast too. That is exactly what the chain rule says, and forgetting it gives an answer that is off by a constant factor.

What is the difference between f prime of x and f prime of a?

The first is a function, which gives the slope at every point. The second is a number, the slope at the point where x equals a exactly. The equation of a tangent line uses the number, never the function, which is why a tangent line is always a straight line whose equation contains no square and no sine.

Is a continuous function always differentiable?

No, and it is a classic mistake. The absolute value function is continuous everywhere but has a corner at the origin, where the slopes on the left and on the right are minus one and plus one. The implication only works the other way: a function that is differentiable at a point is necessarily continuous there.

How do you differentiate when y is not isolated in the equation?

You use implicit differentiation. Differentiate both sides with respect to x, treating y as a function of x, which produces a factor y prime every time a y is differentiated, then collect the terms and solve for y prime. The slope you get then depends on both coordinates of the point.

How do you differentiate x to the power x?

Neither the power rule nor the exponential rule applies, because the base and the exponent both vary. Use logarithmic differentiation: take the logarithm of both sides, differentiate, then multiply by y. The result is x to the power x times the quantity natural log of x plus one.

Practise it

Corrected exercises: The derivative and its rules, 201-SN2-RE

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.

17 corrected exercises
170 points
225 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc1-derivative-rules. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: the derivative and its rules (201-SN2-RE)

The essentials

The derivative is a limit, and which one

The rules, in the order you look for them

What is NOT a rule

Reading non-differentiability on an exam graph

The rules in table form

The basic derivatives

The five operations, and what they produce

The mistakes that cost marks

1. Differentiating a product factor by factor

2. Losing the inside derivative in the chain rule

3. Reversing the order in the numerator of the quotient rule

4. Writing the tangent line with the derivative function instead of the derivative at the point

5. Believing that a continuous function is differentiable

6. Applying the power rule to a variable exponent

7. Forgetting the factor y′y'y′ in implicit differentiation

8. Differentiating a sine with an argument in degrees

9. Differentiating a constant because it contains a symbol

10. Believing that a tangent line touches the curve at only one point

Which method to choose

Which rule, according to the last operation in the expression

Should you simplify before differentiating

How the answer is expected to be written

The equation of a tangent line, written up for the marker

The derivative from the definition, when the question requires it

Check before you hand in

The typical problem, taken apart

Tangent line to an implicitly defined curve, as on the exam

Learn by heart

Frequently asked questions

How do you know which differentiation rule to use?

Why do you have to multiply by the inside derivative?

What is the difference between f prime of x and f prime of a?

Is a continuous function always differentiable?

How do you differentiate when y is not isolated in the equation?

How do you differentiate x to the power x?

Corrected exercises: The derivative and its rules, 201-SN2-RE

See also

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7. Forgetting the factor $y'$ in implicit differentiation