Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: derivatives of transcendental functions (201-SN2-RE)

Q: How do you differentiate a function like two to the power x?

You do not use the power rule: the exponent is the variable, not the base. The derivative of two to the power x is two to the power x multiplied by the natural logarithm of two. More generally, the derivative of a to the power u is a to the power u, times the natural logarithm of a, times the derivative of u. With base e, this factor equals one and disappears.

Q: When should you use logarithmic differentiation?

In two cases. First, when the base and the exponent both contain the variable, as in x to the power x: no other rule exists then. Second, for convenience, when the expression is a product or a quotient of at least three factors with roots or powers: the logarithm turns these products into sums, which are much easier to differentiate.

Q: Why is the derivative of arccosine negative?

Because the arccosine function decreases: it starts at pi when x equals minus one and goes down to zero when x equals one. A decreasing function has a negative derivative everywhere, hence the minus sign in front of one over the square root of one minus x squared. Arcsine, on the other hand, increases, so its derivative is positive.

Q: How do you find the domain of the derivative of a logarithm?

First write the domain of the function, by requiring the argument of the logarithm to be strictly positive, then differentiate. The formula you get can often still be computed outside that domain, but it has no meaning there. This is the most costly trap of the chapter: the domain is written before the derivative, never after.

Q: What does a growth rate of three percent per year mean?

It means that the derivative divided by the quantity equals zero point zero three, not that the derivative equals zero point zero three. It is a relative rate, which leads to exponential growth of the form y equals y zero times e to the power zero point zero three t. An absolute rate, on the other hand, would give a straight line.

This sheet does not rewrite the table of derivatives: you already have it in your notes, and copying a table neither loses nor earns marks. It deals with what really costs them, namely the factor $u'$ that disappears as soon as there are two layers, the power rule applied to $2^{x}$ , $\cos(x^{2})$ confused with $\cos^{2}x$ , the absolute value forgotten in $\arccos\left(\frac{3}{x}\right)$ , and the domain of the logarithm discovered after differentiating. It also connects with the tg, cotg and cosec notation of French-language Quebec textbooks, and names the trigonometric logarithms that simplify, such as $\ln(\sec x+\tan x)$ , whose derivative is $\sec x$ .

It is written for CEGEP students in the Science program (Sciences de la nature) taking Differential Calculus, 201-SN2-RE (formerly 201-NYA). Once you have read the sheet, the corrected exercise set of 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

A transcendental derivative is never read straight off the table: there is always a factor $u'$ behind it, and a domain to check before differentiating. Almost every mark lost in this chapter comes from one or the other.

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

Derivatives of trigonometric functions
Chain rule
Derivative of the exponential function
Derivative of the natural logarithm
Change of base
Logarithmic differentiation
Inverse trigonometric functions
Hyperbolic functions
Relative rate of change
Exponential growth
Logistic model
Newton's law of cooling

The essentials

The six trigonometric derivatives, and the sign that separates them

• $(\sin x)'=\cos x$ , $(\tan x)'=\sec^{2}x=1+\tan^{2}x$ , $(\sec x)'=\sec x\tan x$ .
• $(\cos x)'=-\sin x$ , $(\cot x)'=-\csc^{2}x$ , $(\csc x)'=-\csc x\cot x$ .
• Reading rule: the three functions whose name starts with “co” carry the MINUS sign. Nothing else needs memorizing about the signs.
• All of this assumes the argument is in RADIANS. In degrees, $(\sin x)'=\frac{\pi}{180}\cos x$ , which is never what the problem wants.
• Notation of French-language Quebec textbooks: $\tan$ is also written tg, $\cot$ cotg and $\csc$ cosec. Same functions, same derivatives: $(\mathrm{cotg}\,x)'=-\mathrm{cosec}^{2}x$ .

The slope of

y=\sin x

(green tangent lines) is read on

y=\cos x

(dashed):

1

0

0

\frac{\pi}{2}

where the sine levels off. In radians only.

The three “co” derivatives can also be recovered by differentiating the cofunction: $\cos x=\sin\left(\frac{\pi}{2}-x\right)$ produces the minus sign through the chain rule, since the inner derivative is $-1$ .

Exponential and logarithm: the base changes a FACTOR, never the form

• $(e^{x})'=e^{x}$ , and more generally $(e^{u})'=u'e^{u}$ .
• $(a^{x})'=a^{x}\ln a$ : the factor $\ln a$ is the only trace of the base.
• $(\ln x)'=\frac{1}{x}$ for $x>0$ , $(\ln u)'=\frac{u'}{u}$ , and $(\ln|x|)'=\frac{1}{x}$ for every $x\neq 0$ .
• $(\log_{a}x)'=\frac{1}{x\ln a}$ , which is just $\left(\frac{\ln x}{\ln a}\right)'$ with the constant $\frac{1}{\ln a}$ in front.
• Three logarithms that simplify, because $u'$ is a multiple of $u$ : $(\ln\sin x)'=\cot x$ , $(\ln\sec x)'=\tan x$ and $\left(\ln(\sec x+\tan x)\right)'=\sec x$ .

The derivative of

2^{x}

, dashed, is the same curve multiplied by

\ln 2

, so it runs BELOW it: only the base

e

gives a curve that coincides with its derivative.

The number $e$ is defined exactly by this: it is the only base whose tangent line at $0$ has slope $1$ , hence the only base for which the factor $\ln a$ equals $1$ and disappears.

The factor $u'$ : one per layer, no exceptions

• $\frac{d}{dx}\big[F(u(x))\big]=F'(u)\cdot u'$ . The table gives $F'$ ; you are the one who writes $u'$ .
• Three layers give three factors: $\left(\sin(e^{3x})\right)'=\cos(e^{3x})\cdot e^{3x}\cdot 3$ .
• Count check: as many multiplied factors as nested functions. Two nested functions and only two factors, one is missing.
• The only case where $u'=1$ is $u(x)=x$ : that is the line of the table, and it is rare on a test.
• A radical is rewritten as a power BEFORE differentiating: $\sqrt[3]{\tan^{2}u}=(\tan u)^{\frac{2}{3}}$ . The exponent $-\frac{1}{3}$ that comes out of the derivative says where it does not exist: where $\tan u=0$ .

Inverse and hyperbolic functions: the two tables where the sign is at stake

• $(\arcsin x)'=\frac{1}{\sqrt{1-x^{2}}}$ and $(\arccos x)'=\frac{-1}{\sqrt{1-x^{2}}}$ , both for $|x|<1$ STRICTLY.
• $(\arctan x)'=\frac{1}{1+x^{2}}$ on all of $\mathbb{R}$ , with no condition.
• $(\sinh x)'=\cosh x$ and $(\cosh x)'=\sinh x$ , WITHOUT a minus sign, with $\cosh^{2}x-\sinh^{2}x=1$ .
• Consequence to remember: $\cos$ and $\sin$ satisfy $y''=-y$ and oscillate; $\cosh$ and $\sinh$ satisfy $y''=y$ and shoot off to infinity.

\cosh

and

\sinh

differentiate into each other, without a minus sign:

y''=y

, and the curves shoot off to infinity. The cosine, with its minus sign, satisfies

y''=-y

and oscillates.

The domain of the derivative is the INTERIOR of the domain of the function. $\arcsin$ is defined at $1$ , its derivative is not: the tangent line there is vertical.

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 six trigonometric derivatives

Three rows without a sign, three rows with one. The reading rule fits in one word: the functions whose name starts with “co” carry the minus.

$f(x)$	Composite version	$f'(x)$
$\sin x$	$u'\cos u$	$\cos x$
Example: $\left(\sin(3x)\right)'=3\cos(3x)$ , which equals $3$ at $x=0$ , not $1$ : the factor $3$ is half the mark.
$\cos x$	$-u'\sin u$	$-\sin x$
Example: $\left(\cos(x^{2})\right)'=-2x\sin(x^{2})$ , which is about $-1.68$ at $x=1$ .
$\tan x$	$u'\sec^{2}u$	$\sec^{2}x=1+\tan^{2}x$
Example: At $x=\frac{\pi}{4}$ , $\sec^{2}x=2$ : the slope of the tangent line there is twice that of $y=x$ .
$\cot x$	$-u'\csc^{2}u$	$-\csc^{2}x$
Example: At $x=\frac{\pi}{4}$ , the slope is $-2$ , exactly the opposite of that of $\tan$ at the same point.
$\sec x$	$u'\sec u\tan u$	$\sec x\tan x$
Example: At $x=\frac{\pi}{3}$ , $\sec x=2$ and $\tan x=\sqrt{3}$ , so the slope is $2\sqrt{3}\approx 3.46$ .
$\csc x$	$-u'\csc u\cot u$	$-\csc x\cot x$
Example: At $x=\frac{\pi}{3}$ , the slope is $-\frac{2}{\sqrt{3}}\cdot\frac{1}{\sqrt{3}}\approx -0.667$ .

This whole table assumes the argument is in RADIANS. In degrees, $(\sin x)'=\frac{\pi}{180}\cos x\approx 0.01745\cos x$ , which the calculator displays without warning when it was left in DEG mode.

Exponentials, logarithms, inverse and hyperbolic functions

Four families, one difficulty per row: the factor the base leaves behind, and the domain that is written down BEFORE differentiating.

$f(x)$	Domain	$f'(x)$
$\mathrm{e}^{x}$	everywhere	$\mathrm{e}^{x}$
Example: $\left(\mathrm{e}^{3x}\right)'=3\mathrm{e}^{3x}$ : the $3$ comes from $u'$ , and it equals $3$ at $x=0$ .
$a^{x}$ , $a>0$	everywhere	$a^{x}\ln a$
Example: $\left(5^{x}\right)'=5^{x}\ln 5$ , so about $8.05$ at $x=1$ . The factor $\ln a$ is the only trace of the base.
$\ln x$	$x>0$	$\frac{1}{x}$
Example: $\left(\ln(x^{2}+1)\right)'=\frac{2x}{x^{2}+1}$ , which equals $1$ at $x=1$ : the $u'$ is in the numerator, never anywhere else.
$\log_{a}x$	$x>0$	$\frac{1}{x\ln a}$
Example: $\left(\log_{10}x\right)'=\frac{1}{x\ln 10}$ , that is about $0.0434$ at $x=10$ .
$\arcsin x$	$\|x\|<1$ strictly	$\frac{1}{\sqrt{1-x^{2}}}$
Example: At $x=0$ the slope is $1$ ; at $x=0.8$ it is already $\frac{1}{0.6}\approx 1.67$ , and it blows up at $\pm 1$ .
$\arccos x$	$\|x\|<1$ strictly	$-\frac{1}{\sqrt{1-x^{2}}}$
Example: Same absolute value as $\arcsin$ , opposite sign: the sum $\arcsin x+\arccos x$ equals $\frac{\pi}{2}$ for every $x$ , so its derivative is zero.
$\arctan x$	everywhere	$\frac{1}{1+x^{2}}$
Example: At $x=1$ , the slope is $0.5$ . No domain condition: it is the only inverse trigonometric function in this case.
$\sinh x$ and $\cosh x$	everywhere	$\cosh x$ and $\sinh x$
Example: No minus sign, unlike the circle: at $x=1$ , $\sinh'(1)=\cosh 1\approx 1.543$ . They satisfy $y''=y$ , while $\sin$ and $\cos$ satisfy $y''=-y$ .

The domain is written before differentiating, not after: $\left(\ln(x-3)\right)'=\frac{1}{x-3}$ is wrong as long as $x>3$ has not been stated, and the test counts that line separately.

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. The factor $u'$ forgotten as soon as there are two layers

1 mark each time, and often 4 or 5 per test

What not to write

“ $(\sin 3x)'=\cos 3x$ ”

What to write

“ $(\sin 3x)'=3\cos 3x$ : the table gives the outer derivative, the chain rule adds $u'=3$ .”

0

\sin 3x

rises three times as fast as

\sin x

: its tangent line has slope

3\cos 0=3

(green). Without the factor

u'=3

, you find slope

1

(red), that of

\sin x

Why: This is the most profitable mistake to fix in the whole chapter, because it repeats: a test that has it once has it everywhere. The check is mechanical: count the nested functions and count the factors.

2. Confusing $\cos(x^{2})$ and $\cos^{2}x$

the whole question, and textbooks often put both in the same sum

What not to write

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

What to write

“ $\cos^{2}x=(\cos x)^{2}$ : the square is OUTSIDE, so $\left(\cos^{2}x\right)'=2\cos x\cdot(-\sin x)=-\sin(2x)$ . It is $\cos(x^{2})$ that gives $-2x\sin(x^{2})$ .”

\cos^{2}x

(orange) stays between

0

and

1

and keeps a fixed period;

\cos(x^{2})

(blue) oscillates faster and faster. Two different functions, two different derivatives.

Why: The notation $\cos^{2}x$ hides the parentheses that say which function is on the outside. Rewrite it before differentiating: the outer function of $(\cos x)^{2}$ is the square, that of $\cos(x^{2})$ is the cosine.

3. Differentiating $2^{x}$ with the power rule

the whole question

What not to write

“ $(2^{x})'=x\,2^{x-1}$ ”

What to write

“ $(2^{x})'=2^{x}\ln 2$ : the exponent is the VARIABLE, so this is not a power but an exponential.”

x=1

, the tangent line to

y=2^{x}

has slope

2\ln 2\approx 1.39

(green). The power rule would give

1\cdot 2^{0}=1

(red): a line that cuts through the curve.

Why: The rule $(x^{n})'=nx^{n-1}$ assumes the exponent is CONSTANT. Here it is the base that is constant: the two rules apply to different objects, and only the position of the $x$ tells them apart.

4. Differentiating the logarithm before writing its domain

2 marks, plus the increasing/decreasing question that follows

What not to write

“ $f(x)=\ln(x^{2}-4)$ so $f'(x)=\frac{2x}{x^{2}-4}$ , and $f'(0)=0$ .”

What to write

“ $f$ is defined only for $x^{2}-4>0$ , that is $|x|>2$ . The formula $f'(x)=\frac{2x}{x^{2}-4}$ holds only there, and $f'(0)$ does not exist.”

\ln(x^{2}-4)

exists only for

|x|>2

: two branches, nothing between

-2

and

2

. The formula

\frac{2x}{x^{2}-4}

can be computed at

0

, but there is no curve at

0

, so no slope.

Why: The formula for the derivative can still be computed where the function does not exist: it does not complain. This is the one chapter where the domain must be written BEFORE differentiating, and the marker looks for that line.

5. Believing that $\ln(5x)$ and $\ln x$ have different derivatives

1 mark

What not to write

“ $(\ln 5x)'=\frac{1}{5x}$ ”

What to write

“ $(\ln 5x)'=\frac{5}{5x}=\frac{1}{x}$ , the same derivative as $\ln x$ , which can also be seen from $\ln 5x=\ln 5+\ln x$ .”

\ln 5x=\ln 5+\ln x

: the curve is that of

\ln x

raised by

\ln 5

. The tangent lines at

x=1

are parallel, with slope

1

(\ln 5x)'=\frac{1}{x}

, not

\frac{1}{5x}

Why: A constant factor inside a logarithm becomes an ADDITIVE constant, so it vanishes when you differentiate. Simplifying the logarithm before differentiating avoids the mistake and saves time.

6. Applying a power rule to $x^{x}$ , where base AND exponent vary

the whole question

What not to write

“ $(x^{x})'=x\cdot x^{x-1}=x^{x}$ ”

What to write

“No rule applies. Set $y=x^{x}$ , so $\ln y=x\ln x$ , and differentiate: $\frac{y'}{y}=\ln x+1$ , hence $y'=x^{x}(\ln x+1)$ .”

x=2

, the tangent line to

y=x^{x}

has slope

4(\ln 2+1)\approx 6.77

(green). The power rule gives

4

(red): a line that is far too flat.

Why: The power rule freezes the exponent, the exponential rule freezes the base: when both move, only logarithmic differentiation is left. The wrong answer $x^{x}$ happens to be right at $x=1$ , where both give $1$ , which is why a check at $x=1$ proves nothing; at $x=e$ it gives $e^{e}$ instead of $2e^{e}$ .

7. Stopping logarithmic differentiation at $\frac{y'}{y}$

2 marks, half the question

What not to write

“ $\frac{y'}{y}=\frac{3}{x}+\frac{1}{2x+1}$ , so $y'=\frac{3}{x}+\frac{1}{2x+1}$ .”

What to write

“ $y'=y\left(\frac{3}{x}+\frac{1}{2x+1}\right)$ , where $y$ is replaced by the original expression.”

Why: $\frac{y'}{y}$ is the RELATIVE rate of change, not the derivative. The final multiplication is the step everyone skips because the interesting computation is already done.

8. The minus sign of $\arccos$ , and the extra one on $\cosh$

1 mark each, and the increasing/decreasing behaviour of the whole analysis

What not to write

“ $(\arccos x)'=\frac{1}{\sqrt{1-x^{2}}}$ ” and “ $(\cosh x)'=-\sinh x$ ”

What to write

“ $(\arccos x)'=\frac{-1}{\sqrt{1-x^{2}}}$ , because $\arccos$ is DECREASING” and “ $(\cosh x)'=+\sinh x$ , there is no minus sign for the hyperbolic functions.”

\arcsin

rises and

\arccos

falls: that is all the minus sign is about. At both edges

x=\pm 1

the tangent lines are vertical, so the derivative does not exist there.

Why: The reflex “co means minus” comes from circular trigonometry and carries over neither to the inverse functions nor to the hyperbolic ones. The check is visual: a decreasing function has a negative derivative, an increasing function a positive one.

9. Simplifying $\sqrt{x^{2}}$ to $x$ in the derivative of $\arccos\left(\frac{3}{x}\right)$

1 mark, and the wrong increasing/decreasing behaviour on the whole left half

What not to write

“ $\left(\arccos\frac{3}{x}\right)'=\frac{3}{x\sqrt{x^{2}-9}}$ ”

What to write

“ $\sqrt{1-\frac{9}{x^{2}}}=\frac{\sqrt{x^{2}-9}}{|x|}$ , so the derivative is $\frac{3}{|x|\sqrt{x^{2}-9}}$ , positive on the whole domain $|x|\ge 3$ .”

Why: $\sqrt{x^{2}}=|x|$ , never $x$ : the square root is positive. The wrong form gives $-0.15$ at $x=-5$ , a negative slope for a function that increases, which the check by increasing/decreasing behaviour catches.

10. Confusing the absolute rate of change with the relative rate

the whole question, since the model is wrong from the first line

What not to write

“The population grows by $3\%$ per year, so $P'=0.03$ .”

What to write

“A percentage is a RELATIVE rate: $\frac{P'}{P}=0.03$ , so $P'=0.03P$ and $P(t)=P_{0}e^{0.03t}$ .”

Starting from

P_{0}=100

: with

P'=0.03P

(green), the population more than quadruples in

50

years; with

P'=0.03

(red), it gains

1.5

individuals. A percentage is a RELATIVE rate.

Why: $P'=0.03$ describes LINEAR growth of three hundredths of an individual per year. The word “percentage” in a problem always means $\frac{P'}{P}$ , and that is what produces the exponential.

Which method to choose

Which rule, depending on the FORM of the expression

Look at where the $x$ is, in the base or in the exponent, before looking at which function it is. The position of the $x$ decides, not the chapter.

Three neighbouring curves and three different rules: constant exponent, constant base, then both variable. It is the position of the

x

in the expression that decides, never the shape of the graph.

If variable base, CONSTANT exponent, like $x^{5}$ or $x^{-1/2}$ → power rule $(x^{n})'=nx^{n-1}$

Example: $(x^{7})'=7x^{6}$
If CONSTANT base, variable exponent, like $2^{x}$ or $e^{3x}$ → exponential rule $(a^{u})'=a^{u}\ln a\cdot u'$

Example: $(5^{x^{2}})'=5^{x^{2}}\ln 5\cdot 2x$

With $a=e$ the factor $\ln a$ equals $1$ , which explains why $e$ is the working base.
If base AND exponent both variable, like $x^{x}$ or $(\sin x)^{x}$ → logarithmic differentiation, no other rule exists

Example: $y=x^{\sin x}$ , $\ln y=\sin x\ln x$
If product or quotient of at least THREE factors, with exponents or roots → logarithmic differentiation too, for convenience: the logarithm turns products into sums

Example: $y=\frac{x^{3}\sqrt{2x+1}}{(x^{2}+1)^{4}}$ , taken apart below

The direct computation is possible, but it stacks two quotient rules and a product rule: that is where sign errors are born.
If a transcendental function applied to something other than $x$ alone → the line of the table, multiplied by $u'$

Example: $(\ln(3x^{2}+1))'=\frac{6x}{3x^{2}+1}$

If no branch applies, the expression is a sum: differentiate term by term and you land on one of the branches for each term.

Which exponential model, depending on the wording of the problem

The Part B problems all look alike. It is a precise phrase in the problem that points to the differential equation, and therefore to the model.

If “grows by $k$ percent per year”, “doubles every $T$ ”, “half-life” → $y'=ky$ , so $y=y_{0}e^{kt}$ , with a constant RELATIVE rate

Example: half-life $T$ gives $k=\frac{\ln 2}{T}$ , up to sign
If “tends to room temperature”, “approaches the value $a$ ” → Newton's law of cooling $y'=k(y-a)$ , so $y=a+(y_{0}-a)e^{kt}$

Example: a coffee at $80$ degrees in a room at $20$ degrees

It is the GAP from $a$ that decays exponentially, not the temperature itself: setting $u=y-a$ brings you back to the previous case.
If “carrying capacity”, “ceiling”, “the environment cannot feed more than $K$ individuals” → logistic model $y'=\frac{r}{K}y(K-y)$ , S-shaped curve

Example: maximum growth rate $\frac{rK}{4}$ , reached when $y=\frac{K}{2}$
If the problem gives two measurements and asks for the rate → do not guess the model, identify it: two points are enough to find $y_{0}$ and $k$ in $y=y_{0}e^{kt}$

Example: $y(0)=1200$ and $y(10)=1620$ give $k=\frac{1}{10}\ln\frac{1620}{1200}$

In every case, write the differential equation BEFORE its solution: it carries the modelling marks, since solving it is taken for granted at this level.

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.

Logarithmic differentiation, written out for the marker

When to use it: Base and exponent both variable, or a product or quotient of at least three factors with exponents.

1 State the function and announce the method: “Let $y=\frac{x^{3}\sqrt{2x+1}}{(x^{2}+1)^{4}}$ , defined and strictly positive for $x>0$ .”
2 Take the logarithm of both sides, justifying that you are allowed to: “For $x>0$ , $y>0$ , so $\ln y$ exists.” If the sign is not guaranteed, write $\ln|y|$ , whose derivative is the same.
3 Break into a sum BEFORE differentiating; this is where the method pays off: “ $\ln y=3\ln x+\frac{1}{2}\ln(2x+1)-4\ln(x^{2}+1)$ .”
4 Differentiate both sides with respect to $x$ , remembering that the left side gives $\frac{y'}{y}$ and not $y'$ .
5 Multiply by $y$ and replace $y$ by its original expression: this is the step that concludes, and the one people skip.

Concluding sentence

For every $x>0$ , $y'=\frac{x^{3}\sqrt{2x+1}}{(x^{2}+1)^{4}}\left(\frac{3}{x}+\frac{1}{2x+1}-\frac{8x}{x^{2}+1}\right)$ .

The trap: Writing $\ln y=\ldots$ then differentiating the left side as $\frac{1}{y}$ instead of $\frac{y'}{y}$ . The $y$ is a function of $x$ , so the chain rule applies on this side of the equals sign too.

Marking: Usually 1 mark for the justified switch to the logarithm, 2 marks for the decomposition into a sum and its differentiation, 1 mark for the final multiplication by $y$ .

Solving an exponential decay problem

When to use it: Half-life, carbon-14 dating, elimination of a drug, cooling: whenever a quantity decreases by a constant percentage.

1 Name the function and its unit: “Let $N(t)$ be the remaining mass, in grams, after $t$ years.”
2 Write the differential equation and justify it from the problem: “The decay is proportional to the quantity present, so $N'=-kN$ with $k>0$ .”
3 Give the general solution $N(t)=N_{0}e^{-kt}$ , then use the half-life data to find $k$ : $N(T)=\frac{N_{0}}{2}$ gives $e^{-kT}=\frac{1}{2}$ , so $k=\frac{\ln 2}{T}$ .
4 Answer the question asked by isolating the unknown with a logarithm, never by trial and error on the calculator.
5 Check the order of magnitude: if the sample has lost a little less than half, the age must be a little less than one half-life.

Concluding sentence

With $T=5730$ years, $k=\frac{\ln 2}{5730}\approx 1.21\times 10^{-4}$ per year; since the sample contains $72\%$ of the initial carbon 14, its age is $t=\frac{-\ln 0.72}{k}\approx 2716$ years.

The trap: Forgetting the minus sign in $N'=-kN$ , then finding a negative age and handing it in as is. A decreasing quantity has a negative derivative: the sign is decided before the computation, not after.

Marking: 1 mark for the differential equation stated and justified, 1 mark for finding $k$ from the half-life, 1 mark for the numerical answer with its unit.

Check before you hand in

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

Have the calculator recompute the derivative at one point

On a TI, MATH then nDeriv $(f(x),x,a)$ gives the numerical value of $f'(a)$ ; on a Casio, the RUN menu and the $\frac{d}{dx}$ command. Compare with the value your expression gives at the same point.

For $f(x)=\sin 3x$ at $x=1$ , the calculator shows $-2.97$ ; an answer $\cos 3x$ would give $-0.99$ , exactly a third of that: this is the missing factor $u'$ showing up.
Count the factors before turning the page

Count the nested functions in the problem, then count the multiplied factors in your answer. The two numbers must match.

$\sin(e^{3x})$ contains three nested functions, so the derivative must contain three factors: $\cos(e^{3x})\cdot e^{3x}\cdot 3$ . Only two factors means a skipped line.
Compare the domain of the derivative with that of the function

The derivative cannot exist where the function does not exist. Reread the domain written at the start, and check that the formula you found is not used anywhere else.

$\frac{2x}{x^{2}-4}$ can be computed at $x=0$ even though $\ln(x^{2}-4)$ is not defined there: the computation never complains, it is up to you to do so.
Check the sign against increasing/decreasing behaviour

An increasing function has a positive derivative on its domain, a decreasing function a negative one. Test the formula at a convenient point.

$\arccos$ decreases from $\pi$ to $0$ , so a derivative found positive is wrong; $e^{-x}$ decreases, so $-e^{-x}$ is indeed negative, while $+e^{-x}$ betrays a lost sign.
Check RADIAN mode before any trigonometric derivative

Every formula in the chapter assumes the argument is in radians. A three-second check: $\sin(0.01)$ must display $0.00999\ldots$ , not $0.000174$ .

In degree mode, $(\sin x)'$ equals $\frac{\pi}{180}\cos x$ : a numerical check would then give $0.0175$ times the expected value, which looks like a computation error and is not one.

The typical problem, taken apart

Differentiating a quotient of three factors, then evaluating the slope at $x=1$

Let $y=\dfrac{x^{3}\sqrt{2x+1}}{(x^{2}+1)^{4}}$ for $x>0$ .

Compute $y'$ by logarithmic differentiation, then give the slope of the tangent line at the point with $x$ -coordinate $1$ .

Step 1

For $x>0$ , $2x+1>0$ and $x^{3}>0$ , so $y>0$ and $\ln y$ exists. Write $\ln y=\ln\left(x^{3}\right)+\ln\left(\sqrt{2x+1}\right)-\ln\left((x^{2}+1)^{4}\right)$ .

Why

Justifying positivity before taking the logarithm is the first mark on the marking scheme: without it, writing $\ln y$ makes no sense and the marker notes it.

Step 2

The exponents come down in front: $\ln y=3\ln x+\dfrac{1}{2}\ln(2x+1)-4\ln(x^{2}+1)$ .

Why

This is where the method saves time: the root becomes a half and the fourth power a coefficient. The direct computation would still have a quotient to differentiate.

Step 3

Differentiate both sides: $\dfrac{y'}{y}=\dfrac{3}{x}+\dfrac{1}{2}\cdot\dfrac{2}{2x+1}-4\cdot\dfrac{2x}{x^{2}+1}=\dfrac{3}{x}+\dfrac{1}{2x+1}-\dfrac{8x}{x^{2}+1}$ .

Why

The left side gives $\frac{y'}{y}$ and not $\frac{1}{y}$ : $y$ is a function of $x$ , so the chain rule applies on both sides of the equals sign. Each term on the right is a $\frac{u'}{u}$ .

Step 4

Multiply by $y$ : $y'=\dfrac{x^{3}\sqrt{2x+1}}{(x^{2}+1)^{4}}\left(\dfrac{3}{x}+\dfrac{1}{2x+1}-\dfrac{8x}{x^{2}+1}\right)$ .

Why

The step skipped on half of the tests. As long as you have not multiplied by $y$ , you hold the RELATIVE rate, which is not the derivative asked for.

Step 5

At $x=1$ : $y(1)=\dfrac{1\cdot\sqrt{3}}{2^{4}}=\dfrac{\sqrt{3}}{16}$ and the parenthesis equals $3+\dfrac{1}{3}-4=-\dfrac{2}{3}$ , so $y'(1)=\dfrac{\sqrt{3}}{16}\cdot\left(-\dfrac{2}{3}\right)=-\dfrac{\sqrt{3}}{24}$ .

Why

Evaluate AFTER differentiating symbolically. Substituting $x=1$ as early as the third step would lose the general formula, which the next question almost always asks for.

Step 6

Check: $-\dfrac{\sqrt{3}}{24}\approx -0.0722$ , and the calculator gives $\text{nDeriv}\approx -0.0722$ . The negative sign is consistent, since the parenthesis is negative and $y$ is positive.

Why

A numerical check at one point catches both a sign error and a forgotten term, for the price of ten seconds on the calculator.

The conclusion, written out

For every $x>0$ , $y'=\dfrac{x^{3}\sqrt{2x+1}}{(x^{2}+1)^{4}}\left(\dfrac{3}{x}+\dfrac{1}{2x+1}-\dfrac{8x}{x^{2}+1}\right)$ , and the tangent line at the point with $x$ -coordinate $1$ has slope $-\dfrac{\sqrt{3}}{24}\approx -0.072$ .

The classic mistake on this problem: Writing $\ln y=3\ln x+\frac{1}{2}\ln(2x+1)-4\ln(x^{2}+1)$ then concluding $y'=\frac{3}{x}+\frac{1}{2x+1}-\frac{8x}{x^{2}+1}$ : the final multiplication by $y$ is missing, and the result would be $-\frac{2}{3}$ instead of $-\frac{\sqrt{3}}{24}$ , that is about nine times too large.

Learn by heart

• The three trigonometric derivatives starting with “co” carry the MINUS sign, the other three do not.
• $(a^{x})'=a^{x}\ln a$ , and $(x^{n})'=nx^{n-1}$ : it is the position of the $x$ , base or exponent, that picks the rule.
• $(\ln u)'=\frac{u'}{u}$ , but the DOMAIN $u>0$ is written before differentiating, not after.
• $\cos^{2}x=(\cos x)^{2}$ , square outside; $\cos(x^{2})$ , square inside. Rewrite before differentiating.
• As many factors in the derivative as nested functions in the problem.
• Logarithmic differentiation: $\ln$ , break into a sum, differentiate, THEN multiply by $y$ .
• $(\arcsin x)'=\frac{1}{\sqrt{1-x^{2}}}$ and $(\arccos x)'=-\frac{1}{\sqrt{1-x^{2}}}$ , for $|x|<1$ strictly.
• $(\cosh x)'=\sinh x$ and $(\sinh x)'=\cosh x$ : no minus sign for the hyperbolic functions.
• A percentage in a problem means $\frac{y'}{y}$ , the RELATIVE rate, hence an exponential.

Frequently asked questions

How do you differentiate a function like two to the power x?

You do not use the power rule: the exponent is the variable, not the base. The derivative of two to the power x is two to the power x multiplied by the natural logarithm of two. More generally, the derivative of a to the power u is a to the power u, times the natural logarithm of a, times the derivative of u. With base e, this factor equals one and disappears.

When should you use logarithmic differentiation?

In two cases. First, when the base and the exponent both contain the variable, as in x to the power x: no other rule exists then. Second, for convenience, when the expression is a product or a quotient of at least three factors with roots or powers: the logarithm turns these products into sums, which are much easier to differentiate.

Why is the derivative of arccosine negative?

Because the arccosine function decreases: it starts at pi when x equals minus one and goes down to zero when x equals one. A decreasing function has a negative derivative everywhere, hence the minus sign in front of one over the square root of one minus x squared. Arcsine, on the other hand, increases, so its derivative is positive.

How do you find the domain of the derivative of a logarithm?

First write the domain of the function, by requiring the argument of the logarithm to be strictly positive, then differentiate. The formula you get can often still be computed outside that domain, but it has no meaning there. This is the most costly trap of the chapter: the domain is written before the derivative, never after.

What does a growth rate of three percent per year mean?

It means that the derivative divided by the quantity equals zero point zero three, not that the derivative equals zero point zero three. It is a relative rate, which leads to exponential growth of the form y equals y zero times e to the power zero point zero three t. An absolute rate, on the other hand, would give a straight line.

Practise it

Corrected exercises: Derivatives of trigonometric, exponential and logarithmic functions, 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.

16 corrected exercises
160 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet The derivative and its rules Next sheet Implicit differentiation

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

Looking for a 201-SN2-RE tutor in Montreal?

Get in touch for a first session. We work on transcendental derivatives at the level of real CEGEP tests, from the trigonometric table all the way to logarithmic differentiation and exponential models.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: derivatives of transcendental functions (201-SN2-RE)

The essentials

The six trigonometric derivatives, and the sign that separates them

Exponential and logarithm: the base changes a FACTOR, never the form

The factor u′u'u′: one per layer, no exceptions

Inverse and hyperbolic functions: the two tables where the sign is at stake

The rules in table form

The six trigonometric derivatives

Exponentials, logarithms, inverse and hyperbolic functions

The mistakes that cost marks

1. The factor u′u'u′ forgotten as soon as there are two layers

2. Confusing cos⁡(x2)\cos(x^{2})cos(x2) and cos⁡2x\cos^{2}xcos2x

3. Differentiating 2x2^{x}2x with the power rule

4. Differentiating the logarithm before writing its domain

5. Believing that ln⁡(5x)\ln(5x)ln(5x) and ln⁡x\ln xlnx have different derivatives

6. Applying a power rule to xxx^{x}xx, where base AND exponent vary

7. Stopping logarithmic differentiation at y′y\frac{y'}{y}yy′​

8. The minus sign of arccos⁡\arccosarccos, and the extra one on cosh⁡\coshcosh

9. Simplifying x2\sqrt{x^{2}}x2​ to xxx in the derivative of arccos⁡(3x)\arccos\left(\frac{3}{x}\right)arccos(x3​)

10. Confusing the absolute rate of change with the relative rate

Which method to choose

Which rule, depending on the FORM of the expression

Which exponential model, depending on the wording of the problem

How the answer is expected to be written

Logarithmic differentiation, written out for the marker

Solving an exponential decay problem

Check before you hand in

The typical problem, taken apart

Differentiating a quotient of three factors, then evaluating the slope at x=1x=1x=1

Learn by heart

Frequently asked questions

How do you differentiate a function like two to the power x?

When should you use logarithmic differentiation?

Why is the derivative of arccosine negative?

How do you find the domain of the derivative of a logarithm?

What does a growth rate of three percent per year mean?

Corrected exercises: Derivatives of trigonometric, exponential and logarithmic functions, 201-SN2-RE

See also

Looking for a 201-SN2-RE tutor in Montreal?

The factor $u'$ : one per layer, no exceptions

1. The factor $u'$ forgotten as soon as there are two layers

2. Confusing $\cos(x^{2})$ and $\cos^{2}x$

3. Differentiating $2^{x}$ with the power rule

5. Believing that $\ln(5x)$ and $\ln x$ have different derivatives

6. Applying a power rule to $x^{x}$ , where base AND exponent vary

7. Stopping logarithmic differentiation at $\frac{y'}{y}$

8. The minus sign of $\arccos$ , and the extra one on $\cosh$

9. Simplifying $\sqrt{x^{2}}$ to $x$ in the derivative of $\arccos\left(\frac{3}{x}\right)$

Differentiating a quotient of three factors, then evaluating the slope at $x=1$