MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: derivatives of logarithms, logarithmic differentiation, growth and decay (MATH 140)

This sheet is not a summary of sections 3.6 and 3.8 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on the derivatives of logarithms and on growth and decay in MATH 140 at McGill University, and which precise gesture avoids each loss.

The chapter looks like three short formulas, and that is the danger: most errors are not in the derivative of $\ln$ but in what comes before it, the domain and the laws of logarithms, and after it, the multiplication by $y$ and the meaning of $k$ . Every value below is exact and done by hand, as on the exam, since no calculator is allowed.

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 logarithm is a translator: it turns products into sums, quotients into differences and powers into products, so rewrite with $\ln$ BEFORE differentiating, keeping $\ln$ on positive inputs only ( $\ln|x|$ , $\ln|y|$ ); and in $y' = ky$ , the constant $k$ is the RELATIVE rate $\frac{y'}{y}$ , never the percentage gained per period.

This chapter is part of MATH 140, Calculus 1 (McGill)

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

Derivative of the natural logarithm
Logarithms in other bases
Laws of logarithms
Logarithmic differentiation
Relative rate of change
Limit definition of e
Exponential growth and decay
Half-life
Doubling time
Newton's law of cooling
Continuous compounding

The essentials

Three derivatives, and the domain that comes first

• $\frac{d}{dx}\ln x = \frac{1}{x}$ for $x > 0$ . With the chain rule, inner function $u$ named: $\frac{d}{dx}\ln u = \frac{u'}{u}$ wherever $u > 0$ .
• $\frac{d}{dx}\ln|x| = \frac{1}{x}$ for EVERY $x \ne 0$ , and $\frac{d}{dx}\ln|u| = \frac{u'}{u}$ wherever $u \ne 0$ : the absolute value extends the formula, it does not change it.
• $\log_b x = \frac{\ln x}{\ln b}$ , so $\frac{d}{dx}\log_b x = \frac{1}{x\ln b}$ : the base survives only in the constant $\ln b$ .
• Laws, for $a, b > 0$ : $\ln(ab) = \ln a + \ln b$ , $\ln\frac{a}{b} = \ln a - \ln b$ , $\ln(a^r) = r\ln a$ . Nothing splits $\ln(a + b)$ , and $(\ln a)^2 \ne 2\ln a$ .
• The domain is read on the FUNCTION, before differentiating: $\ln(\ln x)$ needs $x > 1$ , $\ln(x - 3)$ needs $x > 3$ . A derivative formula says nothing outside that set.

y = \ln(5x)

y = \ln x

shifted up by

\ln 5

: at

x = 1

the two dashed tangents are parallel, both of slope

1

, not

5

Since $\ln(cx) = \ln c + \ln x$ , multiplying inside the logarithm by a constant shifts the graph vertically and changes no slope. This single fact kills the most frequent error of the chapter.

Relative rates: logarithmic differentiation and y' = ky are the same idea

• The relative rate of $y$ is $\frac{y'}{y} = \frac{d}{dx}\ln|y|$ . Taking $\ln$ turns a product into a sum, so relative rates ADD under a product and SUBTRACT under a quotient.
• Logarithmic differentiation: take $\ln|y|$ , expand with the laws, differentiate (the left side gives $\frac{y'}{y}$ ), multiply by $y$ .
• Variable in the base AND in the exponent: $\ln y = (\text{exponent})\ln(\text{base})$ is the only way in. $\frac{d}{dx}x^x = x^x(\ln x + 1)$ .
• Growth and decay: $y' = ky$ holds exactly when $y = y_0e^{kt}$ (given in the course). $k = \frac{y'}{y}$ is the constant relative rate, in inverse time units.
• Doubling time $\frac{\ln 2}{k}$ ( $k > 0$ ), half-life $\frac{\ln 2}{|k|}$ ( $k < 0$ ). Cooling: it is the GAP $T - T_s$ that decays, $T = T_s + (T_0 - T_s)e^{kt}$ .

Keep $k$ exact, as $\ln 2$ over something: every later answer then reduces to powers of $2$ by hand, while $k = 0.139$ makes the exam impossible without a calculator.

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 rule for which function: look where the x is

Read a line as: a function of the first column, recognized by the second, has the derivative of the third. The red lines are not answers: one is a rule that does not exist, one is a form no single rule covers.

Function	Where x sits	Derivative
$x^n$	base only	$nx^{n-1}$
Example: $\frac{d}{dx}x^e = ex^{e-1}$ , which equals $e^e$ at $x = e$ .
$b^x$	exponent only	$b^x\ln b$
Example: $\frac{d}{dx}2^x = 2^x\ln 2$ , equal to $8\ln 2$ at $x = 3$ .
$e^e$ , $\ln 5$	nowhere	$0$
Example: $e^e$ is a number, a little above $15$ : its derivative is $0$ .
$\log_b u$	inside the log	$\frac{u'}{u\ln b}$
Example: $\frac{d}{dx}\log_2(3x - 1) = \frac{3}{(3x - 1)\ln 2}$ , equal to $\frac{3}{8\ln 2}$ at $x = 3$ .
$f(x)^{g(x)}$	base and exponent	no single rule no rule applies
Example: $\frac{d}{dx}x^x = x^x(\ln x + 1)$ , equal to $2e^e$ at $x = e$ , not $e^e$ . What to do: Take $\ln y = g(x)\ln f(x)$ , differentiate with the product rule, multiply by $y$ .
$\ln(u + v)$	inside a sum	$\ln u + \ln v$ rule that does not exist
Example: $\ln(1 + 1) = \ln 2$ , while $\ln 1 + \ln 1 = 0$ . What to do: Differentiate $\ln(u + v)$ as it is: $\frac{u' + v'}{u + v}$ . The laws only split products, quotients and powers.

Two questions before differentiating any power: does the base contain $x$ ? Does the exponent? One yes picks the power rule or the exponential rule; two yes answers mean logarithmic differentiation.

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 ln(5x) as 5 over x

1 to 2 marks, and every tangent line computed from it

What not to write

“ $\frac{d}{dx}\ln(5x) = \frac{5}{x}$ .”

What to write

“ $\ln(5x) = \ln 5 + \ln x$ , so $\frac{d}{dx}\ln(5x) = \frac{1}{x}$ .” (Or by the chain rule: $\frac{5}{5x} = \frac{1}{x}$ .)

Why: The chain rule divides by the WHOLE inside, $5x$ , not by $x$ . The figure of the first box shows it: $\ln(5x)$ is $\ln x$ shifted up, so no slope changes.

2. Forgetting the chain rule inside ln

1 mark

What not to write

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

What to write

“With inner function $u = x^2 + 1$ , $u' = 2x$ : $\frac{d}{dx}\ln(x^2 + 1) = \frac{2x}{x^2 + 1}$ .”

Why: $\frac{1}{x}$ is the derivative of $\ln$ at the point $x$ only; at the point $u(x)$ it must be multiplied by $u'(x)$ . Naming the inner function on the copy prevents the omission.

3. Differentiating log base b without ln b

1 mark

What not to write

“ $\frac{d}{dx}\log_3 x = \frac{1}{3x}$ ” or “ $= \frac{1}{x}$ .”

What to write

“ $\log_3 x = \frac{\ln x}{\ln 3}$ , so $\frac{d}{dx}\log_3 x = \frac{1}{x\ln 3}$ .”

Why: Only the natural logarithm has derivative exactly $\frac{1}{x}$ ; every other base is $\ln$ divided by a constant, and the constant stays.

4. Confusing the square of ln with the ln of a square

2 marks, and the whole logarithmic differentiation of

x^{\ln x}

What not to write

“ $(\ln x)^2 = 2\ln x$ , so its derivative is $\frac{2}{x}$ .”

What to write

“By the chain rule, outer function the square: $\frac{d}{dx}(\ln x)^2 = 2\ln x \cdot \frac{1}{x} = \frac{2\ln x}{x}$ .”

y = (\ln x)^2

and

y = 2\ln x

are different curves that cross only at

x = 1

and

x = e^2

: at

x = 1

the first is flat, the second has slope

2

Why: $\ln(a^r) = r\ln a$ moves an exponent of the ARGUMENT; $(\ln x)^2$ squares the OUTPUT. The two curves meet only at $x = 1$ and $x = e^2$ .

5. Evaluating a derivative outside the domain

the whole question

What not to write

“ $f(x) = \ln(x - 3)$ , $f'(x) = \frac{1}{x - 3}$ , so $f'(1) = -\frac{1}{2}$ .”

What to write

“ $f$ is defined for $x > 3$ only, so $f'(1)$ does not exist.”

Why: The formula $\frac{1}{x - 3}$ is only the derivative ON the domain of $f$ . Write the domain as the first line of every logarithm question.

6. Taking ln y where y is negative

1 method mark

What not to write

“ $y = \frac{(x - 2)^3}{(x + 1)^2}$ , $\ln y = 3\ln(x - 2) - 2\ln(x + 1)$ ”, then evaluating at $x = 0$ .

What to write

“ $\ln|y| = 3\ln|x - 2| - 2\ln|x + 1|$ , valid wherever $y \ne 0$ ; at $x = 0$ , $\frac{y'}{y} = -\frac{7}{2}$ , $y' = -8 \cdot (-\frac{7}{2}) = 28$ .”

Why: At $x = 0$ , $y = -8$ and $\ln(x - 2) = \ln(-2)$ does not exist. With absolute values the method is valid everywhere $y \ne 0$ , and the resulting formula is the same.

7. Stopping at y'/y

1 to 2 marks

What not to write

“ $\ln y = x\ln x$ , so $y' = \ln x + 1$ .”

What to write

“ $\frac{y'}{y} = \ln x + 1$ , so $y' = y(\ln x + 1) = x^x(\ln x + 1)$ .”

Why: The left side of the differentiated equation is $\frac{y'}{y}$ , not $y'$ : the chain rule applied to $\ln y$ produces the division. The last line multiplies back, and writes $y$ in terms of $x$ .

8. Treating x to the x as a power or as an exponential

the whole question

What not to write

“ $\frac{d}{dx}x^x = x \cdot x^{x-1} = x^x$ ” or “ $\frac{d}{dx}x^x = x^x\ln x$ .”

What to write

“ $\ln y = x\ln x$ , so $\frac{y'}{y} = \ln x + 1$ and $y' = x^x(\ln x + 1)$ .”

Why: The power rule needs a constant exponent, the exponential rule a constant base. The correct derivative is the SUM of the two wrong ones: one term per place where $x$ appears.

9. Taking k for the percentage gained per period

every numerical answer of the problem, often 5 marks or more

What not to write

“The population doubles every hour, so it grows by $100\%$ per hour and $k = 1$ .”

What to write

“Doubling in one hour means $e^{k} = 2$ , so $k = \ln 2$ per hour.”

Why: $k = \frac{P'}{P}$ is the INSTANTANEOUS relative rate. With $k = 1$ the population would be multiplied by $e$ , not $2$ , each hour.

10. Writing Newton's law of cooling without the room temperature

the whole problem

What not to write

“The soup goes from $85$ °C to $53$ °C in $5$ minutes, so $T = 85e^{kt}$ with $e^{5k} = \frac{53}{85}$ .”

What to write

“The GAP with the room decays: $T = 21 + 64e^{kt}$ , $64e^{5k} = 32$ , $k = -\frac{\ln 2}{5}$ .”

Why: The law is $T' = k(T - T_s)$ : the quantity with a constant relative rate is $T - T_s$ . The wrong model even predicts a soup colder than the room after $15$ minutes.

Which method to choose

Which method, by the FORM of what you must differentiate or solve

Look at the shape of the expression, or at the verb of the problem, before writing anything

If $\ln$ of a product, a quotient, a power or a root → expand with the laws of logarithms first, then differentiate term by term

Example: $\ln\frac{x^3(2x + 1)^4}{\sqrt{x^2 + 1}} = 3\ln x + 4\ln(2x + 1) - \frac{1}{2}\ln(x^2 + 1)$
If a product or quotient of three factors or more, with powers → logarithmic differentiation on $\ln|y|$ , then multiply by $y$

Example: $y = \frac{\sqrt x(x + 3)^2}{(x + 1)^3}$ : $y'(1) = 2 \cdot (-\frac{1}{2}) = -1$
If $x$ in the base AND in the exponent → $\ln y = (\text{exponent})\ln(\text{base})$ , product rule, multiply by $y$

Example: $(\sin x)^x$ : $y' = (\sin x)^x(\ln\sin x + x\cot x)$
If $\log_b$ of anything → rewrite as $\frac{\ln}{\ln b}$ and keep the constant

Example: $\frac{d}{dx}\log_{10}x = \frac{1}{x\ln 10}$
If a limit $\lim_{x\to 0}(1 + ax)^{1/x}$ or $\frac{\ln(1 + h)}{h}$ → take the logarithm, recognize the derivative of $\ln$ at $1$ , exponentiate

Example: $(1 + 3x)^{1/x} \to e^3$ , $(1 - 2x)^{1/x} \to e^{-2}$
If the rate is proportional to the amount (growth, decay, interest) → $y = y_0e^{kt}$ , $k$ from a second datum by $\ln$ , rates by $y' = ky$

Example: $500 \to 2000$ in $2$ h: $k = \ln 2$
If the rate is proportional to a difference with the surroundings → the gap $T - T_s$ decays exponentially, never $T$ itself

Example: $T = 21 + 64e^{kt}$ , $k = -\frac{\ln 2}{5}$

No L'Hospital in this chapter: every limit here is a derivative read backwards. And a simple power like $x^5$ or exponential like $e^{3x}$ does not deserve logarithmic differentiation: the method is for when the direct rules become long or do not apply.

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.

A logarithmic differentiation the marker can follow

When to use it: A heavy product or quotient, or a function with x in the base and in the exponent

1 State the domain, and whether $y$ can be negative there; if it can, write $\ln|y|$ .
2 Expand the right side with the laws of logarithms, one term per factor, the exponents in front.
3 Differentiate both sides with respect to $x$ , writing $\frac{y'}{y}$ on the left and naming the chain or product rule on the right.
4 Multiply by $y$ , and replace $y$ by its expression in $x$ .
5 If a value is asked, evaluate $\frac{y'}{y}$ and $y$ separately, then multiply.

Concluding sentence

“For $x > 0$ , $y > 0$ and $\ln y = x\ln x$ . Differentiating, $\frac{y'}{y} = \ln x + 1$ by the product rule, so $y' = x^x(\ln x + 1)$ .”

The trap: Stopping at $\frac{y'}{y}$ , or dividing by $y$ at a point where $y = 0$ : a horizontal tangent at a zero of $y$ is found only after simplifying the final formula.

Marking: Typically 1 mark for the logarithm and its domain, 2 for the expansion, 2 for the differentiation, 1 for the multiplication by y.

A growth or decay problem

When to use it: The rate of change is proportional to the amount, or a half-life or a doubling time is given

1 Name the variable and its units, and write the law: $\frac{dy}{dt} = ky$ , hence $y = y_0e^{kt}$ by the result of the course.
2 Find $k$ EXACTLY from the second datum, by taking $\ln$ of both sides; check its sign against growth or decay.
3 Answer each question with exact values, using $e^{t\ln 2} = 2^t$ to compute by hand.
4 For a rate, use $y' = ky$ and give its unit (amount per unit of time), with its sign.

Concluding sentence

“Since the rate of decay is proportional to the mass, $m(t) = 80e^{kt}$ . From $m(30) = 40$ , $e^{30k} = \frac{1}{2}$ , so $k = -\frac{\ln 2}{30}$ per year.”

The trap: Giving $k$ as a percentage per period, or rounding it to three decimals: the rest of the problem then needs a calculator.

Marking: Typically 2 marks for the model, 3 for k, 3 for the answers, 2 for the rate with its unit.

Check before you hand in

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

The domain before the value

Before evaluating a derivative, check that the point is in the domain of the FUNCTION. A logarithm of a negative number or of zero means the question has no answer at that point.

$\ln(x - 3)$ at $x = 1$ : $\ln(-2)$ does not exist, so neither does $f'(1)$ , whatever $\frac{1}{x - 3}$ gives.
Plug x = 1 into a derivative of ln

At x = 1, ln x = 0 and many terms vanish, which makes a quick test of a formula against a direct computation.

$\frac{d}{dx}(\ln x)^2$ at $x = 1$ must be $0$ (the curve is flat at its minimum $0$ ); a formula that gives $2$ there is wrong.
The sign of k and the direction of change

Growth needs k > 0, decay and cooling need k < 0. A decaying sample whose predicted mass exceeds its initial mass has the wrong sign.

Half-life $30$ years: $k = -\frac{\ln 2}{30} < 0$ , and $m(90) = 10 < 80$ .
Substitute back one datum

Plug the time of the second measurement into the model you found: it must return the measured value exactly.

$T(5) = 21 + 64e^{-\ln 2} = 21 + 32 = 53$ : the cooling model passes. $85e^{5k}$ with the wrong $k$ would also pass, so also test a limit: the model must tend to $21$ .

The typical problem, taken apart

A tangent line to a function with x in the base and in the exponent

Let $y = x^{\sin x}$ for $x > 0$ . Find the equation of the tangent line to its graph at $x = \frac{\pi}{2}$ .

No calculator. Every step must be justified as on a MATH 140 final.

The curve

y = x^{\sin x}

touches the line

y = x

x = \frac{\pi}{2}

, the tangent found by the computation. It also meets that line at

x = 1

, since

1^{\sin 1} = 1

Step 1

Domain: $x > 0$ , so $y = e^{\sin x\ln x} > 0$ and $\ln y = \sin x \cdot \ln x$ .

Why

The variable is in the base AND in the exponent: neither the power rule nor the exponential rule applies. The logarithm turns the power into a product, and positivity makes $\ln y$ legitimate without absolute values.

Step 2

Differentiate both sides: $\frac{y'}{y} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}$ , by the chain rule on the left and the product rule on the right.

Why

Naming the two rules is where the method marks are. The left side is $\frac{y'}{y}$ , not $y'$ , which the next step must not forget.

Step 3

$y' = x^{\sin x}\left(\cos x\ln x + \frac{\sin x}{x}\right)$ .

Why

Multiplying by $y$ , written in terms of $x$ , gives the derivative itself. Stopping at $\frac{y'}{y}$ gives a relative rate, and a wrong slope.

Step 4

At $x = \frac{\pi}{2}$ : $\sin\frac{\pi}{2} = 1$ and $\cos\frac{\pi}{2} = 0$ , so $y = \left(\frac{\pi}{2}\right)^1 = \frac{\pi}{2}$ and $\frac{y'}{y} = 0 + \frac{1}{\pi/2} = \frac{2}{\pi}$ . Hence $y' = \frac{\pi}{2} \cdot \frac{2}{\pi} = 1$ .

Why

Evaluate $y$ and $\frac{y'}{y}$ separately, then multiply: each piece is simple, and the exact values of $\sin$ and $\cos$ do all the work.

Step 5

Tangent: $y = \frac{\pi}{2} + 1 \cdot \left(x - \frac{\pi}{2}\right) = x$ . Check: the point $\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$ is on the line $y = x$ .

Why

The point-slope form with the exact point and slope. The check costs five seconds and catches a wrong $y$ value, the most common slip after forgetting to multiply by $y$ .

The conclusion, written out

“The tangent line to $y = x^{\sin x}$ at $x = \frac{\pi}{2}$ is $y = x$ .”

The classic mistake on this problem: Writing $y' = \sin x \cdot x^{\sin x - 1}$ (the power rule with a variable exponent), which gives $1 \cdot \left(\frac{\pi}{2}\right)^0 = 1$ at $\frac{\pi}{2}$ by pure luck: $\cos\frac{\pi}{2} = 0$ kills the missing term $x^{\sin x}\cos x\ln x$ . At $x = 2$ that term is not zero, and a marker checks the method, not the number.

Learn by heart

• $\frac{d}{dx}\ln|u| = \frac{u'}{u}$ , $\frac{d}{dx}\log_b x = \frac{1}{x\ln b}$ . The domain is read on the function, before differentiating.
• $\ln(ab) = \ln a + \ln b$ , $\ln\frac{a}{b} = \ln a - \ln b$ , $\ln(a^r) = r\ln a$ . NEVER $\ln(a + b)$ , NEVER $(\ln a)^2 = 2\ln a$ .
• Logarithmic differentiation: $\ln|y|$ , expand, differentiate, MULTIPLY BY $y$ .
• $x$ in the base and in the exponent: $\ln y = (\text{exponent})\ln(\text{base})$ . $\frac{d}{dx}x^x = x^x(\ln x + 1)$ .
• $\lim_{h\to 0}\frac{\ln(1 + h)}{h} = 1$ , $\lim_{x\to 0}(1 + ax)^{1/x} = e^a$ .
• $y' = ky \iff y = y_0e^{kt}$ ; $k = \frac{y'}{y}$ is a relative rate; doubling time $\frac{\ln 2}{k}$ .
• Cooling: the gap decays, $T = T_s + (T_0 - T_s)e^{kt}$ .

Frequently asked questions

When should I use logarithmic differentiation in MATH 140?

Use it in two situations. First, when the variable appears both in the base and in the exponent, like x to the power x, because no other rule applies. Second, when the function is a long product or quotient of powers, because the logarithm turns it into a sum that differentiates term by term. Take the log of the absolute value, differentiate, then multiply by y.

Why is the derivative of ln of 5x equal to 1 over x and not 5 over x?

Because ln of 5x equals ln 5 plus ln x, and ln 5 is a constant whose derivative is zero. With the chain rule you get the same thing: the derivative of the inside, 5, divided by the whole inside, 5x, which simplifies to 1 over x. Multiplying the input of a logarithm by a constant only shifts its graph up or down.

What is the derivative of x to the power x?

It is x to the power x, times the quantity ln x plus 1. Write y equals x to the x, take the natural log of both sides to get ln y equals x ln x, differentiate with the product rule to get y prime over y equals ln x plus 1, then multiply by y. Neither the power rule nor the exponential rule gives this answer.

What does the constant k mean in exponential growth y' = ky?

The constant k is the relative growth rate, the rate of change divided by the current amount. It is measured per unit of time. It is not the percentage gained over one period: a population that doubles every hour has k equal to ln 2, about 0.69 per hour, not 1. Find k from two data points by taking the natural log.

How do I solve Newton's law of cooling problems without a calculator?

Write the temperature as the room temperature plus the initial gap times e to the kt. Use the second reading to find e to the power k times the time, which usually comes out as a simple fraction like one half. Then every later temperature is the room temperature plus the gap multiplied by powers of one half, all done by hand.

Practise it

Corrected exercises: Logarithmic differentiation and growth, MATH 140 at McGill

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 Implicit differentiation Next sheet Related rates

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math140-logarithmic-differentiation. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Get in touch for a first session. The logarithm chapter closes the differentiation rules: after it, every function of the course can be differentiated, and every growth problem of the final reads the same way.

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Revision sheet: derivatives of logarithms, logarithmic differentiation, growth and decay (MATH 140)

The essentials

Three derivatives, and the domain that comes first

Relative rates: logarithmic differentiation and y' = ky are the same idea

The rules in table form

Which rule for which function: look where the x is

The mistakes that cost marks

1. Differentiating ln(5x) as 5 over x

2. Forgetting the chain rule inside ln

3. Differentiating log base b without ln b

4. Confusing the square of ln with the ln of a square

5. Evaluating a derivative outside the domain

6. Taking ln y where y is negative

7. Stopping at y'/y

8. Treating x to the x as a power or as an exponential

9. Taking k for the percentage gained per period

10. Writing Newton's law of cooling without the room temperature

Which method to choose

Which method, by the FORM of what you must differentiate or solve

How the answer is expected to be written

A logarithmic differentiation the marker can follow

A growth or decay problem

Check before you hand in

The typical problem, taken apart

A tangent line to a function with x in the base and in the exponent

Learn by heart

Frequently asked questions

When should I use logarithmic differentiation in MATH 140?

Why is the derivative of ln of 5x equal to 1 over x and not 5 over x?

What is the derivative of x to the power x?

What does the constant k mean in exponential growth y' = ky?

How do I solve Newton's law of cooling problems without a calculator?

Corrected exercises: Logarithmic differentiation and growth, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?