MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: derivatives of inverse functions and logarithms (MATH 203)

This sheet is not a summary of section 3.8 of Thomas' Calculus: you have the course notes. It answers one question only, what makes students lose marks on the derivatives of inverse functions and of logarithms in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The rules of the chapter fit in four lines. The marks go elsewhere: in the point where $f'$ is evaluated for an inverse, and in the ALGEBRA done before differentiating, a root turned into a fractional power, a denominator turned into a minus sign, a law of logarithms that does not exist. A scientific calculator is allowed on the exams, so exact answers come first and the calculator checks them.

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The thread of the chapter

Every derivative of section 3.8 comes from undoing: the slope of $f^{-1}$ at $b$ is the reciprocal of the slope of $f$ at the SWAPPED point $a = f^{-1}(b)$ , and $\ln$ undoes products, quotients and powers, so the laws of logarithms, the true ones only, are applied BEFORE differentiating, and every base other than $e$ leaves its constant $\ln a$ .

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

What is on this sheet

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

Topics covered

Derivative of an inverse function
One-to-one functions
Derivative of the natural logarithm
Logarithms in other bases
General exponential functions
Laws of logarithms
Logarithmic differentiation
Limit definition of e

The essentials

The inverse: a reciprocal slope, at the swapped point

• If $f$ is differentiable on an interval with $f' \ne 0$ there, then at $b = f(a)$ : $(f^{-1})'(b) = \frac{1}{f'(a)}$ , with $a = f^{-1}(b)$ .
• The point $(a, b)$ on $f$ is the point $(b, a)$ on $f^{-1}$ : mirror images in $y = x$ , and a reflected line has its rise and run swapped.
• No formula for $f^{-1}$ is needed: find $a$ by inspection (try $0$ , $1$ , $-1$ ), check $f(a) = b$ , then invert $f'(a)$ .
• Where $f'(a) = 0$ the theorem fails and $f^{-1}$ has a VERTICAL tangent at $b$ .
• Applied to $e^x$ , the theorem gives $\frac{d}{dx}\ln x = \frac{1}{e^{\ln x}} = \frac{1}{x}$ : this is how Thomas obtains the derivative of $\ln$ .

y = e^x

and

y = \ln x

are mirror images in

y = x

: the tangent of slope

e

(1, e)

becomes the tangent of slope

\frac{1}{e}

(e, 1)

Write the pair $(a, b)$ on your copy before any derivative. It is the one line that prevents evaluating $f'$ at $b$ , a loss worth the whole question.

Logarithms and other bases: rewrite first, then differentiate

• $\frac{d}{dx}\ln u = \frac{u'}{u}$ where $u > 0$ , and $\frac{d}{dx}\ln|u| = \frac{u'}{u}$ where $u \ne 0$ : no absolute value in the answer.
• $\log_a u = \frac{\ln u}{\ln a}$ , so $\frac{d}{dx}\log_a u = \frac{u'}{u\ln a}$ ; $a^u = e^{u\ln a}$ , so $\frac{d}{dx}a^u = a^u\ln a \cdot u'$ .
• Laws, for $a, b > 0$ : $\ln(ab) = \ln a + \ln b$ , $\ln\frac{a}{b} = \ln a - \ln b$ , $\ln(a^r) = r\ln a$ , $\ln e^u = u$ . A root is a power: $\ln\sqrt[3]{u} = \frac{1}{3}\ln u$ .
• The domain is read on the function BEFORE expanding: $\ln\left((x - 1)(x - 4)\right)$ exists at $x = 0$ , $\ln(x - 1)$ does not.
• Logarithmic differentiation: $\ln|y|$ , expand, differentiate, multiply by $y$ . It is REQUIRED only when $x$ sits in the base and the exponent of the same power.

The base $a$ never disappears from a derivative: it survives as $\ln a$ , negative when $0 < a < 1$ . Only $e$ has $\ln e = 1$ , which is why $e^x$ and $\ln x$ have derivatives with no constant.

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 rewriting is legitimate before differentiating

Read a line as: the expression of the first column, under the condition of the second, may be replaced by the third. The red lines are rewritings that students write and that do not exist.

Expression	Condition	Rewrite as
$\ln\frac{\sqrt[3]{x}\,(x^2 + 3)^2}{(5x + 1)^4}$	$x > 0$	$\frac{1}{3}\ln x + 2\ln(x^2 + 3) - 4\ln(5x + 1)$
Example: Derivative at $x = 1$ : $\frac{1}{3} + 1 - \frac{10}{3} = -2$ .
$\ln\left((x - 1)(x - 4)\right)$	signs unknown	$\ln\|x - 1\| + \ln\|x - 4\|$
Example: At $x = 0$ : derivative $-1 - \frac{1}{4} = -\frac{5}{4}$ , where $\ln(x - 1)$ alone does not exist.
$\ln(9x^2)$	$x \ne 0$	$\ln 9 + 2\ln\|x\|$
Example: Derivative $\frac{2}{x}$ , equal to $-\frac{2}{3}$ at $x = -3$ .
$a^x$	$a > 0$	$e^{x\ln a}$
Example: $\frac{d}{dx}9^x = 9^x\ln 9$ , equal to $3\ln 9 = 6\ln 3$ at $x = \frac{1}{2}$ .
$\log_2 x \cdot \log_x 8$	$x > 0$ , $x \ne 1$	$\frac{\ln 8}{\ln 2} = 3$
Example: A constant: its derivative is $0$ everywhere on its domain.
$\ln(x^2 + 9)$	none	$2\ln x + \ln 9$ rule that does not exist
Example: At $x = 3$ : $\ln 18 \approx 2.89$ , while $2\ln 3 + \ln 9 = \ln 81 \approx 4.39$ . Same form, other result: $\ln(1 + 1) = \ln 2 \approx 0.69$ but $\ln 1 + \ln 1 = 0$ . What to do: Keep the sum inside: $\frac{d}{dx}\ln(x^2 + 9) = \frac{2x}{x^2 + 9}$ , which is $\frac{1}{3}$ at $x = 3$ .
$\ln u - \ln v$	none	$\frac{\ln u}{\ln v}$ rule that does not exist
Example: $\ln 2 - \ln 4 = -\ln 2 < 0$ , while $\frac{\ln 2}{\ln 4} = \frac{1}{2} > 0$ . Same form, other result: $\ln 8 - \ln 2 = \ln 4 \approx 1.39$ , while $\frac{\ln 8}{\ln 2} = 3$ . What to do: The true law is $\ln u - \ln v = \ln\frac{u}{v}$ ; or differentiate the difference term by term.

Every legal rewriting keeps the function and changes its form. Test any rewriting at one value of $x$ with the calculator before differentiating: a fake law fails the test at once.

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. Evaluating f' at b for the derivative of an inverse

the whole question

What not to write

“ $f(x) = e^x + 2x$ , so $(f^{-1})'(1) = \frac{1}{f'(1)} = \frac{1}{e + 2}$ .”

What to write

“ $f(0) = 1$ , so $a = 0$ ; $f'(0) = 3$ and $(f^{-1})'(1) = \frac{1}{3}$ .”

Why: The theorem pairs $(b, a)$ on $f^{-1}$ with $(a, b)$ on $f$ . The number $\frac{1}{e + 2}$ is a true slope of $f^{-1}$ , at $b = e + 2$ : another point.

2. Splitting the logarithm of a sum

2 marks, and every tangent computed from it

What not to write

“ $\ln(x^2 + 9) = 2\ln x + \ln 9$ , so its derivative is $\frac{2}{x}$ .”

What to write

“Chain rule, inner function $x^2 + 9$ : $\frac{d}{dx}\ln(x^2 + 9) = \frac{2x}{x^2 + 9}$ .”

y = \ln(x^2 + 9)

and

y = 2\ln x + \ln 9

are different curves: they cross once, near

x = 1.06

, and the red one even goes negative.

Why: $2\ln x + \ln 9 = \ln(9x^2)$ , another function. The laws split products, quotients and powers, never sums.

3. Losing the domain when a logarithm is expanded

1 method mark

What not to write

“ $\ln\left((x - 1)(x - 4)\right) = \ln(x - 1) + \ln(x - 4)$ ”, then evaluating at $x = 0$ .

What to write

“ $= \ln|x - 1| + \ln|x - 4|$ on the whole domain; $F'(0) = -1 - \frac{1}{4} = -\frac{5}{4}$ .”

Why: $\ln(ab) = \ln a + \ln b$ needs $a > 0$ AND $b > 0$ . At $x = 0$ both factors are negative and their product is $4$ : the function exists, the expansion does not.

4. Forgetting ln a in a base other than e

1 mark

What not to write

“ $\frac{d}{dx}\log_4(x^2 + 12) = \frac{2x}{4(x^2 + 12)}$ ” or “ $\frac{d}{dx}\log_4 u = \frac{u'}{u}$ .”

What to write

“ $\log_4 u = \frac{\ln u}{\ln 4}$ , so the derivative is $\frac{2x}{(x^2 + 12)\ln 4}$ , equal to $\frac{1}{8\ln 2}$ at $x = 2$ .”

Why: The base enters through the constant $\ln 4$ , never through a factor $4$ . Only the natural logarithm differentiates with no constant.

5. Applying the power rule to an exponential

the whole question

What not to write

“ $\frac{d}{dx}10^{1 - x^2} = (1 - x^2)\,10^{-x^2}$ .”

What to write

“ $\frac{d}{dx}10^{u} = 10^u\ln 10 \cdot u'$ with $u = 1 - x^2$ : $-2x\ln 10 \cdot 10^{1 - x^2}$ , equal to $-2\ln 10$ at $x = 1$ .”

Why: The power rule needs a constant EXPONENT; here the exponent is the variable. Rewrite $10^u = e^{u\ln 10}$ if in doubt.

6. Writing a root or a denominator with the wrong coefficient

2 marks, before any derivative is taken

What not to write

“ $\ln\frac{(2x + 1)^5}{\sqrt[3]{x^2 + 7}} = 5\ln(2x + 1) + 3\ln(x^2 + 7)$ .”

What to write

“ $= 5\ln(2x + 1) - \frac{1}{3}\ln(x^2 + 7)$ : the cube root is the power $\frac{1}{3}$ , and it sits in the denominator.”

Why: Two separate algebra facts, $\sqrt[3]{u} = u^{1/3}$ and $\ln\frac{a}{b} = \ln a - \ln b$ . Each one, missed, changes every later line.

7. Differentiating ln|x| as 1 over |x|

1 mark, and the sign of every conclusion

What not to write

“ $\frac{d}{dx}\ln|2x - 5| = \frac{2}{|2x - 5|}$ , so the slope at $x = 2$ is $2$ .”

What to write

“ $\frac{d}{dx}\ln|u| = \frac{u'}{u}$ : $\frac{2}{2x - 5}$ , equal to $-2$ at $x = 2$ .”

Why: On the branch where $u < 0$ , $\ln|u| = \ln(-u)$ , and the chain rule gives $\frac{-u'}{-u} = \frac{u'}{u}$ . The absolute value extends the domain; it never enters the derivative.

8. Using logarithmic differentiation on a sum

the whole question

What not to write

“ $y = (x + 1)^4 + \sqrt{2x + 7}$ , so $\ln y = 4\ln(x + 1) + \frac{1}{2}\ln(2x + 7)$ .”

What to write

“ $y' = 4(x + 1)^3 + \frac{1}{\sqrt{2x + 7}}$ , so $y'(1) = \frac{97}{3}$ .”

Why: The student's right side is the logarithm of the PRODUCT. The method is for products, quotients and powers; a sum is differentiated term by term.

9. Taking a yearly percentage for the relative rate

every numerical answer of the problem

What not to write

“The car loses $15\%$ per year, so $\frac{V'}{V} = -0.15$ .”

What to write

“ $V = V_0(0.85)^t$ , so $\frac{V'}{V} = \ln 0.85 \approx -0.1625$ per year.”

Why: The relative rate of $a^t$ is $\ln a$ , not $a - 1$ . The $15\%$ is the loss over a whole year; the instantaneous rate must be larger to lose it.

Which method to choose

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

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

If the derivative of $f^{-1}$ at a point, no formula for $f^{-1}$ → find $a$ with $f(a) = b$ , check $f'(a) \ne 0$ , answer $\frac{1}{f'(a)}$

Example: $f(x) = 2x + \ln x$ , $b = 2$ : $a = 1$ , $(f^{-1})'(2) = \frac{1}{3}$
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: $\ln\frac{e^{2x}}{1 + e^{2x}} = 2x - \ln(1 + e^{2x})$
If $\log_a u$ or $a^u$ with a constant base $a \ne e$ → the rule for $\ln$ or $e^u$ , times or divided by $\ln a$

Example: $\frac{d}{dx}10^{1 - x^2} = -2x\ln 10 \cdot 10^{1 - x^2}$
If $x$ in the base AND the exponent of the same power → $\ln y = (\text{exponent})\ln(\text{base})$ , product rule, multiply by $y$

Example: $x^{1/x}$ : $y' = x^{1/x}\,\frac{1 - \ln x}{x^2}$
If a product or quotient of many powers → logarithmic differentiation on $\ln|y|$ , then multiply by $y$

Example: $y = \frac{x^3}{(x - 2)^2}$ : $\frac{y'}{y} = \frac{x - 6}{x(x - 2)}$
If a limit of the form $(1 + kx)^{m/x}$ → take $\ln$ , bring out $\frac{\ln(1 + u)}{u} \to 1$ , exponentiate

Example: $\left(1 + \frac{x}{2}\right)^{3/x} \to e^{3/2}$
If a limit $\frac{g(c + h) - g(c)}{h}$ with $g$ a log or an exponential → read it as $g'(c)$

Example: $\frac{5^h - 1}{h} \to \ln 5$

No L'Hôpital's Rule in this chapter: every limit here is a derivative read backwards or a logarithm brought out. And $3^x x^3$ needs no logarithm: $x$ is in two DIFFERENT powers, and the product rule is shorter.

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 derivative of an inverse at a point

When to use it: A one-to-one function is given by a formula that cannot be inverted by algebra

1 Justify that $f$ is one-to-one (a sum of increasing functions, a restricted domain), in one line.
2 Find $a = f^{-1}(b)$ by inspection and write the pair: $f(a) = b$ .
3 Compute $f'(a)$ and check that it is not $0$ .
4 Conclude $(f^{-1})'(b) = \frac{1}{f'(a)}$ ; for a tangent, use the point $(b, a)$ .

Concluding sentence

“Since $f(1) = 2$ , the point $(2, 1)$ is on the graph of $f^{-1}$ , and $(f^{-1})'(2) = \frac{1}{f'(1)} = \frac{1}{3}$ .”

The trap: Evaluating $f'$ at $b$ , or using the point $(a, b)$ for the tangent to $f^{-1}$ : both swap the pair back.

Marking: Typically 1 mark for one-to-one, 2 for the pair (a, b), 1 for f'(a), 1 for the reciprocal and the tangent.

A logarithmic differentiation

When to use it: A heavy product or quotient of powers, or x in the base and the exponent of one power

1 State where $y > 0$ ; if $y$ can be negative, write $\ln|y|$ .
2 Expand: exponents in front, roots as fractional powers, denominators with minus signs, $\ln e^u = u$ .
3 Differentiate both sides, $\frac{y'}{y}$ on the left, naming the chain or product rule on the right.
4 Multiply by $y$ ; for a value, compute $y$ and $\frac{y'}{y}$ separately, then multiply.

Concluding sentence

“For $x > 0$ , $\ln y = \frac{\ln x}{x}$ . Differentiating, $\frac{y'}{y} = \frac{1 - \ln x}{x^2}$ , so $y' = x^{1/x}\,\frac{1 - \ln x}{x^2}$ .”

The trap: Stopping at $\frac{y'}{y}$ : that is a relative rate, not the slope.

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

Check before you hand in

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

Check the pair before inverting

For an inverse, plug a into f: it must return b exactly. If it does not, the reciprocal is taken at the wrong point.

$f(x) = 3x + \ln x$ , $b = 3$ : $f(1) = 3 + 0 = 3$ , so $a = 1$ and the slope of $f^{-1}$ at $3$ is $\frac{1}{f'(1)} = \frac{1}{4}$ .
Test a rewriting at one value

Before differentiating an expanded logarithm, compute the original and the expansion at one convenient x with the calculator. They must agree.

At $x = 3$ : $\ln(x^2 + 9) = \ln 18 \approx 2.89$ , but $2\ln 3 + \ln 9 \approx 4.39$ : the expansion is false.
The sign of the base

A base below 1 gives a negative ln a, so a decaying quantity has a negative derivative. A positive rate for a car losing value is a lost sign.

$\ln 0.85 \approx -0.1625 < 0$ , so $V'(t) < 0$ for every $t$ .
Compare with a difference quotient

A calculator cannot differentiate, but it can compute (f(a + 0.001) - f(a))/0.001, which must be close to your f'(a).

For $\log_4(x^2 + 12)$ at $x = 2$ : the quotient is about $0.1804$ , the formula gives $\frac{1}{8\ln 2} \approx 0.1803$ .

The typical problem, taken apart

A tangent line to the graph of an inverse function

Let $f(x) = 3x + \ln x$ for $x > 0$ . Find the equation of the tangent line to the graph of $f^{-1}$ at the point where $x = 3$ .

The equation $3x + \ln x = y$ cannot be solved for $x$ by algebra. Every step must be justified as on a MATH 203 final.

The point

(1, 3)

f

and its mirror image

(3, 1)

f^{-1}

: the tangent is asked at the second one, and its slope comes from the first.

Step 1

$3x$ and $\ln x$ are strictly increasing on $x > 0$ , so $f$ is strictly increasing, hence one-to-one: $f^{-1}$ exists.

Why

The theorem is about an inverse that exists. A sum of increasing functions settles it in one line, with no derivative test, which comes later in the course.

Step 2

Find $a$ with $f(a) = 3$ : try $a = 1$ , $f(1) = 3 + \ln 1 = 3$ . So $a = 1$ , the pair is $(1, 3)$ on $f$ and $(3, 1)$ on $f^{-1}$ .

Why

Inspection is the method: $\ln 1 = 0$ makes $x = 1$ the first value to try. Writing the pair is the step that prevents evaluating $f'$ at $3$ .

Step 3

$f'(x) = 3 + \frac{1}{x}$ , so $f'(1) = 4 \ne 0$ .

Why

The hypothesis $f'(a) \ne 0$ is part of the theorem, and checking it is part of the marks.

Step 4

$(f^{-1})'(3) = \frac{1}{f'(1)} = \frac{1}{4}$ .

Why

The reciprocal of the slope at the SWAPPED point. The wrong line $\frac{1}{f'(3)} = \frac{1}{10/3} = \frac{3}{10}$ is what the marker looks for first.

Step 5

Tangent at $(3, 1)$ : $y = 1 + \frac{1}{4}(x - 3) = \frac{1}{4}x + \frac{1}{4}$ .

Why

The point is $(b, a) = (3, 1)$ , not $(1, 3)$ . Check: at $x = 3$ the line gives $\frac{3}{4} + \frac{1}{4} = 1$ .

The conclusion, written out

“The tangent line to the graph of $f^{-1}$ at $x = 3$ is $y = \frac{1}{4}x + \frac{1}{4}$ .”

The classic mistake on this problem: Using $f'(3) = \frac{10}{3}$ gives the slope $\frac{3}{10}$ , and using the point $(1, 3)$ gives a line through the wrong point: two independent losses, and a copy can make both.

Learn by heart

• $(f^{-1})'(b) = \frac{1}{f'(a)}$ with $f(a) = b$ : write the pair $(a, b)$ first.
• $\frac{d}{dx}\ln|u| = \frac{u'}{u}$ , $\frac{d}{dx}\log_a u = \frac{u'}{u\ln a}$ , $\frac{d}{dx}a^u = a^u\ln a \cdot u'$ .
• Laws only for products, quotients and powers of POSITIVE quantities; with unknown signs, absolute values. NEVER $\ln(a + b) = \ln a + \ln b$ .
• Roots are fractional powers, denominators come out with a minus sign, $\ln e^u = u$ .
• Logarithmic differentiation: $\ln|y|$ , expand, differentiate, MULTIPLY BY $y$ ; required when $x$ is in the base and exponent of one power.
• $\lim_{h\to 0}\frac{\ln(1 + h)}{h} = 1$ and $e = \lim_{x\to 0}(1 + x)^{1/x}$ .
• The relative rate of $Q_0a^t$ is $\ln a$ , not $a - 1$ .

Frequently asked questions

How do I find the derivative of an inverse function at a point in MATH 203?

You do not need a formula for the inverse. Find the number a such that f(a) equals the given b, usually by trying simple values like 0 or 1. Check that f prime of a is not zero. The derivative of the inverse at b is then one over f prime of a. The most common mistake is to evaluate f prime at b instead of at a.

Why does the derivative of 2 to the x have a ln 2 in it?

Because 2 to the x equals e to the power x times ln 2, and the chain rule brings down the constant ln 2. Every base other than e leaves its natural logarithm in the derivative. The power rule does not apply, since the exponent is the variable. The same constant appears for log base 2, but in the denominator.

Can I split ln of a sum into a sum of logarithms?

No. The laws of logarithms turn the log of a product into a sum, the log of a quotient into a difference and the log of a power into a product. Nothing splits the log of a sum. To differentiate ln of x squared plus 9, keep it whole and use the chain rule: the answer is 2x over x squared plus 9.

When do I need logarithmic differentiation?

It is required when the variable appears in both the base and the exponent of the same power, like x to the power 1 over x. It is also the fastest route for long products and quotients of powers. It is useless for sums, and for a product like 3 to the x times x cubed, where the product rule is shorter. Always multiply by y at the end.

Why is the limit of (1 + x) to the power 1/x equal to e and not 1?

The base tends to 1 but the exponent grows without bound, so the form tells you nothing. Take the natural log: you get ln of 1 plus x, divided by x, which is a difference quotient of ln at 1 and tends to 1. Since the log tends to 1, the expression itself tends to e to the power 1, which is e.

Practise it

Corrected exercises: Derivatives of inverse functions and logarithms, MATH 203 at Concordia

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

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Implicit differentiation Next sheet Inverse trigonometric functions

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

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Revision sheet: derivatives of inverse functions and logarithms (MATH 203)

The essentials

The inverse: a reciprocal slope, at the swapped point

Logarithms and other bases: rewrite first, then differentiate

The rules in table form

Which rewriting is legitimate before differentiating

The mistakes that cost marks

1. Evaluating f' at b for the derivative of an inverse

2. Splitting the logarithm of a sum

3. Losing the domain when a logarithm is expanded

4. Forgetting ln a in a base other than e

5. Applying the power rule to an exponential

6. Writing a root or a denominator with the wrong coefficient

7. Differentiating ln|x| as 1 over |x|

8. Using logarithmic differentiation on a sum

9. Taking a yearly percentage for the relative rate

Which method to choose

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

How the answer is expected to be written

The derivative of an inverse at a point

A logarithmic differentiation

Check before you hand in

The typical problem, taken apart

A tangent line to the graph of an inverse function

Learn by heart

Frequently asked questions

How do I find the derivative of an inverse function at a point in MATH 203?

Why does the derivative of 2 to the x have a ln 2 in it?

Can I split ln of a sum into a sum of logarithms?

When do I need logarithmic differentiation?

Why is the limit of (1 + x) to the power 1/x equal to e and not 1?

Corrected exercises: Derivatives of inverse functions and logarithms, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?