Source: Midterm 2 Review Sheet, Math 15200 Section 42
Tags: one-to-one functions, inverse functions, natural logarithm, ln, logarithmic differentiation, logarithmic integration, trig integrals, calculus II, MATH 152
These sections build up from the idea that only one-to-one functions have inverses, define the natural logarithm as an integral, and then develop all the differentiation and integration rules that follow from that definition. The payoff is a set of new integral formulas, especially for trigonometric functions that weren't integrable before.
One-to-one function
A function where each output comes from exactly one input. Formally: f(x₁) = f(x₂) implies x₁ = x₂. Graphically, it passes the horizontal line test.
Inverse function (f⁻¹)
If f is one-to-one, then f⁻¹ is the unique function satisfying f(f⁻¹(x)) = x for all x in the range of f. The graph of f⁻¹ is the reflection of f across the line y = x.
Natural logarithm (ln)
Defined as the integral ln(x) = ∫ from 1 to x of (1/t) dt. This is not just a convenient formula; it is the definition of ln in this course.
Euler's number (e)
The unique number satisfying ln(e) = 1. Approximately 2.718281828459.
u-substitution (for logarithmic integrals)
The technique used to show that ∫ u'(x)/u(x) dx = ln|u(x)| + C, which underpins many of the trig integration results.
Testing for one-to-one
A function that is strictly increasing (f'(x) > 0 for all x in its domain) is one-to-one.
A function that is strictly decreasing (f'(x) < 0 for all x in its domain) is one-to-one.
If f is not strictly monotonic, it fails the horizontal line test and has no inverse.
Properties of inverse functions
If f is continuous, then f⁻¹ is continuous.
The domain of f⁻¹ is the range of f, and vice versa.
The graph of f⁻¹ is obtained by reflecting the graph of f across y = x. This means you swap every (a, b) point to (b, a).
Derivative of an inverse function
If f is differentiable, f(a) = b, and f'(a) ≠ 0, then f⁻¹ is differentiable at b and:
(f⁻¹)'(b) = 1 / f'(a)
Equivalently, if y = f(x), then dx/dy = 1 / (dy/dx).
The condition f'(a) ≠ 0 is essential. If f has a horizontal tangent at a, then f⁻¹ has a vertical tangent at b and is not differentiable there.
Definition as an integral
ln(x) = ∫ from 1 to x of (1/t) dt
This means ln(x) is the net signed area under the curve y = 1/t from t = 1 to t = x.
Basic properties of ln
Domain: (0, ∞). You cannot take the log of zero or a negative number.
Range: (−∞, ∞). The logarithm can produce any real number.
ln is increasing on its entire domain because its derivative 1/x is positive for all x > 0.
Sign behaviour
ln(x) < 0 when 0 < x < 1
ln(x) = 0 when x = 1
ln(x) > 0 when x > 1
This follows directly from the integral definition: when x < 1, you are integrating "backwards" from 1, which gives a negative value.
Computational properties (log rules)
ln(1) = 0
ln(e) = 1
ln(ab) = ln a + ln b
ln(a/b) = ln a − ln b
ln(1/b) = −ln b
ln(a^(p/q)) = (p/q) · ln a
These rules are used constantly when simplifying expressions before differentiating or integrating.
Differentiation
d/dx [ln(x)] = 1/x, for x > 0
d/dx [ln|x|] = 1/x, for all x ≠ 0
The absolute-value version extends the result to negative x, which matters when integrating functions whose argument can be negative.
Integration
∫ (1/x) dx = ln|x| + C
More generally, by u-substitution:
∫ u'(x) / u(x) dx = ln|u(x)| + C
This pattern (derivative of the denominator sitting in the numerator) appears everywhere. Train yourself to spot it quickly.
Trigonometric integrals using logarithms
Previously known results:
∫ sin x dx = −cos x + C
∫ cos x dx = sin x + C
∫ sec²x dx = tan x + C
∫ csc²x dx = −cot x + C
∫ sec x tan x dx = sec x + C
∫ csc x cot x dx = −csc x + C
New results enabled by the logarithm:
∫ tan x dx = −ln|cos x| + C = ln|sec x| + C
∫ cot x dx = ln|sin x| + C
∫ sec x dx = ln|sec x + tan x| + C
∫ csc x dx = ln|csc x − cot x| + C
The tan x and cot x integrals come from rewriting them as sin/cos or cos/sin and applying the u-substitution pattern. The sec x and csc x integrals require the "multiply by 1" trick (multiplying by (sec x + tan x)/(sec x + tan x) for sec x, for instance).
Inverse function derivative: (f⁻¹)'(b) = 1 / f'(a), where f(a) = b
Natural log definition: ln(x) = ∫ from 1 to x of (1/t) dt
Core differentiation rule: d/dx [ln|x|] = 1/x
Core integration rule: ∫ u'/u dx = ln|u| + C
All six trig log integrals (see list in Core Content above)
⚠️ The derivative-of-inverse formula (f⁻¹)'(b) = 1/f'(a) requires you to find the right value of a first. Given b, solve f(a) = b to get a, then compute f'(a). A common error is plugging b into f' directly.
⚠️ ln(x) is only defined for x > 0. When integrating 1/x, the result is ln|x| + C (with absolute value), which covers both positive and negative x.
⚠️ The log rules (product, quotient, power) are heavily tested, both on their own and as a simplification step before differentiation. Simplify first, then differentiate; it is almost always easier.
⚠️ For ∫ tan x dx, both forms (−ln|cos x| + C and ln|sec x| + C) are correct. Know how to convert between them using the identity sec x = 1/cos x and the log quotient rule.
⚠️ The integral of sec x is notoriously tricky to derive but easy to apply once memorised. Many students confuse the sec x integral with the sec²x integral. sec²x gives tan x; sec x gives ln|sec x + tan x|.
Q: If f(2) = 5 and f'(2) = 3, what is (f⁻¹)'(5)?
A: (f⁻¹)'(5) = 1/f'(2) = 1/3. You evaluate f' at the input that maps to 5, which is x = 2.
Q: Why is the natural logarithm defined as an integral rather than as "the inverse of eˣ"?
A: In this course's development, ln comes first. The exponential function is defined later as the inverse of ln. Defining ln via the integral ∫(1/t)dt gives it a rigorous foundation that does not depend on first defining irrational exponents.
Q: Evaluate ∫ (3x²)/(x³ + 1) dx.
A: Let u = x³ + 1, so du = 3x² dx. The integral becomes ∫ du/u = ln|u| + C = ln|x³ + 1| + C.
Q: What is ∫ tan(x) dx, and how do you derive it?
A: Write tan x = sin x / cos x. Let u = cos x, so du = −sin x dx. Then ∫ sin x / cos x dx = −∫ du/u = −ln|u| + C = −ln|cos x| + C, which equals ln|sec x| + C.
Q: A function f satisfies f'(x) > 0 for all x in its domain. Does f have an inverse?
A: Yes. f'(x) > 0 everywhere means f is strictly increasing, which makes it one-to-one. Every one-to-one function has an inverse.
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