Calculus II ep07: The natural logarithm (Sep 23, 2026)

Overview

Prof Staecker defines the natural logarithm by ln x = ∫₁ˣ 1/t dt, uses its derivative to evaluate integrals, and derives the product, quotient, and power rules from the integral-based definition. The lecture concludes with logarithmic differentiation of y = x²(7 + x)¹⁰/(5x² − 7x + 4), replacing product and quotient rules with simpler logarithm properties.

Key takeaways

The integral definition ln x = ∫₁ˣ 1/t dt immediately gives ln 1 = 0 because the integral has identical bounds.
For x ≠ 0, the antiderivative of 1/x is ln|x| + C; absolute value allows the formula to apply on both positive and negative intervals.
When a rational integrand’s numerator is proportional to the derivative of its denominator, substituting the denominator reduces the integral to ∫du/u and produces a logarithm.
The identity ln(ab) = ln a + ln b follows by differentiating ln(ab) and ln b with respect to b, then determining their constant difference using b = 1.
Logarithms convert quotients to differences and powers to coefficients: ln(a/b) = ln a − ln b and ln(aⁿ) = n ln a, provided the expressions are defined.
Logarithmic differentiation can simplify derivatives of products and quotients: take ln of both sides, expand with log rules, differentiate implicitly, and multiply through by y.

Chapters

0:00 Defining ln x by an Integral and Establishing Its Derivative

Prof Staecker recalls the definition ln x = ∫₁ˣ 1/t dt, interpreting the logarithm as signed area under 1/t.
The fundamental calculus fact is d(ln x)/dx = 1/x, so an antiderivative of 1/x is ln|x| + C.
The chain rule gives d(ln(−x))/dx = 1/x, motivating d(ln|x|)/dx = 1/x for x ≠ 0.

5:14 A Denominator Substitution Produces a Natural Log

For ∫ 3x/(4 − x²) dx, choose the entire denominator u = 4 − x², giving du = −2x dx.
Rewriting x dx as −du/2 reduces the integral to −3/2 ∫ du/u.
Substitution back gives −(3/2) ln|4 − x²| + C, illustrating how a logarithm can emerge from a rational integrand.

10:06 Practice with Tangent and a Logarithm-over-x Integral

Rewrite tan x as sin x/cos x, then set u = cos x to evaluate ∫ tan x dx = −ln|cos x| + C.
For ∫ (ln x)/x dx, setting u = ln x and du = dx/x gives ∫u du = ½(ln x)² + C.
The second substitution works because the derivative of ln x appears in the integrand; choosing u = x would leave the original integral unchanged.

20:26 Why the Natural Logarithm of One Is Zero

Using the integral definition, ln 1 = ∫₁¹ 1/t dt.
An integral with identical upper and lower bounds has zero width, so ln 1 = 0.
This establishes the identity directly from the area interpretation rather than from ln being the inverse of the exponential.

23:42 Deriving the Product Rule for Natural Logarithms

Treat a as constant and differentiate ln(ab) with respect to b; the chain rule gives (1/(ab))·a = 1/b.
Since ln b also has derivative 1/b, ln(ab) and ln b differ by a constant.
Setting b = 1 and using ln 1 = 0 identifies the constant as ln a, proving ln(ab) = ln a + ln b.
Prof Staecker connects this multiplication-to-addition property to the historical usefulness of logarithms and the slide rule.

32:45 Logarithm Rules for Reciprocals, Quotients, and Powers

From ln(a/a) = ln 1 = 0 and the product rule, ln(1/a) = −ln a.
Writing a/b as a·(1/b) gives ln(a/b) = ln a − ln b.
Repeated products yield ln(aⁿ) = n ln a; the rule extends beyond whole-number exponents when the logarithms are defined.

38:42 Logarithmic Differentiation Replaces Product and Quotient Rules

Prof Staecker introduces logarithmic differentiation as a way to avoid applying product and quotient rules directly to complicated expressions.
For y = x²(7 + x)¹⁰/(5x² − 7x + 4), take ln of both sides, simplify with the logarithm rules, differentiate, and solve for dy/dx.
Taking the logarithm turns the numerator product into a sum and the overall quotient into a subtraction.

42:03 Simplifying the Logarithm of the Worked Function

The quotient rule for logarithms separates the denominator: ln y = ln[x²(7 + x)¹⁰] − ln(5x² − 7x + 4).
The product and power rules give ln y = 2 ln x + 10 ln(7 + x) − ln(5x² − 7x + 4).
The sum inside the denominator polynomial cannot be split into separate logarithms; only products and quotients split this way.

46:16 Implicit Differentiation and the Final Derivative

Differentiating ln y introduces (1/y)(dy/dx) on the left by the chain rule.
Differentiating the simplified right side gives 2/x + 10/(7 + x) − (10x − 7)/(5x² − 7x + 4).
Multiplying by y and substituting the original expression yields dy/dx = [x²(7 + x)¹⁰/(5x² − 7x + 4)]·[2/x + 10/(7 + x) − (10x − 7)/(5x² − 7x + 4)].

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Prof Staecker.