Calculus II ep08: The exponential function (Sep 24, 2026)

Overview

Prof Staecker uses logarithmic differentiation to simplify derivatives of products, quotients, and powers, then reviews the graph and integral definition of ln x to establish its domain, range, and limiting behavior. He defines e by ln e = 1, identifies e^x as the inverse of ln x, and practices switching between logarithmic and exponential equations to solve for x.

Key takeaways

Logarithmic differentiation changes products into sums and quotients into differences, making y = x²(2x + 1)(3x − 7)¹⁰/(1 + x²) manageable without repeated product and quotient rules.
When differentiating ln(g(x)), use the chain rule: the derivative is g′(x)/g(x), as in d/dx[ln(3x² − 7x)] = (6x − 7)/(3x² − 7x).
The integral definition ln x = ∫₁ˣ (1/t) dt gives ln x domain (0, ∞), range all real numbers, and limits −∞ at 0 from the right and ∞ as x grows without bound.
Continuity and ln 2 < 1 < ln 4 guarantee an input e between 2 and 4 for which ln e = 1; e is approximately 2.71828.
The inverse relationship ln(eˣ) = x means ln(A) = B can be rewritten as eᴮ = A, while eˣ = A can be rewritten as x = ln A.
To solve equations involving ln or eˣ, first isolate the logarithm or exponential; for example, 4eˣ⁻² = 7 leads to x = ln(7/4) + 2.

Chapters

0:00 Logarithmic Differentiation Simplifies a Product-and-Quotient Derivative

For y = x²(2x + 1)(3x − 7)¹⁰/(1 + x²), taking ln of both sides turns products into sums and the denominator into a subtraction.
The resulting derivative is y′ = y[2/x + 2/(2x + 1) + 30/(3x − 7) − 2x/(1 + x²)].
Substitute the original expression for y so the final derivative is written entirely in terms of x.

7:20 Practice Logarithmic Differentiation with Two Denominator Factors

For y = x⁴/[(2x − 1)(3x² − 7x)], logarithms produce 4 ln x − ln(2x − 1) − ln(3x² − 7x).
Applying the chain rule gives y′ = y[4/x − 2/(2x − 1) − (6x − 7)/(3x² − 7x)].
Both denominator factors contribute negative logarithm terms, and each inner derivative supplies a chain-rule factor.

13:42 Graphing ln x from Its Integral Definition

The definition ln x = ∫₁ˣ (1/t) dt explains why ln 1 = 0 and why ln x increases as x moves right.
For x < 1, reversing the integral’s limits makes ln x negative; the function is undefined at x = 0 and for negative x.
The domain is (0, ∞), the range is all real numbers, limₓ→∞ ln x = ∞, and limₓ→0⁺ ln x = −∞.

22:06 Using Logarithm Values and the Intermediate Value Theorem to Define e

ln 2 is approximately 0.693, and the log power rule gives ln 4 = ln(2²) = 2 ln 2, approximately 1.386.
Because ln x is continuous and its values lie below and above 1 between x = 2 and x = 4, it must equal 1 at an intermediate input.
That unique input is e, approximately 2.71828; by definition, ln e = 1.

27:50 Establishing e^x as the Inverse of ln x

The graph of the inverse of ln x is its reflection across the line y = x, exchanging the roles of inputs and outputs.
Using ln(aˣ) = x ln a and ln e = 1 gives ln(eˣ) = x.
Therefore the inverse function of ln x is eˣ, the natural exponential function.

35:00 Translating Logarithmic Equations into Exponential Form

Because ln and eˣ are inverse functions, ln(A) = B is equivalent to eᴮ = A.
For example, ln(2x + 7) = 5 converts to e⁵ = 2x + 7.
Writing the same relationship in logarithmic or exponential form can make a variable easier to isolate.

38:15 Solving for a Variable Inside a Natural Logarithm

For 12 = 2 ln(10x + 1), first divide by 2 to isolate the logarithm: ln(10x + 1) = 6.
Convert to exponential form to obtain e⁶ = 10x + 1, then solve x = (e⁶ − 1)/10.
Taking e to both sides is another way to describe the inverse operation that removes ln.

41:50 Solving Exponential Equations by Applying ln

For 4eˣ⁻² = 7, isolate the exponential first: eˣ⁻² = 7/4.
Convert to logarithmic form, x − 2 = ln(7/4), yielding x = ln(7/4) + 2.
The exponential should be isolated before taking ln; dividing by e does not remove a variable exponent.

43:47 Practice Switching Forms and Isolating x

For e²ˣ⁺¹ = 5, take ln to get 2x + 1 = ln 5, so x = (ln 5 − 1)/2.
For the final exercise, isolate ln(3 + x) = 2 before converting to exponential form, giving x = e² − 3.
The examples reinforce a consistent order: isolate the logarithm or exponential, apply its inverse, then solve the remaining linear equation.

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.