Calculus II ep09: Other exponential functions (Sep 28, 2026)

Overview

Prof Staecker connects exponential and logarithmic functions to derive exponent laws, the derivative and antiderivative of e^x, and the definition of exponential functions with arbitrary positive bases. The lecture also derives e = limₙ→∞(1 + 1/n)ⁿ from the derivative of ln x, then uses b^x = e^(x ln b) to establish exponent rules and the derivative formula for b^x.

Key takeaways

The identity e^(x+y) = e^x e^y follows directly from the logarithm product rule: take ln of the product, simplify to x + y, then convert back to exponential form.
Implicitly differentiating ln y = x proves that y = e^x satisfies dy/dx = e^x, explaining why e^x is its own derivative.
For ∫e^(kx) dx, a constant multiplier k in the exponent produces the factor 1/k, giving e^(kx)/k + C when k ≠ 0.
The limit e = limₙ→∞(1 + 1/n)ⁿ can be derived from f′(1) for f(x) = ln x by rewriting limₕ→₀ ln(1+h)/h = 1 in exponential form.
Defining b^x as e^(x ln b) gives real meaning to irrational powers such as 2^π and requires b > 0.
For any positive base b, the derivative is (b^x)′ = b^x ln b; with an inner function g(x), it becomes b^g(x) ln b · g′(x).

Chapters

0:00 Using Logarithmic Form to Derive the Exponent-Sum Rule

The equivalence y = ln x and e^y = x lets equations switch between logarithmic and exponential forms.
Taking ln of e^x e^y turns the product into ln(e^x) + ln(e^y) = x + y.
Converting back to exponential form proves e^(x+y) = e^x e^y.

7:30 Implicit Differentiation Shows That the Derivative of e^x Is e^x

Rewrite y = e^x as ln y = x before differentiating implicitly.
The chain rule gives (1/y)(dy/dx) = 1, so dy/dx = y = e^x.
For e^g(x), preserve e^g(x) and multiply by g′(x); for example, differentiating e^(3x²+7x) adds the factor 6x + 7.

13:10 Applying the Chain, Quotient, and Product Rules to e-Based Functions

For e^(sin x), the chain rule gives e^(sin x) cos x.
Treating √x as x^(1/2), the derivative of e^(√x) is e^(√x)/(2√x).
For e^(3x/(x²+1)), differentiate the exponent with the quotient rule, then multiply by the unchanged exponential.
For x e^(5x), combine the product rule and chain rule to get e^(5x)(5x + 1).

21:20 Integrating e^(x³−1) with a u-Substitution

The antiderivative of e^u is e^u + C because the derivative of e^u is itself.
For ∫x²e^(x³−1) dx, choose u = x³ − 1, giving du = 3x² dx.
Substitution yields (1/3)∫e^u du, so the result is (1/3)e^(x³−1) + C.

26:10 A Shortcut for Integrating e^(kx)

Differentiating e^(kx) produces k e^(kx) by the chain rule.
Therefore, for nonzero constant k, ∫e^(kx) dx = e^(kx)/k + C.
For example, ∫e^(7x) dx = (1/7)e^(7x) + C; a nonconstant exponent generally requires u-substitution.

30:00 Deriving the Limit Definition of e from the Derivative of ln x

Start with f(x) = ln x, for which f′(x) = 1/x and f′(1) = 1.
The derivative definition at x = 1 gives limₕ→₀ ln(1+h)/h = 1.
Rewrite the quotient as ln((1+h)^(1/h)) and exponentiate to obtain e = limₕ→₀(1+h)^(1/h).
Setting h = 1/n converts the result to e = limₙ→∞(1 + 1/n)ⁿ.

42:00 Why Real Exponents Need More Than Repeated Multiplication

Integer exponents such as 2³ mean repeated multiplication, while 2⁻³ means 1/2³.
Rational exponents use roots: 2^(1/3) is the cube root of 2, and 2^(5/4) is the fourth root of 2 raised to the fifth power.
An irrational exponent such as 2^π cannot be defined by repeated multiplication or a rational root alone.

45:00 Defining 2^π by Rational Approximation and Continuity

Approximate π with rational numbers such as 3.1, 3.14, and 3.1415, whose powers of 2 can be interpreted using roots.
Define 2^π as the limit of 2 raised to rational exponents that approach π.
This interpretation matches the idea that the graph of 2^x is continuous, rather than containing holes at irrational inputs.

49:00 Defining Any Positive-Base Exponential as e^(x ln b)

Use the identity b^x = e^(x ln b) to define real powers for any base b > 0.
For example, 2^π means e^(π ln 2), an expression defined using the natural exponential.
The base must be positive because ln b is not defined for zero or negative real b.

53:30 Four Exponent Laws for the Function b^x

The laws include b^(x+y) = b^x b^y and b^(x−y) = b^x/b^y.
A power raised to a power satisfies (b^x)^r = b^(xr).
For positive a and b, a product in the base distributes as (ab)^x = a^x b^x.

56:00 Proving b^(x+y) = b^x b^y from the Definition

Substitute the definition into the left side: b^(x+y) = e^((x+y)ln b).
Distribute to get e^(x ln b + y ln b), then use the exponent-sum rule to split it into a product.
Recognize e^(x ln b) and e^(y ln b) as b^x and b^y, respectively.

1:00:30 Beginning a Definition-Based Proof of (ab)^x

The exercise asks students to prove (ab)^x = a^x b^x using the definition of real exponentiation.
The proof begins by rewriting (ab)^x as e^(x ln(ab)).
The key next step is to apply ln(ab) = ln a + ln b.

1:06:10 Completing the Product-Base Exponent Proof

Expand e^(x ln(ab)) to e^(x(ln a + ln b)), then distribute x across the logarithmic sum.
Use e^(A+B) = e^A e^B to split the expression into e^(x ln a)e^(x ln b).
By definition, the factors are a^x and b^x, proving (ab)^x = a^x b^x.

1:09:00 Differentiating b^x and Applying the Chain Rule

Differentiate b^x = e^(x ln b), treating ln b as a constant, to obtain (b^x)′ = b^x ln b.
When b = e, ln e = 1, so the formula reduces to (e^x)′ = e^x.
For 4^(3x²−2x), multiply 4^(3x²−2x) by ln 4 and by the inner derivative 6x − 2.

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.