Tags: general exponential, a^x, derivative of a^x, log base a, logarithmic differentiation, variable base and exponent, L'Hopital's Rule, indeterminate forms, 0/0, infinity/infinity, Calculus, Texas A&M, Chapter 5
Section 5.5 extends differentiation and integration to exponential and logarithmic functions with bases other than e (such as 2^x, 9^(10t), log base 6). Section 5.6 covers L'Hopital's Rule for evaluating limits that produce indeterminate forms like 0/0, ∞/∞, 0·∞, 1^∞, and ∞^0. Together these sections round out the toolkit for handling any exponential, logarithmic, or limit problem on the exam.
General exponential function, a^x
Any exponential function with a positive base a ≠ 1. Can be rewritten as e^(x ln a).
General logarithmic function, log_a(x)
The inverse of a^x. Related to ln by the change-of-base formula: log_a(x) = ln(x)/ln(a).
L'Hopital's Rule
If lim f(x)/g(x) gives 0/0 or ±∞/±∞, then the limit equals lim f'(x)/g'(x), provided that second limit exists (or is ±∞). Can be applied repeatedly.
Indeterminate form
An expression whose limit cannot be determined from the form alone. The seven classic types: 0/0, ∞/∞, 0·∞, ∞ − ∞, 0⁰, 1^∞, ∞⁰.
d/dx [a^x] = a^x · ln(a)
d/dx [a^u] = a^u · ln(a) · u'
d/dx [log_a(x)] = 1/(x ln a)
d/dx [log_a(u)] = u'/(u ln a)
Product rule with general exponentials
For f(t) = t³ · 9^(10t):
Use the product rule
f'(t) = 3t² · 9^(10t) + t³ · 9^(10t) · 10 ln(9)
f'(t) = 3t² · 9^(10t) + 10 ln(9) · t³ · 9^(10t)
Logarithm of a quotient
For f(x) = log_6((x² − 7)/(x − 3)):
Rewrite using change of base: f(x) = ln((x² − 7)/(x − 3)) / ln(6)
Use the quotient rule on the inside
f'(x) = (x² − 6x + 7) / [(x² − 7)(x − 3) ln(6)]
Tangent line to y = 2^(−x) at (−1, 2)
dy/dx = 2^(−x) · (−ln 2)
At x = −1: slope = 2^(1) · (−ln 2) = −2 ln 2
Tangent line: y − 2 = −2 ln 2 (x + 1), i.e. y = 2(1 − ln 2)(x + 1)
Logarithmic differentiation for variable base and exponent
For y = x^(2x):
Take ln: ln y = 2x ln x
Differentiate: (1/y) dy/dx = 2 ln x + 2x · (1/x) = 2 ln x + 2
dy/dx = x^(2x) · 2(ln x + 1)
∫ a^x dx = a^x / ln(a) + C
Integral of 8^(3x)
Let u = 3x, du = 3 dx
(1/3) ∫ 8^u du = (1/3) · 8^u / ln(8) + C = 8^(3x) / (3 ln 8) + C
Integral of x³ · 7^(−x⁴)
Let u = −x⁴, du = −4x³ dx
(−1/4) ∫ 7^u du = (−1/4) · 7^u / ln(7) + C = −7^(−x⁴) / (4 ln 7) + C
The rule applies when direct substitution gives 0/0 or ∞/∞. For other indeterminate forms, you must first rewrite the expression into one of those two forms.
Simple algebraic limit (0/0)
lim as x→6 of −5(x − 6)/(x² − 36):
Factor: x² − 36 = (x − 6)(x + 6)
Cancel: −5/(x + 6), then substitute x = 6
Result: −5/12
L'Hopital gives the same: differentiate top and bottom to get −5/(2x), evaluate at 6: −5/12
Repeated application
lim as x→0⁺ of 9(e^x − 1 − x)/(4x³):
Form is 0/0 at x = 0
First application: 9(e^x − 1)/(12x²), still 0/0
Second application: 9e^x/(24x), still 0/0... wait, at x→0⁺ the denominator→0⁺ and numerator→9, so this is 9/0⁺ = +∞
Result: ∞
Logarithmic growth versus polynomial growth
lim as x→∞ of ln(x⁶)/x⁸:
Form is ∞/∞
Rewrite ln(x⁶) = 6 ln x
L'Hopital: (6/x)/(8x⁷) = 6/(8x⁸) → 0
Result: 0. Logarithms always lose to polynomials at infinity.
0·∞ form (rewrite as a fraction)
lim as x→∞ of 10x sin(7/x):
Form is ∞ · 0
Rewrite as 10 sin(7/x)/(1/x), which is 0/0 as x→∞
L'Hopital: 10 cos(7/x) · (−7/x²) / (−1/x²) = 70 cos(7/x)
As x→∞: 70 cos(0) = 70
1^∞ form (use logarithms)
lim as x→0⁺ of (e^x + 2x)^(10/x):
Let L = lim of (10/x) · ln(e^x + 2x)
As x→0⁺: ln(e^x + 2x) → ln(1) = 0, and 10/x → ∞, so L is ∞ · 0
Rewrite L = 10 ln(e^x + 2x)/x, form 0/0
L'Hopital: 10 · (e^x + 2)/(e^x + 2x), evaluate at 0: 10 · 3/1 = 30
Final answer: e^30
Expression | Result |
|---|---|
d/dx [a^x] | a^x ln(a) |
d/dx [log_a(x)] | 1/(x ln a) |
∫ a^x dx | a^x / ln(a) + C |
d/dx [x^(g(x))] | Use logarithmic differentiation |
L'Hopital's Rule: if lim f/g is 0/0 or ∞/∞, then lim f/g = lim f'/g' (when the right side exists).
Rewriting strategies for other indeterminate forms:
0 · ∞: rewrite as f/(1/g) to get 0/0 or ∞/∞
1^∞, 0⁰, ∞⁰: take the natural log, evaluate the limit, then exponentiate
⚠️ The factor of ln(a) appears in every derivative and integral involving a^x. Forgetting it is one of the most common mistakes on this section.
⚠️ For y = x^(g(x)) (variable base AND exponent), you cannot use the power rule or the exponential rule alone. You must use logarithmic differentiation.
⚠️ L'Hopital's Rule only applies to 0/0 or ∞/∞. Before differentiating, always verify the form. If it is not indeterminate, L'Hopital does not apply.
⚠️ For the 1^∞ form, the answer is e^(limit of the log), not just the limit of the log. Students sometimes forget to exponentiate at the end.
⚠️ ln(x) grows more slowly than any positive power of x. This means lim ln(x^n)/x^k = 0 as x→∞ for any positive n and k.
Q: What is d/dx [a^x]?
A: a^x · ln(a).
Q: Find the derivative of f(t) = t³ · 9^(10t).
A: f'(t) = 3t² · 9^(10t) + 10 ln(9) · t³ · 9^(10t).
Q: Evaluate ∫ 8^(3x) dx.
A: 8^(3x) / (3 ln 8) + C.
Q: How do you differentiate y = x^(2x)?
A: Take ln of both sides: ln y = 2x ln x. Differentiate implicitly, then multiply by y. Result: dy/dx = 2x^(2x)(ln x + 1).
Q: What is lim as x→∞ of ln(x⁶)/x⁸?
A: 0. Logarithms grow slower than any polynomial.
Q: Evaluate lim as x→0⁺ of (e^x + 2x)^(10/x).
A: Take ln, apply L'Hopital to get the exponent limit of 30, then exponentiate: e^30.
Q: What are the indeterminate forms that require rewriting before L'Hopital applies?
A: 0·∞, ∞ − ∞, 0⁰, 1^∞, and ∞⁰. Rewrite them as fractions (0/0 or ∞/∞) first. For exponential forms, take the natural log.
general exponential, a^x derivative, a^x integral, log base a, change of base formula, logarithmic differentiation, variable base and exponent, x^x type, L'Hopital's Rule, L'Hospital's Rule, indeterminate forms, 0/0, infinity over infinity, 1 to the infinity, zero times infinity, limits, Calculus Chapter 5, Section 5.5, Section 5.6, Texas A&M Calculus