Tags: natural log, ln, derivative of ln x, implicit differentiation, logarithmic differentiation, exponential function, e^x, chain rule, u-substitution, inverse functions, Calculus, Texas A&M, Chapter 5
Sections 5.1–5.4 cover how to differentiate and integrate functions involving ln(x) and e^x. You will need the derivative formulas for natural log and exponential functions, implicit differentiation with logarithms, logarithmic differentiation for complicated products/quotients, u-substitution for log and exponential integrals, the Fundamental Theorem of Calculus with variable upper limits, and how to find and differentiate inverse functions.
Natural logarithm, ln(x)
The inverse of e^x. Defined for x > 0. Its derivative is 1/x.
Implicit differentiation
A technique for finding dy/dx when y is not isolated. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
Logarithmic differentiation
Take ln of both sides of y = f(x), use log rules to simplify, then differentiate implicitly. Especially useful when the function is a product/quotient of many factors or has a variable in both the base and the exponent.
Inverse function, f⁻¹(x)
The function that "undoes" f. If f(a) = b, then f⁻¹(b) = a. Only exists when f is one-to-one.
Derivative of an inverse function
(f⁻¹)'(a) = 1 / f'(f⁻¹(a)). You find what x-value maps to a under f, then take the reciprocal of f' at that x-value.
u-substitution
A method for evaluating integrals by substituting u = g(x), so du = g'(x) dx. Converts a complicated integral into a simpler one in terms of u.
Fundamental Theorem of Calculus (with variable limits)
If F(x) = ∫ from a to g(x) of f(t) dt, then F'(x) = f(g(x)) · g'(x). The chain rule applies when the upper limit is a function of x.
d/dx [ln(x)] = 1/x
d/dx [ln(u)] = u'/u (chain rule version)
d/dx [ln|x|] = 1/x
Tangent lines involving ln
To find the tangent line to y = ln(x⁵) at (1, 0):
Rewrite as y = 5 ln(x), so y' = 5/x
At x = 1: slope = 5
Tangent line: y = 5(x − 1)
Derivative of y = ln√(x² − 15)
Rewrite as y = (1/2) ln(x² − 15)
dy/dx = (1/2) · 2x/(x² − 15) = x/(x² − 15)
Implicit differentiation with logarithms
For 6x² + 5 ln(xy) = 13:
Differentiate: 12x + 5 · (1/(xy)) · (y + x·dy/dx) = 0
Solve for dy/dx
Answer: dy/dx = −y(12x² + 5)/(5x)
Relative extrema and inflection points
For y = x⁶ · ln(x/6):
Find y', set it to zero for critical points
Find y'', set it to zero for inflection points
Relative minimum at x = 6e^(−1/6)
Inflection point at x = 6e^(−11/30)
Logarithmic differentiation (complex products/quotients)
For y = x⁵² · √(99x − 24) / (x − 1)⁴³:
Take ln of both sides
ln y = 52 ln x + (1/2) ln(99x − 24) − 43 ln(x − 1)
Differentiate, then multiply both sides by y
dy/dx = y · [52/x + 99/(2(99x − 24)) − 43/(x − 1)]
∫ (1/x) dx = ln|x| + C
∫ (u'/u) dx = ln|u| + C
Integral of x²/(8x³ + 5)
Let u = 8x³ + 5, du = 24x² dx
(1/24) ∫ du/u = (1/24) ln|8x³ + 5| + C
Polynomial division before integrating
For ∫ (x² − 6x + 18)/(x + 8) dx:
Perform long division: x² − 6x + 18 = (x + 8)(x − 14) + 130
∫ (x − 14 + 130/(x + 8)) dx = (1/2)x² − 14x + 130 ln|x + 8| + C
Nested logarithmic integrals
For ∫ 1/(x ln(x⁹)) dx:
Note ln(x⁹) = 9 ln(x), so the integral becomes (1/9) ∫ 1/(x ln x) dx
Let u = ln x, du = (1/x) dx
(1/9) ln|ln(x⁹)| + C
Definite integral with substitution
∫ from 1 to e of (1 + ln x)³/x dx:
Let u = 1 + ln x, du = (1/x) dx
Bounds: x = 1 gives u = 1, x = e gives u = 2
∫ from 1 to 2 of u³ du = [u⁴/4] from 1 to 2 = 16/4 − 1/4 = 15/4
Fundamental Theorem with variable upper limit
For F(x) = ∫ from 1 to 2x of (7/t) dt:
F'(x) = (7/(2x)) · 2 = 7/x
Finding f⁻¹(x)
For f(x) = x³ − 3:
Set y = x³ − 3, swap x and y: x = y³ − 3
Solve: y = (x + 3)^(1/3)
f⁻¹(x) = (x + 3)^(1/3)
Derivative of an inverse at a point
For f(x) = (x + 9)/(x + 4), x > −4, find (f⁻¹)'(2):
First find x such that f(x) = 2: (x + 9)/(x + 4) = 2, so x + 9 = 2x + 8, giving x = 1
f'(x) = [(x + 4) − (x + 9)]/(x + 4)² = −5/(x + 4)²
f'(1) = −5/25 = −1/5
(f⁻¹)'(2) = 1/f'(1) = −5
d/dx [e^x] = e^x
d/dx [e^u] = e^u · u'
∫ e^x dx = e^x + C
∫ e^u du = e^u + C
Solving logarithmic equations
For ln(x⁻⁸) = 6:
−8 ln x = 6, so ln x = −3/4
x = e^(−3/4), equivalently the 8th root of e^(−6)
Chain rule with exponentials
For y = e^(9x⁸):
dy/dx = e^(9x⁸) · 72x⁷ = 72x⁷ e^(9x⁸)
Quotient rule with exponentials
For f(x) = (e^x + 4)/(e^x − 4):
f'(x) = [e^x(e^x − 4) − e^x(e^x + 4)]/(e^x − 4)²
= −8e^x/(e^x − 4)²
Fundamental Theorem (exponential upper limit)
For G(x) = ∫ from 0 to e^(4x) of ln(t + 7) dt:
G'(x) = ln(e^(4x) + 7) · 4e^(4x)
Implicit differentiation with exponentials
For 5e^(xy) − y² = 7:
5e^(xy)(y + x·dy/dx) − 2y·dy/dx = 0
dy/dx = −5ye^(xy)/(5xe^(xy) − 2y)
Exponential integrals with u-substitution
For ∫ 3x·e^(−2x²) dx:
Let u = −2x², du = −4x dx
(3/(−4)) ∫ e^u du = −(3/4)e^(−2x²) + C
Definite exponential integral
∫ from ln 2 to ln 3 of e^(−x) dx:
= [−e^(−x)] from ln 2 to ln 3
= −e^(−ln 3) + e^(−ln 2) = −1/3 + 1/2 = 1/6
Function | Derivative |
|---|---|
ln(x) | 1/x |
ln(u) | u'/u |
e^x | e^x |
e^u | e^u · u' |
Function | Integral |
|---|---|
1/x | ln|x| + C |
e^x | e^x + C |
e^u · u' | e^u + C |
Inverse function derivative formula: (f⁻¹)'(a) = 1/f'(f⁻¹(a))
FTC with variable limit: d/dx [∫ from a to g(x) of f(t) dt] = f(g(x)) · g'(x)
⚠️ When differentiating ln(something), the chain rule produces a fraction with "something" in the denominator and its derivative in the numerator. Students often forget the chain rule step.
⚠️ Logarithmic differentiation is the go-to method when you see products/quotients with many factors raised to powers. Take ln first, simplify with log rules, then differentiate.
⚠️ For integrals like ∫ u'/u dx, the answer is ln|u| + C. Train yourself to spot this pattern quickly.
⚠️ When the upper limit of an integral is a function of x (not just x itself), you must multiply by the derivative of that upper limit (chain rule on the FTC).
⚠️ For (f⁻¹)'(a), the most common mistake is evaluating f' at a instead of at f⁻¹(a). Always find the x-value first, then take the reciprocal of f' there.
⚠️ Polynomial long division is required when the degree of the numerator is ≥ the degree of the denominator. Do the division first, then integrate term by term.
Q: What is the derivative of y = ln√(x² − 15)?
A: Rewrite as (1/2) ln(x² − 15). Then dy/dx = x/(x² − 15).
Q: How do you find dy/dx for 6x² + 5 ln(xy) = 13?
A: Differentiate implicitly. Use the chain rule on ln(xy) with the product rule inside. The result is dy/dx = −y(12x² + 5)/(5x).
Q: Find ∫ x²/(8x³ + 5) dx.
A: Let u = 8x³ + 5, du = 24x² dx. The integral becomes (1/24) ln|8x³ + 5| + C.
Q: If F(x) = ∫ from 1 to 2x of (7/t) dt, what is F'(x)?
A: By the FTC with chain rule: F'(x) = (7/(2x)) · 2 = 7/x.
Q: Find the inverse of f(x) = x³ − 3.
A: f⁻¹(x) = (x + 3)^(1/3).
Q: Evaluate ∫ from ln 2 to ln 3 of e^(−x) dx.
A: The antiderivative is −e^(−x). Evaluating: −1/3 + 1/2 = 1/6.
natural logarithm, ln x, derivative of ln, log differentiation, logarithmic differentiation, implicit differentiation with ln, chain rule, e^x derivative, exponential integral, u-substitution, integral of 1/u, Fundamental Theorem of Calculus, FTC variable limits, inverse function, derivative of inverse, polynomial long division before integrating, Calculus Chapter 5, Section 5.1, Section 5.2, Section 5.3, Section 5.4, Texas A&M Calculus