Source: AB Calculus Unit 3 Notes, Texas A&M University
Tags: derivative of ln x, derivative of e^x, natural log rules, logarithmic properties, base a exponential, log base a derivative, linear approximation, tangent line approximation, local linearisation, overestimate, underestimate, concavity, L'Hopital's Rule, indeterminate form, 0/0, infinity/infinity
This section covers the derivative rules for ln x, e^x, and general bases (a^x, log_a x), then applies tangent lines to approximate function values (linear approximation), and finally introduces L'Hopital's Rule for evaluating limits that produce indeterminate forms like 0/0 or ±∞/∞.
Natural logarithm (ln x)
The logarithm base e. Its derivative is 1/x (for x > 0). With the chain rule: d/dx [ln u] = u'/u.
Euler's number (e)
The irrational constant approximately equal to 2.71828. The function e^x is its own derivative.
Logarithmic properties
The identities from algebra that simplify log expressions before differentiating: ln 1 = 0, ln(ab) = ln a + ln b, ln(aⁿ) = n ln a, ln(a/b) = ln a − ln b, ln e = 1, e^(ln a) = a.
Linear approximation (tangent line approximation, local linearisation)
Using the tangent line at a known point to estimate the value of a function at a nearby point. The formula is L(x) = f(a) + f'(a)(x − a).
Overestimate vs underestimate
If the curve is concave up (f''(x) > 0) near the point of tangency, the tangent line lies below the curve, so the approximation is an underestimate. If concave down (f''(x) < 0), the tangent line lies above, giving an overestimate.
Indeterminate form
An expression like 0/0 or ∞/∞ whose value cannot be determined without further work. L'Hopital's Rule applies to these.
L'Hopital's Rule
If lim(x→a) f(x)/g(x) yields 0/0 or ±∞/∞, then lim(x→a) f(x)/g(x) = lim(x→a) f'(x)/g'(x), provided the latter limit exists.
Starting from y = ln x, rewrite as e^y = x, then implicitly differentiate: e^y · dy/dx = 1, so dy/dx = 1/e^y = 1/x.
d/dx [ln x] = 1/x (x > 0)
Chain rule version: d/dx [ln u] = (1/u) · du/dx = u'/u
ln 1 = 0
ln(ab) = ln a + ln b
ln(aⁿ) = n · ln a
ln(a/b) = ln a − ln b
ln(e) = 1
e^(ln a) = a
These let you simplify before differentiating. For example, d/dx [ln x³] = d/dx [3 ln x] = 3/x, which is easier than applying the chain rule to ln(x³) directly (though both give the same result).
d/dx [e^x] = e^x
The exponential function base e is the only function (up to constant multiples) that is its own derivative.
Chain rule version: d/dx [e^u] = e^u · du/dx
Examples: d/dx [e^(2x³ − 3x)] = e^(2x³ − 3x) · (6x² − 3). For e^(1/x), the derivative is e^(1/x) · (−1/x²).
d/dx [a^x] = (ln a) · a^x
Chain rule version: d/dx [a^u] = (ln a) · a^u · du/dx
d/dx [log_a x] = 1 / ((ln a) · x)
Chain rule version: d/dx [log_a u] = 1 / ((ln a) · u) · du/dx
The factor ln a is what distinguishes base-a from base-e. When a = e, ln a = 1 and you recover the standard formulas.
Example: d/dx [2^(3x)] = (ln 2) · 2^(3x) · 3 = 3(ln 2) · 2^(3x).
The tangent line at x = a gives a local linear model for f:
L(x) = f(a) + f'(a)(x − a)
For values of x close to a, L(x) ≈ f(x).
Worked example: f(x) = √x, approximate f(4.1).
At a = 4: f(4) = 2, f'(x) = 1/(2√x), so f'(4) = 1/4.
L(x) = 2 + (1/4)(x − 4).
L(4.1) = 2 + (1/4)(0.1) = 2.025.
The true value √4.1 ≈ 2.0248, so the approximation is very close.
The concavity of f at the point of tangency determines the answer:
Concave up (f'' > 0): tangent line sits below the curve, so the linear approximation is an underestimate.
Concave down (f'' < 0): tangent line sits above the curve, so the linear approximation is an overestimate.
For √x at x = 4: f''(x) = −1/(4x^(3/2)) < 0, so the graph is concave down, and 2.025 is an overestimate.
Statement: If lim(x→a) f(x)/g(x) gives 0/0 or ±∞/∞, then:
lim(x→a) f(x)/g(x) = lim(x→a) f'(x)/g'(x)
provided the right-hand limit exists (or is ±∞). The rule can be applied repeatedly if the new limit is still indeterminate.
You must verify the indeterminate form before applying the rule. If the limit is not 0/0 or ±∞/∞, L'Hopital's Rule does not apply and will give a wrong answer.
lim(x→0) (e^(3x) − 1)/x: Direct substitution gives 0/0. Differentiate top and bottom: lim(x→0) 3e^(3x)/1 = 3.
lim(x→0) (√(1+x) − 1)/x: Direct substitution gives 0/0. Differentiate: lim(x→0) [1/(2√(1+x))] / 1 = 1/2.
lim(x→0) (2 tan x)/(5x): Gives 0/0. Differentiate: lim(x→0) (2 sec²x)/5 = 2/5.
lim(x→∞) (ln x)/(2√x): Gives ∞/∞. Differentiate: lim(x→∞) (1/x)/(1/√x) = lim(x→∞) 1/√x = 0.
L'Hopital's Rule works directly only for 0/0 and ±∞/∞. Other indeterminate forms (0^0, 1^∞, ∞ − ∞, 0 · ∞, ∞^0) must first be rewritten into a fraction that produces 0/0 or ∞/∞.
Common technique for 0 · ∞: rewrite f · g as f / (1/g) or g / (1/f) to create a fraction.
Common technique for ∞ − ∞: combine into a single fraction with a common denominator.
Logarithmic and exponential derivative formulas
d/dx [ln x] = 1/x
d/dx [ln u] = u'/u
d/dx [e^x] = e^x
d/dx [e^u] = e^u · u'
d/dx [a^x] = (ln a) · a^x
d/dx [a^u] = (ln a) · a^u · u'
d/dx [log_a x] = 1/((ln a) · x)
d/dx [log_a u] = u'/((ln a) · u)
Linear approximation formula
L(x) = f(a) + f'(a)(x − a)
L'Hopital's Rule
If 0/0 or ±∞/∞: lim f(x)/g(x) = lim f'(x)/g'(x)
⚠️ The derivative of ln x is 1/x, not 1/(ln x). This is a very common slip.
⚠️ Remember to use log properties to simplify before differentiating when possible. d/dx [ln x³] is easier as d/dx [3 ln x] = 3/x.
⚠️ For bases other than e, the ln a factor is easy to forget. d/dx [2^x] = (ln 2) · 2^x, not just 2^x.
⚠️ On the AP exam, linear approximation questions almost always ask whether the result is an overestimate or underestimate. Check the sign of f'' at the point of tangency.
⚠️ L'Hopital's Rule requires you to show the indeterminate form first. Writing "by L'Hopital's Rule" without demonstrating 0/0 or ∞/∞ will lose points on the free-response section.
⚠️ L'Hopital's Rule differentiates numerator and denominator separately. You are not using the quotient rule.
Q: Find dy/dx for y = x · ln(sin x).
A: Product rule: dy/dx = ln(sin x) + x · (cos x / sin x) = ln(sin x) + x cot x.
Q: Find dy/dx for y = (ln x)³.
A: Chain rule: dy/dx = 3(ln x)² · (1/x) = 3(ln x)² / x.
Q: Find the derivative of y = 2^(3x).
A: dy/dx = (ln 2) · 2^(3x) · 3 = 3(ln 2) · 2^(3x).
Q: The function f is twice differentiable with f(2) = 1, f'(2) = 4, and f''(2) = 3. Approximate f(1.9) using the tangent line at x = 2. Is this an overestimate or underestimate?
A: L(1.9) = 1 + 4(1.9 − 2) = 1 + 4(−0.1) = 0.6. Since f''(2) = 3 > 0, the graph is concave up at x = 2, so the tangent line is below the curve and 0.6 is an underestimate.
Q: Evaluate lim(x→∞) (ln x) / (2√x).
A: Direct substitution gives ∞/∞. Apply L'Hopital's Rule: lim (1/x) / (1/√x) = lim √x / x = lim 1/√x = 0. The logarithm grows slower than any positive power of x.
Q: Evaluate lim(x→0) (e^(3x) − 1) / x.
A: Direct substitution gives 0/0. L'Hopital's: lim 3e^(3x) / 1 = 3.
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