Difficulty: Intermediate | Prerequisites: Basic differentiation rules (power, product, quotient, chain rule), familiarity with ln(x) and e^x.
Exponential and logarithmic functions sit at the heart of calculus because they model growth, decay, and compounding processes. This review covers differentiating expressions built from ln(x) and e^x, including composite functions via the chain rule, products involving logarithms, implicit differentiation when ln(y) appears, and the Second Fundamental Theorem of Calculus applied to integrals with variable upper limits. You should already be comfortable with the chain rule, product rule, and basic properties of logarithms before diving in. These techniques appear repeatedly in later calculus topics (integration by parts, differential equations, series), so solid fluency here pays dividends all term.
To differentiate logarithmic and exponential expressions, you lean heavily on the chain rule: d/dx[ln(u)] = u'/u and d/dx[e^u] = e^u · u'. When a product involves ln(x), use the product rule. When ln(y) appears inside an implicit equation, differentiate both sides with respect to x, solve for dy/dx, and express the result in terms of x and y.
Chain rule (for logarithms)
If y = ln(u), then dy/dx = (1/u) · du/dx. In simple terms, differentiate the inside, then divide by the inside.
Product rule
If y = f(x) · g(x), then y' = f'(x)g(x) + f(x)g'(x). Think of it as: each factor takes a turn being differentiated while the other stays put.
Implicit differentiation
A technique for finding dy/dx when y is not isolated on one side of the equation. You differentiate every term with respect to x, attach dy/dx to every term that contains y, then solve algebraically for dy/dx.
Second Fundamental Theorem of Calculus (2nd FTC)
If g(x) = integral from a to h(x) of f(t) dt, then g'(x) = f(h(x)) · h'(x). In simple terms, plug the upper limit into the integrand and multiply by the derivative of the upper limit.
d/dx[ln(x² + 1)] = 2x / (x² + 1). The inside function is u = x² + 1, so u' = 2x, and the result is u'/u.
d/dx[ln(∛(x² + 1))] = d/dx[(1/3) ln(x² + 1)] = (1/3) · 2x/(x² + 1) = 2x / [3(x² + 1)]. Use log properties to simplify before differentiating. The cube root becomes a 1/3 exponent, which pulls out front of the logarithm.
For y = x³ ln(x), apply the product rule: y' = 3x² ln(x) + x³ · (1/x) = 3x² ln(x) + x². This factors as x²(1 + 3 ln x).
The key move: the derivative of ln(x) is 1/x, which often simplifies nicely when multiplied by a power of x.
For g(x) = integral from 5 to ln(2x) of (3t + 4) dt, the 2nd FTC gives g'(x) = [3 ln(2x) + 4] · d/dx[ln(2x)].
d/dx[ln(2x)] = 1/(2x) · 2 = 1/x.
So g'(x) = [3 ln(2x) + 4] / x.
The pattern: substitute the upper limit into the integrand, then multiply by the derivative of that upper limit.
Given x³ + 5 ln(y) + y³ = 2, differentiate both sides with respect to x.
3x² + 5 · (1/y) · dy/dx + 3y² · dy/dx = 0.
Factor out dy/dx: dy/dx · (5/y + 3y²) = -3x².
Solve: dy/dx = -3x² / (5/y + 3y²) = -3x²y / (5 + 3y³).
Remember that d/dx[ln(y)] = (1/y) · dy/dx. The chain rule still applies because y is a function of x.
\frac{d}{dx}[\ln(u)] = \frac{u'}{u}\frac{d}{dx}[e^u] = e^u \cdot u'\frac{d}{dx}\left[\int_a^{h(x)} f(t)\,dt\right] = f(h(x)) \cdot h'(x)\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x)Log property shortcut: ln(u^n) = n ln(u). Rewriting before differentiating often makes the chain rule cleaner.
Logarithmic differentiation is how engineers and scientists analyse exponential growth and decay models: population dynamics, radioactive half-life calculations, and continuously compounded interest all rely on d/dx[ln(u)] and d/dx[e^u]. The implicit differentiation technique extends to any situation where two quantities are tangled together in an equation, which is standard in thermodynamics (PV = nRT) and economics (supply-demand equilibria).
Students often write d/dx[ln(x² + 1)] = 1/(x² + 1) and forget the chain rule. The correct answer requires multiplying by the derivative of the inside: 2x/(x² + 1).
Confusing the product rule with the chain rule when differentiating x³ ln(x). This is a product of two functions, not a composition, so the product rule applies.
Forgetting to multiply by dy/dx when differentiating terms containing y in implicit differentiation. Every time you differentiate a y-term, dy/dx must appear.
Neglecting to multiply by h'(x) when applying the 2nd FTC. If the upper limit is anything other than plain x, its derivative is not 1 and cannot be ignored.
⚠️ Chain rule with ln(u) is tested in nearly every exam on this material. Expect at least one problem where the inside function is not just x.
⚠️ Product rule combined with ln(x) is a classic exam question. Know how to factor the result neatly.
⚠️ Implicit differentiation with logarithms is a common "show your work" problem. Partial credit depends on correctly attaching dy/dx to every y-term.
⚠️ The 2nd FTC with a non-trivial upper limit (like ln(2x)) regularly appears on exams. The extra chain rule factor is the part students drop.
True or False: d/dx[ln(3x)] = 1/(3x). (False: d/dx[ln(3x)] = 3/(3x) = 1/x.)
Fill in the blank: d/dx[x² ln(x)] = 2x ln(x) + ______. (x)
True or False: When differentiating x³ + 5 ln(y) + y³ = 2 implicitly, the term 5 ln(y) produces 5/y. (False: it produces (5/y) · dy/dx.)
Fill in the blank: If g(x) = integral from 0 to x² of sin(t) dt, then g'(x) = ______. (sin(x²) · 2x)
True or False: ln(√u) = (1/2) ln(u). (True.)
Q: Find dy/dx if y = ln(x² + 1).
A: y' = 2x / (x² + 1). Apply the chain rule: derivative of ln(u) is u'/u, where u = x² + 1 and u' = 2x.
Q: Differentiate f(x) = ln(∛(x² + 1)).
A: Rewrite as f(x) = (1/3) ln(x² + 1). Then f'(x) = (1/3) · 2x/(x² + 1) = 2x / [3(x² + 1)].
Q: Differentiate y = x³ ln(x).
A: Product rule: y' = 3x² ln(x) + x³(1/x) = 3x² ln(x) + x² = x²(1 + 3 ln x).
Q: If g(x) = integral from 5 to ln(2x) of (3t + 4) dt, find g'(x).
A: By the 2nd FTC, g'(x) = [3 ln(2x) + 4] · (1/x) = [3 ln(2x) + 4] / x.
Q: Given x³ + 5 ln(y) + y³ = 2, find dy/dx.
A: Differentiate implicitly: 3x² + (5/y)(dy/dx) + 3y²(dy/dx) = 0. Factor: dy/dx(5/y + 3y²) = -3x². So dy/dx = -3x²y / (5 + 3y³).
These derivative techniques feed directly into integration: recognising that 1/u · u' is the derivative of ln|u| is the basis for many integral substitutions you will see in the integrals portion of this review. Implicit differentiation with logarithms extends to logarithmic differentiation, a technique used when differentiating products of many factors or variable-exponent expressions like x^x. The 2nd FTC reappears in accumulation-function problems and is tested heavily on AP Calculus exams.
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