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calculusintegrationdifferentiationconvergencevolumespedagogySat Apr 25

Calculus II: Step-by-Step Derivations of Core Techniques

Abstract

This article presents rigorous step-by-step derivations of three foundational Calculus II techniques: logarithmic differentiation, convergence testing for improper integrals, and volume calculation for solids of revolution. Each derivation is motivated by the underlying mathematical structure and illustrated with worked examples. The goal is to provide a reference for students and instructors seeking transparent, technically sound explanations of these essential methods.

Background

Calculus II extends single-variable calculus into more sophisticated territory. Where Calculus I focuses on basic differentiation and integration, Calculus II demands fluency with advanced integration techniques, infinite series, and geometric applications. Three topics recur across most curricula: handling derivatives of complex functions via logarithmic methods, determining whether improper integrals converge, and computing volumes of three-dimensional solids formed by rotation.

This article treats each as a self-contained derivation, building from first principles and showing the algebraic steps that justify the final formulas.

Key Results

1. Logarithmic Differentiation

Problem: Differentiate a function y=f(x)y = f(x) where ff is a product, quotient, or involves variable exponents—cases where the product or quotient rule becomes unwieldy.

Derivation [logarithmic-differentiation]

Begin with y=f(x)y = f(x). Take the natural logarithm of both sides: ln(y)=ln(f(x))\ln(y) = \ln(f(x))

Differentiate both sides with respect to xx using the chain rule. On the left: ddx[ln(y)]=1ydydx\frac{d}{dx}[\ln(y)] = \frac{1}{y} \frac{dy}{dx}

On the right: ddx[ln(f(x))]=f(x)f(x)\frac{d}{dx}[\ln(f(x))] = \frac{f'(x)}{f(x)}

Equating these: 1ydydx=f(x)f(x)\frac{1}{y} \frac{dy}{dx} = \frac{f'(x)}{f(x)}

Solve for dydx\frac{dy}{dx}: dydx=yf(x)f(x)\frac{dy}{dx} = y \cdot \frac{f'(x)}{f(x)}

Substitute back y=f(x)y = f(x): dydx=f(x)f(x)f(x)=f(x)\frac{dy}{dx} = f(x) \cdot \frac{f'(x)}{f(x)} = f'(x)

More usefully, for a composite function, this yields: dydx=f(x)f(x)f(x)\frac{dy}{dx} = f(x) \cdot \frac{f'(x)}{f(x)}

Why this works: Logarithms convert products into sums and exponents into coefficients, transforming a multiplicative problem into an additive one. This is especially powerful when the variable appears in both base and exponent.

2. Convergence of Improper Integrals

Problem: Determine whether an integral abf(x)dx\int_a^b f(x) \, dx converges (yields a finite value) or diverges (approaches infinity or fails to exist) when the limits are infinite or the integrand is unbounded.

Definition [convergence-of-integrals]

An integral abf(x)dx\int_a^b f(x) \, dx is said to converge if the limit of the integral exists and is finite as the bounds approach their limits. It diverges if the limit does not exist or is infinite.

For an improper integral with an infinite upper limit: af(x)dx=limtatf(x)dx\int_a^{\infty} f(x) \, dx = \lim_{t \to \infty} \int_a^t f(x) \, dx

The integral converges if and only if this limit is finite.

Comparison Test: If 0f(x)g(x)0 \le f(x) \le g(x) for all xax \ge a, then:

  • If ag(x)dx\int_a^{\infty} g(x) \, dx converges, so does af(x)dx\int_a^{\infty} f(x) \, dx.
  • If af(x)dx\int_a^{\infty} f(x) \, dx diverges, so does ag(x)dx\int_a^{\infty} g(x) \, dx.

Limit Comparison Test: If f(x),g(x)>0f(x), g(x) > 0 and limxf(x)g(x)=L\lim_{x \to \infty} \frac{f(x)}{g(x)} = L where 0<L<0 < L < \infty, then af(x)dx\int_a^{\infty} f(x) \, dx and ag(x)dx\int_a^{\infty} g(x) \, dx either both converge or both diverge.

Intuition: These tests allow us to determine convergence without computing the integral explicitly, by comparing the given function to a known benchmark (often a power function or exponential).

3. Volume of a Solid of Revolution

Problem: Find the volume of a three-dimensional solid formed by rotating a two-dimensional region around an axis.

Disk Method (Rotation about the x-axis) [volume-of-solid-of-revolution]

Consider a region bounded by y=f(x)y = f(x) (above) and y=g(x)y = g(x) (below), from x=ax = a to x=bx = b. Rotate this region about the x-axis.

At each position xx, the cross-section perpendicular to the x-axis is a disk (or washer) with:

  • Outer radius: R(x)=f(x)R(x) = f(x)
  • Inner radius: r(x)=g(x)r(x) = g(x)

The area of this washer is: A(x)=π[R(x)2r(x)2]=π[f(x)2g(x)2]A(x) = \pi [R(x)^2 - r(x)^2] = \pi [f(x)^2 - g(x)^2]

Integrate along the axis of rotation: V=abA(x)dx=πab[f(x)2g(x)2]dxV = \int_a^b A(x) \, dx = \pi \int_a^b [f(x)^2 - g(x)^2] \, dx

Shell Method (Rotation about the y-axis)

For rotation about the y-axis, consider cylindrical shells. At height yy, a thin vertical strip of width dydy traces out a cylindrical shell with:

  • Radius: x(y)x(y) (distance from the y-axis)
  • Height: h(y)k(y)h(y) - k(y) (the width of the region at height yy)
  • Circumference: 2πx(y)2\pi x(y)

The volume element is: dV=2πx(y)[h(y)k(y)]dydV = 2\pi x(y) [h(y) - k(y)] \, dy

Integrate: V=2πcdx(y)[h(y)k(y)]dyV = 2\pi \int_c^d x(y) [h(y) - k(y)] \, dy

Worked Examples

Example 1: Logarithmic Differentiation

Find dydx\frac{dy}{dx} for y=xxy = x^x.

Solution: Take the natural logarithm: ln(y)=ln(xx)=xln(x)\ln(y) = \ln(x^x) = x \ln(x)

Differentiate both sides: 1ydydx=ln(x)+x1x=ln(x)+1\frac{1}{y} \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1

Solve for dydx\frac{dy}{dx}: dydx=y[ln(x)+1]=xx[ln(x)+1]\frac{dy}{dx} = y[\ln(x) + 1] = x^x[\ln(x) + 1]

Example 2: Convergence of an Improper Integral

Determine whether 11x2+1dx\int_1^{\infty} \frac{1}{x^2 + 1} \, dx converges.

Solution: For large xx, 1x2+11x2\frac{1}{x^2 + 1} \approx \frac{1}{x^2}.

We know 11x2dx\int_1^{\infty} \frac{1}{x^2} \, dx converges (p-integral with p=2>1p = 2 > 1).

Since 0<1x2+1<1x20 < \frac{1}{x^2 + 1} < \frac{1}{x^2} for all x1x \ge 1, by the Comparison Test, 11x2+1dx\int_1^{\infty} \frac{1}{x^2 + 1} \, dx converges.

Example 3: Volume of a Solid of Revolution

Find the volume of the solid obtained by rotating the region bounded by y=xy = \sqrt{x}, y=0y = 0, and x=4x = 4 about the x-axis.

Solution: Using the disk method with f(x)=xf(x) = \sqrt{x} and g(x)=0g(x) = 0: V=π04(x)2dx=π04xdxV = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx

Evaluate: V=π[x22]04=π162=8πV = \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi \cdot \frac{16}{2} = 8\pi

References

AI Disclosure

This article was drafted with the assistance of an AI language model. The mathematical content is derived from class notes and standard Calculus II pedagogy; all formulas and derivations have been verified against the source materials. The AI was used to organize the notes into a coherent narrative structure, paraphrase content, and format the mathematical exposition. A human author reviewed the final text for accuracy and clarity.

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References

AI disclosure: Generated from personal class notes with AI assistance. Every factual claim cites a note. Model: claude-haiku-4-5-20251001.