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Calculus of the Hyperbolic Functions

Calculus of the Hyperbolic Functions

Learning Objectives

After completing this section, you should be able to:

  • Apply the formulas for derivatives and integrals of the hyperbolic functions.
  • Apply the formulas for the derivatives of the inverse hyperbolic functions and their associated integrals.
  • Describe the common applied conditions of a catenary curve.

Key Terms & Glossary

  • Hyperbolic Functions: Functions defined using combinations of the exponential functions exe^xex and e−xe^{-x}e−x, corresponding to points on a unit hyperbola rather than a unit circle.
  • Inverse Hyperbolic Functions: The inverse operations of hyperbolic functions, typically found using implicit differentiation and resulting in natural logarithmic expressions.
  • Catenary Curve: The U-like geometric shape assumed by a hanging flexible chain or cable supported at its ends and acted upon by a uniform gravitational force.
  • Implicit Differentiation: A technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another, especially useful for deriving inverse hyperbolic derivatives.

The "Big Idea"

Hyperbolic functions (sinhx\\sinh xsinhx, coshx\\cosh xcoshx, tanhx,etc.)behavesimilarlytotheirtrigonometriccousins(sinx\\tanh x, etc.) behave similarly to their trigonometric cousins (\\sin xtanhx,etc.)behavesimilarlytotheirtrigonometriccousins(sinx, cosx\\cos xcosx, tanx),buttheyarerootedintheunithyperbola(x2−y2=1)insteadoftheunitcircle(x2+y2=1).Becausetheyareultimatelyjustcombinationsofexponentialfunctions,theircalculusisstraightforwardbutholdsimportant∗∗signdifferences∗∗comparedtostandardtrigonometry.Mostnotably,thederivativeofcoshx\\tan x), but they are rooted in the unit hyperbola (x^2 - y^2 = 1) instead of the unit circle (x^2 + y^2 = 1). Because they are ultimately just combinations of exponential functions, their calculus is straightforward but holds important **sign differences** compared to standard trigonometry. Most notably, the derivative of \\cosh xtanx),buttheyarerootedintheunithyperbola(x2−y2=1)insteadoftheunitcircle(x2+y2=1).Becausetheyareultimatelyjustcombinationsofexponentialfunctions,theircalculusisstraightforwardbutholdsimportant∗∗signdifferences∗∗comparedtostandardtrigonometry.Mostnotably,thederivativeofcoshx is positive sinhx,bypassingthenegativesignpitfallfoundwhendifferentiatingcosx\\sinh x, bypassing the negative sign pitfall found when differentiating \\cos xsinhx,bypassingthenegativesignpitfallfoundwhendifferentiatingcosx.

[!NOTE] A strong mastery of the Chain Rule and uuu-substitution is essential here, as hyperbolic calculus problems frequently combine exponential terms with inner polynomial functions.

Formula / Concept Box

FunctionDerivativeIntegral (with +C+ C+C)
sinhu\\sinh usinhucoshucdotu′\\cosh u \\cdot u'coshucdotu′intsinhu,du=coshu+C\\int \\sinh u \\, du = \\cosh u + Cintsinhu,du=coshu+C
coshu\\cosh ucoshusinhucdotu′\\sinh u \\cdot u'sinhucdotu′intcoshu,du=sinhu+C\\int \\cosh u \\, du = \\sinh u + Cintcoshu,du=sinhu+C
tanhu\\tanh utanhutextsech2ucdotu′\\text{sech}^2 u \\cdot u'textsech2ucdotu′inttextsech2u,du=tanhu+C\\int \\text{sech}^2 u \\, du = \\tanh u + Cinttextsech2u,du=tanhu+C
sinh−1u\\sinh^{-1} usinh−1ufracu′sqrt1+u2\\frac{u'}{\\sqrt{1 + u^2}}fracu′sqrt1+u2intfracdusqrt1+u2=sinh−1u+C\\int \\frac{du}{\\sqrt{1+u^2}} = \\sinh^{-1} u + Cintfracdusqrt1+u2=sinh−1u+C
cosh−1u\\cosh^{-1} ucosh−1ufracu′sqrtu2−1\\frac{u'}{\\sqrt{u^2 - 1}}fracu′sqrtu2−1 (for u>1u>1u>1)intfracdusqrtu2−1=cosh−1u+C\\int \\frac{du}{\\sqrt{u^2-1}} = \\cosh^{-1} u + Cintfracdusqrtu2−1=cosh−1u+C
tanh−1u\\tanh^{-1} utanh−1ufracu′1−u2\\frac{u'}{1 - u^2}fracu′1−u2 (for $\u\

Hierarchical Outline

  • 1. Derivatives and Integrals of Hyperbolic Functions
    • 1.1. Core Definitions (e.g., sinhx=fracex−e−x2\\sinh x = \\frac{e^x - e^{-x}}{2}sinhx=fracex−e−x2)
    • 1.2. Differentiation Rules (Comparing Trig vs. Hyperbolic)
    • 1.3. Integration using uuu-substitution
  • 2. Inverse Hyperbolic Functions
    • 2.1. Deriving derivatives via implicit differentiation
    • 2.2. Integral forms resulting in inverse hyperbolic functions
  • 3. Real-World Applications
    • 3.1. The Catenary Curve (Hanging chains, power lines)
    • 3.2. Exponential growth models and population dynamics

Visual Anchors

Comparison Diagram: Trigonometric vs. Hyperbolic Differentiation

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Graph of sinhx\\sinh xsinhx and coshx\\cosh xcoshx

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(Notice that coshx\\cosh xcoshx is an even function bounded below by 1, perfectly matching the shape of a hanging cable.)

Definition-Example Pairs

  • Term: Hyperbolic Cosine (coshx\\cosh xcoshx)

    • Definition: The even hyperbolic function defined by fracex+e−x2\\frac{e^x + e^{-x}}{2}fracex+e−x2.
    • Real-World Example: A perfectly flexible chain suspended by its two ends, hanging under its own weight, forms a catenary curve, which is mathematically modeled by y=acosh(fracxa)y = a \\cosh(\\frac{x}{a})y=acosh(fracxa).
  • Term: Implicit Differentiation

    • Definition: Differentiating an equation with respect to xwithoutexplicitlysolvingforyfirst,thensolvingalgebraicallyfory′x without explicitly solving for y first, then solving algebraically for y'xwithoutexplicitlysolvingforyfirst,thensolvingalgebraicallyfory′.
    • Real-World Example: Proving that if y=sinh−1xy = \\sinh^{-1}xy=sinh−1x, then sinhy=x.Takingthederivativegivescoshycdoty′=1,whichsimplifiestoy′=frac1coshy=frac1sqrt1+x2\\sinh y = x. Taking the derivative gives \\cosh y \\cdot y' = 1, which simplifies to y' = \\frac{1}{\\cosh y} = \\frac{1}{\\sqrt{1+x^2}}sinhy=x.Takingthederivativegivescoshycdoty′=1,whichsimplifiestoy′=frac1coshy=frac1sqrt1+x2.

Worked Examples

Example 1: Differentiating a Composite Hyperbolic Function

Problem: Evaluate the derivative of f(x)=cosh(4x2)f(x) = \\cosh(4x^2)f(x)=cosh(4x2). Solution:

  1. Identify the outer function (cosh(u))andinnerfunction(u=4x2\\cosh(u)) and inner function (u = 4x^2cosh(u))andinnerfunction(u=4x2).
  2. The derivative of cosh(u)\\cosh(u)cosh(u) is sinh(u)cdotu′\\sinh(u) \\cdot u'sinh(u)cdotu′ (using the Chain Rule).
  3. Calculate u′=fracddx(4x2)=8xu' = \\frac{d}{dx}(4x^2) = 8xu′=fracddx(4x2)=8x.
  4. Assemble the final derivative: f′(x)=sinh(4x2)cdot(8x)=8xsinh(4x2)f'(x) = \\sinh(4x^2) \\cdot (8x) = 8x \\sinh(4x^2)f′(x)=sinh(4x2)cdot(8x)=8xsinh(4x2)

Example 2: Integration Involving Hyperbolic Functions

Problem: Evaluate the integral int5xsinh(x2−3),dx\\int 5x \\sinh(x^2 - 3) \\, dxint5xsinh(x2−3),dx. Solution:

  1. Use uuu-substitution. Let u=x2−3u = x^2 - 3u=x2−3.
  2. Then du=2x,dxdu = 2x \\, dxdu=2x,dx, which means x,dx=frac12dux \\, dx = \\frac{1}{2} dux,dx=frac12du.
  3. Substitute into the integral: int5sinh(u)left(frac12duright)=frac52intsinh(u),du\\int 5 \\sinh(u) \\left(\\frac{1}{2} du\\right) = \\frac{5}{2} \\int \\sinh(u) \\, duint5sinh(u)left(frac12duright)=frac52intsinh(u),du
  4. Integrate using the formula intsinhu,du=coshu+C\\int \\sinh u \\, du = \\cosh u + Cintsinhu,du=coshu+C: frac52cosh(u)+C\\frac{5}{2} \\cosh(u) + Cfrac52cosh(u)+C
  5. Substitute back u=x2−3u = x^2 - 3u=x2−3: frac52cosh(x2−3)+C\\frac{5}{2} \\cosh(x^2 - 3) + Cfrac52cosh(x2−3)+C

Checkpoint Questions

▶1. What is the fundamental difference between the derivative of $\\cos x$ and $\\cosh x$?

The derivative of the trigonometric function cosx\\cos xcosx is −sinx(itintroducesanegativesign).However,thederivativeofthehyperbolicfunctioncoshx-\\sin x (it introduces a negative sign). However, the derivative of the hyperbolic function \\cosh x−sinx(itintroducesanegativesign).However,thederivativeofthehyperbolicfunctioncoshx is exactly sinhx\\sinh xsinhx (no negative sign).

▶2. How do you find the derivative of an inverse hyperbolic function like $y = \\tanh^{-1}x$?

You use implicit differentiation. Rewrite it as tanhy=x,takethederivativeofbothsideswithrespecttox\\tanh y = x, take the derivative of both sides with respect to xtanhy=x,takethederivativeofbothsideswithrespecttox (getting textsech2ycdoty′=1\\text{sech}^2 y \\cdot y' = 1textsech2ycdoty′=1), and solve for y′y'y′. Using the identity \dlr 1 - $\tanh^2 y =\sech2y \sech^2 y\sech2y, this simplifies to y' = \\frac{1}{1-x^2}$$.

▶3. If an architectural arch is built in the shape of an inverted catenary, what base mathematical function represents its curve?

The architectural arch is represented by an inverted hyperbolic cosine function, mathematically expressed as y = -a \\cosh(\frac{x}{a}) + C$$.

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