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Section 5.8 Hyperbolic Functions:
Hyperbolic function arose from comparison of the area of a semicircular regions with the area of a region under a hyperbola.
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Graph of the hyperbola f(x) = √(1+x2) and the circle g(x) = √(1 – x2)
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Definitions of the hyperbolic functions
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Graph of f(x) = sinh (x) Domain: (-, ) Range: (-, )
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Graph of f(x) = cosh (x) Domain: (-, ) Range: [1, )
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Graph of f(x) = tanh (x) Domain: (-, ) Range: (-1, 1)
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Graph of f(x) = csch (x) Domain: (-, 0) (0,) Range: (-, 0) (0,)
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Graph of f(x) = sech (x) Domain: (-, ) Range: (0, 1]
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Graph of f(x) = coth (x) Domain: (-, 0) (0,) Range: (-, -1) (1,)
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Hyperbolic Identities:
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Theorem 5.18: Derivatives and integrals of Hyperbolic functions.
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Theorem 5.18: Derivatives and integrals of Hyperbolic functions.
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Find the derivative of the function.
1. f(x) = cosh (x - 2)
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Find the derivative of the function.
1. f(x) = cosh (x - 2) f’(x) = 1∙ sinh (x -2)
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Find the derivative of the function.
2. y = tanh(3x2 - 1)
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Find the derivative of the function.
2. y = tanh(3x2 - 1) y’ = 6x∙ sech2 (3x2 -1)
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Find the derivative of the function.
3. g(x) = x cosh x –sinh x
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Find the derivative of the function.
3. g(x) = x cosh x –sinh x g’(x) = x ∙ sinh x + cosh x – cosh x g’(x) = x ∙ sinh x
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Find the integral
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Find the integral
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Find the integral 2. ∫ sech2 (2x - 1) dx
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Find the integral 2. ∫ sech2 (2x - 1) dx Let u = 2x – 1 du = 2 dx
Now Substiture ∫sech2 u du/2 = ½ ∫sech2 u du = ½ tan u + C = ½ tan (2x – 1) + C
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Find the integral
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Find the integral
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Inverse Hyperbolic Functions
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Differentiation involving inverse hyperbolic functions.
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Integration involving inverse hyperbolic functions.
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