Hyperbolic Functions

Question 1

sinh(x)

Answer 1

“shine(x)”

(e^x - e^-x) /2

Question 2

cosh(x)

Answer 2

“cosh(x)”

(e^x + e^-x) /2

Question 3

tanh(x)

Answer 3

“tanch(x)”

e^2x - 1)/(e^2x + 1

Question 4

Reciprocal hyperbolics

Answer 4

“cosetch(x)”
“setch(x)”
“coth(x)”
1/the original like in trig

Question 5

Graph of y = sinh(x)

Answer 5

Steep, flattens to a diagonal through (0,0) and then rises steeply again

Question 6

Graph of y = cosh(x)

Answer 6

Parabolic curve with a base at (0,1)

Question 7

Graph of y = tanh(x)

Answer 7

Flat before rising diagonally through (0,0) and flattening again
Asymptotes at y = 1 and y = -1

Question 8

Graph of y = cosech(x)

Answer 8

Like a positive reciprocal graph, asymptotes at x,y = 0

Question 9

Graph of y = sech(x)

Answer 9

Flat asymptotically to the x-axis before rising and flattening to a peak at (0,1) and falling at an increasing rate then flattening asymptotically to the x-axis
Asymptote at y = 0

Question 10

Graph of y = coth(x)

Answer 10

Similar to cosech(x) but starting from y = 1 and -1

Asymptotes at x = 0, y = 1, y = -1

Question 11

Proving inverse hyperbolic formulae

Answer 11

1. Use the hyperbolic on both sides to write in terms of x
2. Replace with the hyperbolic formula
3. Form a hidden quadratic by multiplying by e^y
4. ln both sides, only use the positive case
5. Replace y with the inverse hyperbolic

Question 12

Graph of y = arsinh(x)

Answer 12

Flat, rises diagonally through the origin and flattens again, similar to I/V characteristics of a filament lamp but slower flattening

Question 13

Graph of y = arccosh(x)

Answer 13

Domain: x>= 1

Almost vertical at x = 1 before flattening

Question 14

Graph of y = artanh(x)

Answer 14

Domain: -1 < x < 1

Almost vertically upwards at those asymptotes, flattens through the origin

Question 15

Graph of y = arsech(x)

Answer 15

Domain x>0

Almost downwards at the asymptote x = 0, flattens and falls again until it reaches the x-axis

Question 16

Graph of y = arcosech(x)

Answer 16

Domain x != 0

Like a positive reciprocal curve

Question 17

Graph of y = arcoth(x)

Answer 17

Domain |x| > 1

Like a positive reciprocal curve but with asymptotes at x = +/- 1

Question 18

Hyperbolic pythagorean identities

Answer 18

cosh^2x - sinh^2x = 1
sech^2x = 1 - tanh^2x
cosech^2x = coth^2x - 1

Question 19

sinh(A+/-B)

Answer 19

sinhAcoshB +/- coshA/sinhB

Question 20

cosh(A+/-B)

Answer 20

coshAcoshB +/- sinhAsinhB

Question 21

tanh(A+/-B)

Answer 21

tanhA -/+ tanhB / 1 +/- tanhAtanhB

Question 22

Osborn’s rule

Answer 22

Replace sin with sinh and cos with cosh

Put a - in front of any multiplication of sinhx (including tanh^2) (sinh^4 cancels out)

Question 23

d/dx(coth(x))

Answer 23

-cosech^2(x)

Question 24

d/dx(sinh(x))

Answer 24

cosh(x)

Question 25

d/dx(cosh(x))

Answer 25

sinh(x)

Question 26

d/dx(tanh(x))

Answer 26

sech^2(x)

Question 27

Proving regular differentials

Answer 27

Put it into exponential form and differentiate

Question 28

Proving inverse derivatives

Answer 28

1. x = sinh/cosh/tanh y
2. Do dx/dy
3. Use identities to substitute in terms of x
4. Find the reciprocal for dy/dx

Question 29

∫sinh(x) dx

Answer 29

cosh(x) + c

Question 30

∫cosh(x) dx

Answer 30

sinh(x) + c

Question 31

∫tanh(x) dx

Answer 31

sech^2(x) + c

Question 32

∫cosech^2(x) dx

Answer 32

-coth(x) + c

Question 33

∫sech^2(x) dx

Answer 33

tanh(x) + c

Question 34

∫sech(x)tanh(x) dx

Answer 34

-sech(x) + c

Question 35

∫cosech(x)coth(x) dx

Answer 35

-cosech(x) + c

Question 36

Multiple terms in the numerator

Answer 36

Split and use standard integrals/reverse chain

Question 37

Proving ∫tanh(x)

Answer 37

use sinh(x)/cosh(x) and use reverse chain rule to ln|cosh(x)|

Question 38

Small odd powers of cosh/sinh

Answer 38

Factor out (cosh/sinh)^power-1 and use identity

Question 39

When to use the exponential definition

Answer 39

When there is an exponential term or no simpler way to integrate

Question 40

∫sech(x) or ∫cosech(x) method

Answer 40

Use exponential form, multiply by e^x and use substitution with e^x

Question 41

∫1/sqrt(a^2 + x^2) substitution

Answer 41

x = asinh(u)

Question 42

∫1/sqrt(x^2 - a^2) substitution

Answer 42

x = acosh(u)

Question 43

Completing the square

Answer 43

With a/quadratic or a/sqrt(quadratic), complete the square and use substitution with u as the thing that is squared

Question 44

cosh^2(x) substitution

Answer 44

1/2 + 1/2 cosh(2x)

Question 45

sinh^2(x) substitution

Answer 45

1/2 cosh(2x) - 1/2

Question 46

Hyperbolics to R formula

Answer 46

Use cosh^2 - sinh^2 = 1 for R rather than +