Hyperbolic functions

Hyperbolic functions
The hyperbolic functions sinh x and cosh x have similar names to the trigonometric functions, but they are defined in terms of the exponential function e^x. They are defined by the formula
sinh??x=(e^x-e^(-x))/2? and cosh??x=(e^x+e^(-x))/2?
The hyperbolic function tanh x is defined by
tanh x=sinh?x/cosh?x =(e^x-e^(-x))/(e^x+e^(-x) )
We have also mentioned the reciprocal functions, and these have special names related to the names of the trigonometric reciprocal functions. They are
coth??x=1/(tanh x)? , sech?x=1/cosh?x and csch?x=1/sinh?x

Graphs of hyperbolic functions


Some identities for hyperbolic functions
Hyperbolic functions have identities which are similar to, but not the same as, the identities for trigonometric functions. In this section we shall prove two of these
identities, and list some others.
The first identity is: cosh^2??x-sinh^2??x=1? ?
To prove this, we start by substituting the definitions for sinh?x and cosh?x:
cosh^2??x-sinh^2??x=? ? ((e^x+e^(-x))/2)^2-((e^x-e^(-x))/2)^2
=(e^2x+2e^x e^(-x)+e^(-2x))/4-(e^2x-2e^x e^(-x)+e^(-2x))/4
=(e^2x+2+e^(-2x)-e^2x+2-e^(-2x))/4=( 4 )/( 4 )=1
Here is another identity involving hyperbolic functions: sinh?2x=2 sinh??x cosh?x ?
On the left-hand side we have sinh?2x so, from the definition,
sinh?2x=(e^2x-e^(-2x))/2
We want to manipulate the right-hand side to achieve this. Sowe shall start by substituting the definitions of sinh?x and cosh?x into the right-hand side:
2 sinh??x cosh?x ?=2×(e^x-e^(-x))/2×(e^x+e^(-x))/2=(e^2x-e^(-2x))/2=sinh?2x
There are several more identities involving hyperbolic functions:
cosh??2x=? ?cosh^2 x???+sinh^2??x ? ?
sinh??(x+y)=? sinh??x cosh??y+sinh??y cosh?x ? ? ?
cosh (x+y)?= cosh??x cosh??y+sinh??y sinh??x ? ? ? ?
If you know the trigonometric identities, you may notice that these hyperbolic identities are very similar, although sometimes plus signs have become minus signs and vice versa. In fact the hyperbolic functions are very closely related to the trigonometric functions, and sinh x , cosh?x are sometimes called the hyperbolic sine
and hyperbolic cosine functions.
Solving equations
Suppose that sinh?x=( 3 )/( 4) and we wish to find the exact value of x.
(e^x-e^(-x))/2=( 3 )/( 4)
{ 2e^x-2e^(-x)=3 }×e^x
2e^2x-3e^x-2=0
(2e^x+1)(e^x-2)=0
e^x=2 or e^x=-( 1 )/( 2 )
But e^x is always positive so e^x=2 ?x=ln?2

Example1: Solve for x: 1. 5 cosh?x+3 sinh??x ?=4
5((e^x+e^(-x))/2)+3((e^x-e^(-x))/2)=4 ? (5e^x+5e^(-x)+3e^x-3e^(-x))/2=4
(8e^x+2e^(-x))/2=4 ? 4e^x+e^(-x)=4
4e^x+( 1 )/e^x =4 ? ( 4e^2x+1 )/e^x =4
4e^2x+1=4e^x ? 4(e^x )^2-4e^x+1=0
(2e^x-1)^2=0 ? e^x=( 1 )/2
x=ln??( 1 )/2?=-ln?2=-0.693
2. 2 sinh??x ?-3 cosh?x=-3 ? (2e^x-2e^(-x)-3e^x-3e^(-x))/2=-3
-e^x-5e^(-x)=-6 ? {e^x+5e^(-x)=6}×e^x ? e^2x-6e^x+5=0
(e^x-1)(e^x-5)=0 ? e^x=1 or e^x=5
x=ln?1=0 or x=ln?5=1.609



Derivatives of hyperbolic functions
If u=u(x), then :
1. d/dx ?(sinh??u)=cosh?u du/dx
2. d/dx ?(cosh??u)=sinh?u du/dx
3. d/dx(tanh??u)?=sech^2??u ? du/dx
4. d/dx (coth?u )=-csch^2?u du/dx
5. d/dx (sech?u )=-sech?u tanh?u du/dx
6. d/dx (csch?u )=-csch?u coth?u du/dx
Example 2: Find derivatives of the functions
1. y=sinh??(x^2+1) ? dy/dx=cosh??(x^2+1)×2x/(2?(x^2+1))=(x cosh??(x^2+1))/?(x^2+1)
2. y=cosh?(x^2-3) ? dy/dx=2x sinh?(x^2-3)
3. y=tanh??x ? dy/dx=sech^2 ?x×1/(2?x)=(sech^2 ?x)/(2?x)
4. y=e^2x cosh?3x ? dy/dx=3e^2x sinh?3x+2e^2x cosh?3x

1. Solve for x:
a. 4 cosh?x+sinh??x ?=4 b. 3 sinh??x ?-cosh?x=1
c. sech?2x=0.25 d. 4 tanh?x-sech?x=1
2. Find derivatives of the functions
a. y=cosh?2x-sinh??3x ? b. y=e^(-2x) sinh??2x ?
c. y=3x sech?2x d. y=?x tanh??x