Hyperbolic Functions.

Question

 Hyperbolic Functions. The hyperbolic cosine is not one-to-one but restricting the domain to non-negative real numbers gives a one-to-one function called the restricted hyperbolic cosine.

f(x) = cosh(x), x >= 0

Algebraically find a formula for the inverse function and calculate its derivative. Give your final answers, then show your work

f^-1 (x) =

d/dx (f^-1(x)) =

Answer 1

correction adding x² completes the square for the quadradic.

y=coshx

for inverse

x=[e^y+e^-y]/2 -- x≥1

2xe^y=e^2y+1

e^2y-2xe^y+x²=x²-1

(e^y-x)²=x²-1

e^y=x+/-√(x²-1)

here is a subtlety, if original domain is x≥0, then in inverse range y≥0

meaning e^y≥1 and only x+√(x²-1) has values≥1

so y=cosh^-1x=ln[x+√(x²-1)]

d(cosh^-1(x))dx

is equal to two expressions so they are equal

d(cosh^-1(x))dx=1/[sinh(cosh^-1(x))]

and d(cosh^-1(x))dx=[1/(x+√(x²-1)]*[1+x/√(x²-1)]

meaning

sinh((cosh^-1(x))=(x+√(x²-1))/[1+x/√(x²-1)]

=√(x²-1)

Answer 2

x=cosh y=(1/2)(e^-x+e^x)=(1/2)(1+e^2y)/e^y

This is a quadratic in e^y with solution

e^y = x + sqrt(x²-1)

and y = ln[x+sqrt(x²-1) is the inverse.

The derivative of inverse cosh can be computed directly from the above expression or by using the chain rule and hyperbolic identities.