calculus help urgent

Question

I have this question I got the answer but I think I am wrong cause idk what it means by [0, infinity). If f(x) = (e ^ x + e ^ (- x))/2 with domain confined to the interval [0, infinity), find the derivative of the inverse function by using the formula (f ^ - 1)' * (x) = 1/(f' * (f ^ - 1 * (x)))

Answer 1

f(x)=(e^x+e^-x)/2 same as f(x)=cosh(x)

h(x)=(e^x-e^-x)/2 same as h(x)=sinh(x)

note d(cosh(x))/dx=sinh(x)

For domain x=[0,+∞) cosh(x) has range [1,+∞) ,,,,and is increasing always as x increases

which implies

cosh^-1(x) is a function and has domain x=(1,+∞) and cosh^-1(x) has range (0,+∞)

For this problem, will need to know d(cosh(cosh^-1(x)))d(cosh^-1(x)=sinh(cosh^-1(x))

and that sinh(cosh^-1x)=±√(x²-1)

because cosh²(x)-sinh²(x)=1

cosh²(cosh^-1(x))-sinh²(cosh^-1(x))=1 ,,,,,,cosh(cosh^-1(x))=x

sinh(cosh^-1x)=±√(x²-1)

Now as cosh(cosh^-1(x))=x

then d(cosh(cosh^-1(x))/dx=d(x)dx=1

[d(cosh(cosh^-1(x)))/d(cosh^-1(x))]*[d(cosh^-1(x))/dx)]=1 ,,, chain rule left side of equation

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

d(cosh^-1(x))/dx=1/sinh(cosh^-1(x))=1/±√(x²-1)

since range is specified as (0,+∞), in this case

d(cosh^-1(x))/dx=1/√(x²-1)