help with this calc question plz :)

Question

If f(x) = (e ^ x + e ^ (- x))/2 with domain confined to the interval [0, 0), find the derivative of the inverse function by using the formula (f ^ - 1)' * (x) = 1/(f' * (f ^ - 1 * (x)))

Answer 1

f(x)=cosh(x)

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

if x=0 then coshx=1

if x=1 then cosh^-1x=0

since domain of f(x) is x=0

the domain of f^-1(x) is f(0)=1 and cosh^-1(1)=0

sinh(0)=0

so d(cosh^-1(1))/dx=1/0

Answer 2

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

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

f '(x) = sinh(x) = (e^x-e^-x)/2

f^-1(x) = cosh^-1(x) = ln(x +√(x²-1)) You can look this up or derive it by swapping x and y and solving for y.

f '(ln(x +√(x²-1))) = sinh(ln(x +√(x²-1))) = (e^ln(x +√(x²-1))-e^-ln(x +√(x²-1)))/2

let u = x +√(x²-1)

(e^u-e^-u)/2 = ((e^ln(u))²-1)/(2e^ln(u)) = (u²-1)/(2u) = (2x²+2x√(x²-1)-2)/(2(x+√(x²-1)) = √(x²-1)

(f^-1)'(x)=1/(f'(f^-1(x)) = 1/√(x²-1)

-----

you can also get the same answer by

d cosh^-1(x)/dx = d ln(x+√(x²-1))/dx