Calculus Help Needed!

[unkempt yarrow]

so for a:
you find cosh(ln(x+√(x²-1)))
so:
½e^(ln(x+√(x²-1)))+½e^(-ln(x+√(x²-1)))
½(x+√(x²-1))+1/2(x+√(x²-1))
which simplified to x

and for b you take the derivative:
which is cosh and you use acosh to find the X for which coshx=1
so ln(1+√(1-1))=0
so yet again you just have y=x

@bold rapids

I find this slightly humorous

[bold rapids]

So for part a)

It is asking to find the inverse

but why is it asking to find the inverse for that specific given domain?

@unkempt yarrow

[unkempt yarrow]

you are given the inverse, you just have to show that it is actually the inverse

and the domain of that inverse is purely because for x<1 you have √(x²-1) as being complex

[bold rapids]

ok so I don't have to mention the domain in the answer to part a)?

[unkempt yarrow]

you probably should explain why the domain is that

[bold rapids]

How do the two lines u have written show this?

[unkempt yarrow]

you have the definition of cosh in the text above the questions, I substituted the inverse that you were given into than and then simplified

you can simplify all the way by hand if you want

the definition of an inverse function of f(x) is a function (let the inverse function be g(x)) such that f(g(x))=x

[bold rapids]

ah ok

So all I gotta do is sub in the given inverse

[unkempt yarrow]

and simplify it down to x

[bold rapids]

to the cosh(x) above

[unkempt yarrow]

and then explain the domain thing

yeah

[bold rapids]

and perform algebra to get x

ok ok nice

and then for b

[unkempt yarrow]

you're told what the derivative of sinh is

substitute 1 into the inverse that you've just proven to be accurate

you get 0

so you have the intercept as 0 and the slope as 1

which means you just have y=x

[bold rapids]

the equation of the tangent line is given by y = f(1) + f'(1) (x-1)

right?

[unkempt yarrow]

yeah

[bold rapids]

but

the derivative of sinh

is simply cosh

which is given above

[unkempt yarrow]

f(1)=0 and f'(1)=1

wait no

no no no

that's not right

[bold rapids]

my f' is just cosh

[unkempt yarrow]

you're given that the slope is 1

so you have f'(x)=1

f'(x)=coshx

coshx=1

acosh(1)=0

[bold rapids]

oh

so I need to find

what x gives me

coshx = 1?

[unkempt yarrow]

yeah, and you find that x=0

[bold rapids]

so

my a is

0

?

this is my tangent line formula

[unkempt yarrow]

yeah

cosh(0)=1
sinh(0)=0

a=0

sinh(0)+cosh(0)x

[bold rapids]

yeah so since we were told f'(a) = 1

we knew

what x gives cosh = 1

that x is 0

so y = f(0) + f'(0) (x - 0)

y = 0 + 1 (x)

y = x

Does that make sense/is correct?

[unkempt yarrow]

yeah

[bold rapids]

appreciate it boss

ty