Final exam - help- Inverse function Theorem

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Homework Statement


Let f(x) = sinh(x) and let g be the inverse function of f. Using inverse function theorem, obtain g'(y) explicitly, a formula in y.

Okay the Inverse function Theorem says (f^-1)'(y) = 1/(f'(x))
If f is continuous on [a, b} and differentiable with f'(x)\neq0 for all x\in[a, b]. Then f is 1-1 and f^-1 is continuous and differentiable on f(a,b).

Homework Equations


The Attempt at a Solution


The derivative of f(x) sinh(x) is cosh = (e^x + e^-x)/2

How do I get g'(y) as a formula in y? Right now, I have 2/(e^x + e^-x)
 
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IntroAnalysis said:

Homework Statement


Let f(x) = sinh(x) and let g be the inverse function of f. Using inverse function theorem, obtain g'(y) explicitly, a formula in y.

Okay the Inverse function Theorem says (f^-1)'(y) = 1/(f'(x))
If f is continuous on [a, b} and differentiable with f'(x)\neq0 for all x\in[a, b]. Then f is 1-1 and f^-1 is continuous and differentiable on f(a,b).


Homework Equations





The Attempt at a Solution


The derivative of f(x) sinh(x) is cosh = (e^x + e^-x)/2

How do I get g'(y) as a formula in y? Right now, I have 2/(e^x + e^-x)

You have ##\frac 1 {f'(x)} = \frac 1 {\cosh x}##, and you have ##y=\sinh x##. It is almost always a mistake to put in exponentials. Use the basic sinh and cosh identity to express ##\cosh x## in terms of ##y##, given what you have.
 
Thanks so much, you save me from messing with exponents! Since I know cosh (x) = √(1 + sinh(x)^2 = √1 + y^2 !
 
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