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IntroAnalysis
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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)[itex]\neq[/itex]0 for all x[itex]\in[/itex][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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