How do you differentiate $y=csc^-1(e^x)$ ?

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Answer 1 · 2017-07-09T22:21:32

$dy/dx = - tan (csc^-1 (e^x))$

$csc y = e^x => ln (1/sin y) = x$

$dx/dy = sin y * (-1/sin^2 y) * cos y$

$dy/dx = (dx/dy)^-1 = -sin y / cos y = - tan (csc^-1 (e^x))$

How do you differentiate y=csc^-1(e^x)y=csc^-1(e^x)?