How do you find the derivative of $y = {sinh}^{- 1} (2 x)$ $y = sinh^-1 (2x)$ ?

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Answer 1 · 2016-08-23T11:42:46

$y'= 2/sqrt(1 + 4x^2)$

you can do it the same as you would with a trig function.

first undo the arcsinh as follows:

$sinh y = 2x qquad star$

then diff implicitly

$cosh y \ y' = 2$

$\ y' = 2/(cosh y)$

the fundamental $cosh-sinh$ hyperbolic identity is a little different from the corresponding trig one. It's:

$cosh^2 z color(red)(-) sin^2 z = 1$

So
$y' = 2/sqrt(1 + sinh^2 y)$

see $star$ for this last bit
$= 2/sqrt(1 + 4x^2)$

How do you find the derivative of y = sinh^-1 (2x)y=sinh−1(2x)y = sinh^-1 (2x)?