How do you find the derivative of #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)#