How do you find #dy/dx# of #y = tanh^(-1) (sinx)#?

I know the process but I'm confused at one step

#d/dx(tanh y) = d/dx(sinx)#

#(sech^2 y).dy/dx = cosx#

#dy/dx = cos x / sech^2 y#

#dy/dx = cosx/(1 - tanh^2 y)#

please can you explain that how #sech^2 y# converts into# (1 - tanh^2y)#

Thanks ....

1 Answer
George C.
Oct 21, 2017

See explanation...

Explanation:

Note that:

#cosh x = (e^x+e^(-x))/2#

#sinh x = (e^x-e^(-x))/2#

Hence:

#cosh^2 x - sinh^2 x = ((e^x+e^(-x))/2)^2-((e^x-e^(-x))/2)^2#

#color(white)(cosh^2 x - sinh^2 x) = ((e^(2x)+2+e^(-2x))/4)-((e^(2x)-2+e^(-2x))/4)^#

#color(white)(cosh^2 x - sinh^2 x) = 1#

Dividing both ends by #cosh^2 x# we get:

#1 - tanh^2 x = sech^2 x#

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