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 ....
I know the process but I'm confused at one step
please can you explain that how
Thanks ....
1 Answer
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
#1 - tanh^2 x = sech^2 x#