find derivative: sin^(-1)((e^(ax)-e^(-ax))/(e^(ax)+e(-ax)))

[mrgthiru]

Could any one help me in finding out this derivative of....?
find the derivative of:

sin^(-1) ( ( e^(ax) - e^(-ax) ) / ( e^(ax) + e^(-ax) ) )

I guess we have to consider this...
( ( e^(ax) - e^(-ax) ) / ( e^(ax) + e^(-ax) ) ) as u

u = ( ( e^(ax) - e^(-ax) ) / ( e^(ax) + e^(-ax) ) )

And have to differentiate du/dx

simulatenously we should also differentiate sin^(-1) which is equal to
(dy/dx) = 1/√(1-x^2)
Where we have to substitute u in place of x in the above formula..

then (dy/dx)*(du/dx)..

I know the methodology.. but not getting the expected solution.. Need help in solving out with steps.. .. Please tell me if iam wrong.... and help me out in this..


Need help in deriving the steps..

Warm Regards
Gthiru

 

[morson]

Your function is:

\(\displaystyle \L\ f(x) = Sin^{-1}(tanh(ax))\)

\(\displaystyle \L\ \frac{d}{dx}\ (tanh(ax)) = -a.cosech^2(ax)\)

Finish.