Derivative of sech-1(x)

We use the fact from the definition of the inverse that 

    sech(sech-1(x))  =  x

and the fact that 

    sech'(x)  =  -tanh(x)sech(x)

Now take the derivative of both sides (using the chain rule on the left hand side) to get

    -tanh(sech-1x)sech(sech-1(x))(sech-1(x))'  =  1


  **      -x tanh(sech-1x)(sech-1(x))'  =  1

We know that 

        cosh2(x) - sinh2(x)  =  1

Dividing by the cosh2(x) gives

        1 - tanh2(x)  =  sech2(x)



so that 


Finally substituting into **  gives



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