Find derivation of #tanh^-1x#, hence differentiate #y= tanh^-1 ((2x)/(1+x^2))# ?

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Answer 1 · 2018-07-13T18:13:34

The answers are #=1/(1-x^2)# and #=2/(1-x^2)#

#cosh^2x-sinh^2x=1#

#1-tan^2x=1/cosh^2x#

#cosh^2x=1/(1-tanh^2x)#

Let

#p=tanh^-1x#

Then

#tanhp=x#

#sinhp/coshp=x#

Differentiate wrt #x#

#(coshp*coshp-sinhp*sinhp)/(cosh^2p)(dp)/dx=1#

#(cosh^2p-sinh^2p)/(cosh^2p)(dp)/dx=1#

#1/cosh^2p(dp)/dx=1#

#(dp)/dx=cosh^2p=1/(1-x^2)#

If #y=tanh^_1((2x)/(1+x^2))#

The derivative is

#dy/dx=1/(1-((2x)/(1+x^2))^2)*((2x)/(1+x^2))'#

The derivative of

#((2x)/(1+x^2))'=(2(1+x^2)-2x(2x))/(1+x^2)^2#

#=(2+2x^2-4x^2)/(1+x^2)^2#

#=(2-2x^2)/(1+x^2)^2#

Finally,

#dy/dx=1/(1-((2x)/(1+x^2))^2)*(2-2x^2)/(1+x^2)^2#

#=(2-2x^2)/((1+x^2)^2-4x^2)#

#=(2-2x^2)/(1+2x^2+x^4-4x^2)#

#=(2(1-x^2))/(1-2x^2+x^4)#

#=(2(1-x^2))/(1-x^2)^2#

#=2/(1-x^2)#