What is the first differential of #y=ln(sqrt(x^3-x))#?

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Answer 1 · 2017-05-12T01:51:35

#dy/dx= (3x^2-1)/(2x(x^2-1))#

#y=ln(sqrt(x^3-x))#

First let's simplify #y#

#y=ln(x^3-x)^(1/2) =1/2ln(x^3-x)#

Applying the chain rule and the standard differential for #lnx#

#dy/dx= 1/2*1/(x^3-x) * (3x^2-1)#

#= (3x^2-1)/(2(x^3-x))#

#= (3x^2-1)/(2x(x^2-1))#