How do you find the derivative of #(ln(x^2-1))/x^2 #?

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Answer 1 · 2017-02-22T20:25:09

See below

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

For #F(x)=lnf(x)#, #F'(x)=(f'(x))/f(x)#

For #F(x)=(g(x))/(h(x))#, #F'(x)=(g(x)h'(x)-h(x)g'(x))/(h(x))^2#

Thus #f'(x)=((2x^2)/(x^2-1)-2xln(x^2-1))/x^4=#

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

Cleaning this up, we get:

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