How do you find the derivative of #ln(x+1/x)#?

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Answer 1 · 2016-08-25T09:50:09

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

#f(x) = ln(x+1/x)#

#f'(x) = 1/(x+1/x) * d/dx(x+1/x)# (Standard differential and Chain rule)

#= 1/(x+1/x) * (1-1/x^2)# (Power rule)

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

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

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

Answer 2 · 2016-08-25T09:50:46

#(x ^2-1)/[x(x^2+1)]#

#y=ln(x +1/x)#
Let #(x +1/x)=z#
so #y=lnz#
#dy/dz=1/z#

#z= x + x ^-1#
#dz/dx=1-x ^-2#

We are using the chain rule

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

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

And we have the answer