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

2 Answers
Aug 25, 2016

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

Explanation:

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)

Aug 25, 2016

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

Explanation:

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

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

And we have the answer