What is the derivative of ln(x+1/x)?

Question

What is the derivative of ln(x+1/x)?

Answer 1

f(x) = ln(x+1) - ln(x)

Now take derivative

f'(x) = 1/(x+1) - 1/x

Answer 2

d/dx ln(x+1/x)

let u = x+1/x and use the chain rule

d/du ln(u) du/dx

d/du ln(u) = 1/u

d/dx(x + 1/x) = d/dx(x) + d/dx(1/x) = 1 - 1/x²

(1 - 1/x²)/(x+1/x) = [(x²-1)/x²] / [(x²+1)/x] = (x²-1)/(x³+x)

Answer: (x²-1)/(x³+x)

Answer 3

ln(x+1/x)=ln((x^2+1)/x)=ln(x^2+1)-ln(x).... x^2+1=u..

so du/dx=2x..if f(x)=ln(u)-ln(x)...so f'(x)=(1/u)du/dx-1/x....putting back the u, we have f'(x)=(1/(x^2+1))2x-1/x=2x/(x^2+1)-1/x=(...

u original question is misleading...according to the general rule, u should multiply, divide and add before substract unless u put a bracket before x+1.....