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

Question

Samantha Adkison · Accepted Answer

First of all, this can be written as #y = xln(1 + 1/x)#.   We now use the chain and power rules to differentiate #ln(1 + 1/x)# and the product rule to differentiate the entire function.  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Differentiating #ln(1 + 1/x)#: Let #y=  lnu# and #u = 1 + 1/x#.  Then #dy/(du) = 1/u# and #(du)/dx = -1/x^2#.   #dy/dx = dy/(du) xx (du)/dx# #dy/dx=  1/u xx -1/x^2# #dy/dx = -1/((1 + 1/x)(x^2))# #dy/dx = -1/(x^2 + x)# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Differentiating #y=  xln(1 +1/x)#: #y' = 1(ln(1 + 1/x)) + (-1/(x^2 + x) xx x)# #y' = ln(1 + 1/x) - x/(x(x + 1))# #y' = ln(1 + 1/x) - 1/(x + 1)# Hopefully this helps!

Evelyn Agard · Answer

undefined To find the derivative of ln(1+1/x) / (1/x), we can use the quotient rule. Let u = ln(1+1/x) and v = (1/x). Then, using the quotient rule (u/v)' = (u'v - uv') / v^2, we find that the derivative is:

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

Simplifying this expression yields the derivative:

= [((-1 + (1/x)) * (1/x^2)) + (ln(1+1/x) * (1/x^2))] / (1/x)^2

= [((1/x) * (1/x^2)) + (ln(1+1/x) * (1/x^2))] / (1/x)^2

= [(1/x^3) + (ln(1+1/x) / x^2)] / (1/x^2)

= x * (1/x^3) + (ln(1+1/x) / x^2)

= 1/x^2 + (ln(1+1/x) / x^2)