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

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Answer 1 · 2016-12-16T18:52:56

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.

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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)#

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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!