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

1 Answer

Aug 30, 2015

#d/dx(ln(1+(1/x))) = (-1)/(x(x+1))#

Explanation:

Although you could use #d/dx (ln(u)) = 1/u (du)/dx#, the algebra will get messy that way.

Let's rewrite using properties of #ln#.

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

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

So

#dy/dx = 1/(x+1) - 1/x = (-1)/(x(x+1))#

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