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

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Answer 1 · 2015-11-22T17:22:09

I like the form: #ln((x+1)/x)-1/(x+1)#

Before we differentiate, let's get rid of that silly "divided by #1/x#"

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

We can now use the product rule, but also note that

#1+1/x = (x+1)/x#,

So we have #x ln((x+1)/x)#

When we differentiate the second factor, we'll need the chain rule and we'll need #d/dx((x+1)/x) = d/dx(1+1/x) = 0-1/x^2 = (-1)/x^2#.

So, here we go:

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

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

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

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

Answer 2 · 2015-11-22T18:13:36

Using the property of logs ...

#y=xln((x+1)/x)=x[ln(x+1)-ln(x)]#

#y=x[ln(x+1)-ln(x)]#

Using only the product rule :

#y'=(1)[ln(x+1)-ln(x)] + x[1/(x+1)-1/x]#

Now simplify ...

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

hope that helped