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

2 Answers
Jim H · George C.
Nov 22, 2015

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

Explanation:

Before we differentiate, let's get rid of that silly "divided by 1/x1x"

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

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

1+1/x = (x+1)/x1+1x=x+1x,

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

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^2ddx(x+1x)=ddx(1+1x)=0−1x2=−1x2.

So, here we go:

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

= [1] ln((x+1)/x) + x[1/((x+1)/x) ((-1)/x^2)]=[1]ln(x+1x)+x[1x+1x(−1x2)]

= ln((x+1)/x)+ x^2/(x+1)((-1)/x^2)=ln(x+1x)+x2x+1(−1x2)

= ln((x+1)/x)-1/(x+1)=ln(x+1x)−1x+1

BeeFree
Nov 22, 2015

Using the property of logs ...

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

Explanation:

y=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

Related questions
See all questions in Differentiating Logarithmic Functions with Base e
Impact of this question
2648 views around the world
You can reuse this answer
Creative Commons License