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

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Answer 1 · 2016-07-16T01:03:27

Simplify and apply the chain rule to find that

$d/dxln(1+1/x)/(1/x)=ln(1+1/x)-1/(x+1)$

To make this a little easier, first we will simplify the expression to

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

Now, using the product rule, chain rule, and the derivatives $d/dxln(x) = 1/x$ and $d/dx1/x = -1/x^2$ , we have

$d/dxln(1+1/x)/(1/x) = d/dxxln(1+1/x)$

(simplification)

$=x(d/dxln(1+1/x)) + ln(1+1/x)(d/dxx)$

(product rule)

$=x(1/(1+1/x)(d/dx(1+1/x)))+ln(1+1/x)*1$

(chain rule and derivatives of $ln(x)$ and $x$ )

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

(derivative of $1/x$ )

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

(simplification)

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