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)