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

Question

Samantha Adkison · Accepted Answer

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)

Adam Adkins · Answer

undefined To find the derivative of ln(1+1/x) / (1/x), you can use the quotient rule. The quotient rule states that if you have a function in the form of u/v, where u and v are functions of x, then the derivative can be found using the formula:

(d/dx)(u/v) = (v * du/dx - u * dv/dx) / v^2

In this case, let u = ln(1+1/x) and v = (1/x). Then, compute the derivatives of u and v with respect to x, and substitute them into the quotient rule formula.

The derivative of ln(1+1/x) is (1 / (1+1/x)) * (d/dx)(1+1/x) using the chain rule, and the derivative of 1/x is -1/x^2.

Substituting these derivatives into the quotient rule formula, you get:

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

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

= (-1/(x^2(1+1/x))) + (ln(1+1/x)/(x^3))

= -1/(x^2 + x) + ln(1+1/x)/(x^3)

Therefore, the derivative of ln(1+1/x) / (1/x) is (-1/(x^2 + x)) + (ln(1+1/x)/(x^3)).