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

Simplify and apply the chain rule to find that

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

(simplification)

(product rule)

(simplification)

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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)).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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