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

Using the property of logs ...

Using only the product rule :

Now simplify ...

hope that helped

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I like the form:

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

So, here we go:

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To find the derivative of ( \frac{\ln(1+\frac{1}{x})}{\frac{1}{x}} ), we can apply the quotient rule:

Let ( u = \ln(1+\frac{1}{x}) ) and ( v = \frac{1}{\frac{1}{x}} ).

Then, using the quotient rule ( \frac{d}{dx}(\frac{u}{v}) = \frac{u'v - uv'}{v^2} ):

( u' = \frac{d}{dx}(\ln(1+\frac{1}{x})) = \frac{1}{1+\frac{1}{x}} \cdot \frac{d}{dx}(1+\frac{1}{x}) = \frac{1}{1+\frac{1}{x}} \cdot (-\frac{1}{x^2}) )

( v' = \frac{d}{dx}(\frac{1}{\frac{1}{x}}) = \frac{d}{dx}(x) = 1 )

Now, plugging these into the quotient rule formula:

( \frac{d}{dx}(\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}}) = \frac{\frac{1}{1+\frac{1}{x}} \cdot \frac{1}{x^2} - \ln(1+\frac{1}{x}) \cdot 1}{(\frac{1}{x})^2} )

Simplify:

( = \frac{\frac{1}{x^2+1} \cdot \frac{1}{x^2} - \ln(1+\frac{1}{x})}{\frac{1}{x^2}} )

( = \frac{\frac{1}{x^2(x^2+1)} - \ln(1+\frac{1}{x})}{\frac{1}{x^2}} )

( = x^2 \left( \frac{1}{x^2(x^2+1)} - \ln(1+\frac{1}{x}) \right) )

So, the derivative of ( \frac{\ln(1+\frac{1}{x})}{\frac{1}{x}} ) is ( x^2 \left( \frac{1}{x^2(x^2+1)} - \ln(1+\frac{1}{x}) \right) ).

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To find the derivative of the given function ln(1 + 1/x) / (1/x), we can use the quotient rule of differentiation.

Let u = ln(1 + 1/x) and v = 1/x.

Then, applying the quotient rule, the derivative of the function with respect to x is given by:

[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} ]

Where ( \frac{du}{dx} ) represents the derivative of u with respect to x and ( \frac{dv}{dx} ) represents the derivative of v with respect to x.

Now, let's find the derivatives of u and v:

[ \frac{du}{dx} = \frac{d}{dx} \left( \ln(1 + \frac{1}{x}) \right) ]

Using the chain rule, we get:

[ \frac{du}{dx} = \frac{1}{1 + \frac{1}{x}} \cdot \frac{d}{dx} \left(1 + \frac{1}{x}\right) ]

[ \frac{du}{dx} = \frac{1}{1 + \frac{1}{x}} \cdot \left( -\frac{1}{x^2} \right) ]

[ \frac{du}{dx} = -\frac{1}{x^2 + x} ]

Now, for ( \frac{dv}{dx} ):

[ \frac{dv}{dx} = \frac{d}{dx} \left( \frac{1}{x} \right) ]

[ \frac{dv}{dx} = -\frac{1}{x^2} ]

Now, substituting these derivatives into the quotient rule formula:

[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{\frac{1}{x} \left( -\frac{1}{x^2 + x} \right) - \ln(1 + \frac{1}{x}) \left( -\frac{1}{x^2} \right)}{\left( \frac{1}{x} \right)^2} ]

Simplifying further:

[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{-\frac{1}{x(x^2 + x)} + \frac{\ln(1 + \frac{1}{x})}{x^2}}{\frac{1}{x^2}} ]

[ \frac{d}{dx} \left( \frac{u}{v} \right) = -\frac{x^2 \ln(1 + \frac{1}{x}) - (x^2 + x) \ln(1 + \frac{1}{x})}{x(x^2 + x)} ]

[ \boxed{\frac{d}{dx} \left( \frac{u}{v} \right) = -\frac{x^2 \ln(1 + \frac{1}{x}) - (x^2 + x) \ln(1 + \frac{1}{x})}{x(x^2 + x)}} ]

So, the derivative of ln(1 + 1/x) / (1/x) with respect to x is given by the expression derived above.

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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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