# How do you differentiate #y=(1+lnx)/(x^2-lnx)# using the quotient rule?

Quotient rule:

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To differentiate the function ( y = \frac{1+\ln(x)}{x^2 - \ln(x)} ) using the quotient rule, we apply the formula:

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

where ( u = 1 + \ln(x) ) and ( v = x^2 - \ln(x) ).

First, find the derivatives of ( u ) and ( v ):

[ u' = \frac{d}{dx}(1 + \ln(x)) = \frac{d}{dx}(1) + \frac{d}{dx}(\ln(x)) = 0 + \frac{1}{x} = \frac{1}{x} ]

[ v' = \frac{d}{dx}(x^2 - \ln(x)) = \frac{d}{dx}(x^2) - \frac{d}{dx}(\ln(x)) = 2x - \frac{1}{x} ]

Now, substitute ( u' ) and ( v' ) into the quotient rule formula:

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

Simplify the expression:

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

[ \frac{x^3 - x\ln(x) - (2x^2 + x\ln(x) - 1 - \ln(x))}{x(x^2 - \ln(x))^2} ]

[ \frac{x^3 - 2x^2 + 1}{x(x^2 - \ln(x))^2} ]

This is the derivative of ( y ) with respect to ( x ).

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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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