# Find y′′ for the curve ln(x) + y = ln(x^2) − y^2 at y = 0? Note that the domain for x is that x > 0. There will only be one point on the curve with y = 0

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For

Given the implicit function:

so that the equation becomes:

Differentiate both sides of the equation:

and again:

substituting:

we have:

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To find ( y'' ) for the given curve at ( y = 0 ), we first need to find ( y' ) and then differentiate it again.

Given the equation ( \ln(x) + y = \ln(x^2) - y^2 ), differentiate both sides with respect to ( x ) to find ( y' ):

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

[ \frac{1}{x} + y' = \frac{2x}{x^2} - 2yy' ]

[ y' = \frac{2x}{x^2} - 2yy' - \frac{1}{x} ]

Now, we substitute ( y = 0 ) into the equation:

[ y' = \frac{2x}{x^2} - 2 \cdot 0 \cdot y' - \frac{1}{x} ]

[ y' = \frac{2}{x} - \frac{1}{x} ]

[ y' = \frac{1}{x} ]

Now, differentiate ( y' ) with respect to ( x ) to find ( y'' ):

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

Thus, ( y'' = -\frac{1}{x^2} ).

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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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