Find #y′′# for the curve #ln(x) + y = ln(x^2) − y^2# at #y = 0#?
Given:
Add
Use the fact that subtraction of logarithms is the as division within the argument:
Write is standard quadratic form:
Here is a graph of equation [1]:
We can use the quadratic formula to obtain two equations of y in terms of x
and
Here are the graphs of equations [1a] and [1b]:
Please observe that equation [1a] contains the point of interest,
Evaluate when
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To find ( y'' ) for the curve ( \ln(x) + y = \ln(x^2)  y^2 ) at ( y = 0 ), we first need to differentiate the given equation twice with respect to ( x ).

Differentiate the equation implicitly once: [ \frac{d}{dx}(\ln(x) + y) = \frac{d}{dx}(\ln(x^2)  y^2) ] [ \frac{1}{x} + \frac{dy}{dx} = \frac{2x}{x^2}  2y \frac{dy}{dx} ]

Solve for ( \frac{dy}{dx} ): [ \frac{dy}{dx} = \frac{2x}{x^2}  \frac{1}{x}  2y \frac{dy}{dx} ]

Differentiate the equation obtained in step 2 implicitly again: [ \frac{d}{dx} \left( \frac{2x}{x^2}  \frac{1}{x}  2y \frac{dy}{dx} \right) ]

Solve for ( \frac{d^2y}{dx^2} ) by plugging in ( y = 0 ): [ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{2x}{x^2}  \frac{1}{x} \right) ] [ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{2}{x}  \frac{1}{x} \right) ] [ \frac{d^2y}{dx^2} = \frac{2}{x^2} + \frac{1}{x^2} ]
Finally, when ( y = 0 ): [ \frac{d^2y}{dx^2} = \frac{2}{x^2} + \frac{1}{x^2} ] [ \frac{d^2y}{dx^2} = \frac{1}{x^2} ]
So, ( y'' ) for the curve ( \ln(x) + y = \ln(x^2)  y^2 ) at ( y = 0 ) is ( \frac{1}{x^2} ).
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To find ( y'' ) for the curve ( \ln(x) + y = \ln(x^2)  y^2 ) at ( y = 0 ), we first differentiate the given equation with respect to ( x ) twice, and then substitute ( y = 0 ) into the resulting expression.

Differentiate the equation with respect to ( x ) once: [ \frac{d}{dx}(\ln(x) + y) = \frac{d}{dx}(\ln(x^2)  y^2) ]

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

Differentiate the equation obtained in step 2 with respect to ( x ) again: [ \frac{d}{dx}\left(\frac{1}{x} + y'\right) = \frac{d}{dx}\left(\frac{2x}{x^2}  2yy'\right) ]

Simplify and substitute ( y = 0 ): [ \frac{1}{x^2} + y'' = \frac{2}{x^2} ]

Substitute ( y = 0 ) into the equation: [ \frac{1}{x^2} + y'' = \frac{2}{x^2} ] [ \frac{1}{x^2} + y'' = \frac{2}{x^2} ] [ y'' = \frac{1}{x^2}  \left(\frac{2}{x^2}\right) ] [ y'' = \frac{1}{x^2} ]
So, ( y'' = \frac{1}{x^2} ) for the given curve at ( y = 0 ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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