# How do you find y'' by implicit differentiation for #4x^2 + 3y^2 = 6#?

On to the second derivative:

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To find ( y'' ) by implicit differentiation for the equation ( 4x^2 + 3y^2 = 6 ), follow these steps:

- Differentiate both sides of the equation with respect to ( x ).
- Apply the chain rule when differentiating ( y^2 ).
- Solve for ( y'' ) in terms of ( x ) and ( y ).

First differentiation: [ \frac{d}{dx}(4x^2) + \frac{d}{dx}(3y^2) = \frac{d}{dx}(6) ]

Simplify: [ 8x + 6y \frac{dy}{dx} = 0 ]

Now differentiate ( 6y \frac{dy}{dx} ) using the product rule: [ 6 \frac{dy}{dx} + 6y \frac{d^2y}{dx^2} = 0 ]

Isolate ( \frac{d^2y}{dx^2} ): [ 6y \frac{d^2y}{dx^2} = -6 \frac{dy}{dx} ]

[ \frac{d^2y}{dx^2} = -\frac{1}{y} \frac{dy}{dx} ]

Substitute the value of ( \frac{dy}{dx} ) from the first derivative: [ \frac{d^2y}{dx^2} = -\frac{1}{6y} (8x) ]

[ \frac{d^2y}{dx^2} = -\frac{4x}{3y} ]

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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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