How do you find the second derivative implicitly of #x^2 + xy = 5#?
Implicit differentiation second time gives,
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To find the second derivative implicitly of ( x^2 + xy = 5 ), follow these steps:
 Differentiate both sides of the equation with respect to ( x ) using the implicit differentiation technique.
 Apply the product rule where necessary.
 Solve for ( \frac{{d^2y}}{{dx^2}} ) by differentiating the resulting expression obtained from step 1.
Let's go through the steps:

Implicit differentiation: [ \frac{d}{dx}(x^2) + \frac{d}{dx}(xy) = \frac{d}{dx}(5) ]

Apply the product rule: [ 2x + x\frac{dy}{dx} + y + x\frac{d^2y}{dx^2} = 0 ]

Solve for ( \frac{d^2y}{dx^2} ): [ x\frac{d^2y}{dx^2} = 2x  y  x\frac{dy}{dx} ] [ \frac{d^2y}{dx^2} = \frac{2x  y  x\frac{dy}{dx}}{x} ]
This expression represents the second derivative implicitly of ( x^2 + xy = 5 ).
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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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