How do you find the second derivative of #y^2 + x + sin y = 9#?
I get:
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The second derivative is
Implicit differentiation gives us
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To find the second derivative of the given function, you need to follow these steps:
 Differentiate the given equation implicitly with respect to (x) to find the first derivative (\frac{dy}{dx}).
 Once you have the first derivative, differentiate it again with respect to (x) to find the second derivative (\frac{d^2y}{dx^2}).
Given equation: (y^2 + x + \sin(y) = 9)

Differentiate implicitly with respect to (x): [ \frac{d}{dx}(y^2) + \frac{d}{dx}(x) + \frac{d}{dx}(\sin(y)) = \frac{d}{dx}(9) ] [ 2yy' + 1 + \cos(y) \cdot y' = 0 ] [ 2yy' + y' + \cos(y) = 0 ] [ y'(2y + 1) = \cos(y) ] [ y' = \frac{\cos(y)}{2y + 1} ]

Now, differentiate (y') with respect to (x) to find the second derivative (\frac{d^2y}{dx^2}): [ \frac{d}{dx}(y') = \frac{d}{dx}\left(\frac{\cos(y)}{2y + 1}\right) ] [ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{\cos(y)}{2y + 1}\right) ] [ \frac{d^2y}{dx^2} = \frac{(\sin(y)y' \cdot y' + \cos(y)(2yy'))}{(2y + 1)^2} ]
Substitute (y') from the first derivative: [ \frac{d^2y}{dx^2} = \frac{(\sin(y)\left(\frac{\cos(y)}{2y + 1}\right) \cdot \left(\frac{\cos(y)}{2y + 1}\right) + \cos(y)(2y\left(\frac{\cos(y)}{2y + 1}\right) + 1))}{(2y + 1)^2} ]
Simplify this expression to get the second derivative.
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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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