# How do you differentiate #cos (y) -( x^2y^3) + 2y = pi#?

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To differentiate the given equation ( \cos(y) - x^2y^3 + 2y = \pi ) with respect to (x), we'll use implicit differentiation:

[ \frac{d}{dx}(\cos(y)) - \frac{d}{dx}(x^2y^3) + \frac{d}{dx}(2y) = \frac{d}{dx}(\pi) ]

Using the chain rule and product rule where necessary, we get:

[ -\sin(y)\frac{dy}{dx} - 2xy^3\frac{dy}{dx} - 3x^2y^2 + 2\frac{dy}{dx} = 0 ]

Grouping terms involving ( \frac{dy}{dx} ), we get:

[ \left(-\sin(y) - 2xy^3 + 2\right)\frac{dy}{dx} = 3x^2y^2 ]

Finally, solving for ( \frac{dy}{dx} ), we have:

[ \frac{dy}{dx} = \frac{3x^2y^2}{-\sin(y) - 2xy^3 + 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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