How do you implicitly differentiate #x y + 2x + 3x^2 = 4#?
You can do it like this:
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To implicitly differentiate the equation (x y + 2x + 3x^2 = 4) with respect to (x), follow these steps:
 Differentiate each term of the equation with respect to (x).
 Apply the product rule when differentiating the term (xy).
 Solve for (\frac{dy}{dx}).
Here are the steps:

The differentiation of each term: [ \begin{align*} \frac{d}{dx}(xy) & = y + x\frac{dy}{dx} \ \frac{d}{dx}(2x) & = 2 \ \frac{d}{dx}(3x^2) & = 6x \end{align*} ]

Using the product rule for (\frac{d}{dx}(xy)): [ \frac{d}{dx}(xy) = y + x\frac{dy}{dx} ]

Putting it all together: [ y + x\frac{dy}{dx} + 2 + 6x = 0 ]

Rearrange to solve for (\frac{dy}{dx}): [ x\frac{dy}{dx} = (y + 2 + 6x) \ \frac{dy}{dx} = \frac{y + 2 + 6x}{x} ]
So, the implicit differentiation of (xy + 2x + 3x^2 = 4) with respect to (x) yields (\frac{y + 2 + 6x}{x}).
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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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