How do you implicitly differentiate #x y + 2x + 3x^2 = 4#?

Answer 1

You can do it like this:

#xy+2x+3x^2=4#
#:.D(xy+2x+3x^2)=D(4)#
#:.xy'+y+2+6x=0#
#:.-xy'=6x+2+y#
#:.y'=-((6x+2+y))/(x)#
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Answer 2

To implicitly differentiate the equation (x y + 2x + 3x^2 = 4) with respect to (x), follow these steps:

  1. Differentiate each term of the equation with respect to (x).
  2. Apply the product rule when differentiating the term (xy).
  3. Solve for (\frac{dy}{dx}).

Here are the steps:

  1. 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*} ]

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

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

  4. 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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Answer from HIX Tutor

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.

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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