How do you implicitly differentiate #7=3x-y+y^2x#?

Answer 1

#dy/dx=(y^2+3)/(1-2xy)#

The challenge in this problem lies in using the product rule for the #y^2x# term.
#d/dx[7=3x-y+y^2x]#
#0=3-dy/dx+d/dx[y^2]*x+d/dx[x]*y^2#
#0=3-dy/dx+2xy*dy/dx+y^2#
#dy/dx-2xy*dy/dx=y^2+3#
#dy/dx(1-2xy)=y^2+3#
#dy/dx=(y^2+3)/(1-2xy)#
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Answer 2

To implicitly differentiate (7 = 3x - y + y^2x), follow these steps:

  1. Differentiate each term of the equation with respect to (x).
  2. Apply the chain rule whenever differentiating terms involving (y).
  3. Collect terms involving (y') on one side and terms not involving (y') on the other side.
  4. Solve for (y') (the derivative of (y) with respect to (x)).

Differentiating each term with respect to (x):

(d/dx [7] = d/dx [3x] - d/dx [y] + d/dx [y^2x])

Simplify the derivatives:

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

Rearrange the equation:

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

Factor out (\frac{dy}{dx}):

(\frac{dy}{dx} (1 - 2xy) = 3 + y^2)

Solve for (\frac{dy}{dx}):

(\frac{dy}{dx} = \frac{3 + y^2}{1 - 2xy})

This is the implicit derivative of (y) with respect to (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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