Answer :

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Answer:

[tex]y'=\frac{-9x^2-2y}{2(x+4y)}[/tex]

General Formulas and Concepts:

Algebra

  • Equality Properties

Calculus

  • Chain Rule: [tex]\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
  • Basic Power Rule: f’(x) = c·nxⁿ⁻¹
  • Derivative of a constant equals 0
  • Implicit Differentiation

Step-by-step explanation:

Step 1: Define equation

3x³ + 2xy + 4y² = 0

Step 2: Find 1st Derivative

  1. Set up Derivative:                         [tex]\frac{dy}{dx} [3x^3 + 2xy + 4y^2 = 0][/tex]
  2. Take Implicit Differentiation:        [tex]9x^2+(2y+2xy')+8yy'=0[/tex]

Step 3: Find Derivative (Solve for y')

  1. Define:                         [tex]9x^2+(2y+2xy')+8yy'=0[/tex]
  2. Move 9x² over:           [tex]2y+2xy'+8yy'=-9x^2[/tex]
  3. Move 2y over:             [tex]2xy'+8yy'=-9x^2-2y[/tex]
  4. Factor y':                      [tex]y'(2x+8y)=-9x^2-2y[/tex]
  5. Isolate y':                     [tex]y'=\frac{-9x^2-2y}{2x+8y}[/tex]
  6. Factor GCF:                 [tex]y'=\frac{-9x^2-2y}{2(x+4y)}[/tex]
Supernova

Answer:

[tex]\frac{dy}{dx}= -\frac{9x^2+2y}{2(x+4y)}[/tex]

Step-by-step explanation:

Find the derivative of [tex]3x^3 + 2xy + 4y^2 = 0[/tex].

In order to find the derivative of this function, [tex]\frac{dy}{dx}[/tex], we can start by noticing that there are two variables in this problem, x and y.

Since we want the derivative with respect to x, every time we encounter the variable “y” in this problem we can change it to [tex]\frac{dy}{dx}[/tex]. We will see this happen later on in the solving process by using implicit differentiation.

Let’s start by taking the derivative of the entire function:

  • [tex]\frac{d}{dx} (3x^3 + 2xy + 4y^2 = 0)[/tex]

Using the Power Rule and the Product Rule, we can perform implicit differentiation on this equation. Let’s take the derivative of each separate piece in this function:

[tex]3x^3 \rightarrow 3(3x^3^-^1) \rightarrow 9x^2[/tex]                                   Power Rule

[tex]2xy \rightarrow (2x \times \frac{dy}{dx} + y \times 2)[/tex]                                Product Rule

[tex]4y^2 \rightarrow 2(4y) \times \frac{dy}{dx} \rightarrow 8y \times \frac{dy}{dx}[/tex]                 Power Rule & Implicit Differentiation

Let’s combine these steps into one comprehensive operation:

  • [tex]9x^2 + (2x \times \frac{dy}{dx} + y \times 2) + 8y \times \frac{dy}{dx} = 0[/tex]

Simplify.

  • [tex]9x^2 + 2x\frac{dy}{dx} + 2y + 8y \frac{dy}{dx} = 0[/tex]

Keep all terms containing [tex]\frac{dy}{dx}[/tex] on the left side of the equation and move everything else to the right side of the equation. This way we can solve for [tex]\frac{dy}{dx}[/tex], which, in this case, is the derivative of the original function.

Subtract [tex]9x^2[/tex] and [tex]2y[/tex] from both sides of the equation.

  • [tex]2x\frac{dy}{dx} +8y \frac{dy}{dx} = -9x^2-2y[/tex]

Factor out dy/dx from the left side of the equation.

  • [tex]\frac{dy}{dx}(2x+8y) = -9x^2-2y[/tex]

Divide both sides of the equation by (2x + 8y).

  • [tex]\frac{dy}{dx}= \frac{-9x^2-2y}{2x+8y}[/tex]

You can leave it in this form, or you can convert this to either:

  • [tex]\frac{dy}{dx}= \frac{-9x^2-2y}{2(x+4y)}[/tex]
  • [tex]\frac{dy}{dx}= -\frac{9x^2+2y}{2(x+4y)}[/tex]

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