Mathematics, 07.11.2020 05:40 unknowntay04

Find derivative for 3x³ + 2xy + 4y²=0.

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Answer from: ht2005

$y'=\frac{-9x^2-2y}{2(x+4y)}$

General Formulas and Concepts:

Algebra

Equality Properties

Calculus

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

Step-by-step explanation:

Step 1: Define equation

3x³ + 2xy + 4y² = 0

Step 2: Find 1st Derivative

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

Step 3: Find Derivative (Solve for y')

Define:

Move 9x² over: 2y+2xy'+8yy'=-9x^2

Move 2y over: 2xy'+8yy'=-9x^2-2y

Factor y': y'(2x+8y)=-9x^2-2y

Isolate y': $y'=\frac{-9x^2-2y}{2x+8y}$ Factor GCF: $y'=\frac{-9x^2-2y}{2(x+4y)}$

Answer from: Keemdadream13

$\frac{dy}{dx}= -\frac{9x^2+2y}{2(x+4y)}$

Step-by-step explanation:

Find the derivative of 3x^3 + 2xy + 4y^2 = 0 .

In order to find the derivative of this function, $\frac{dy}{dx}$ , 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 $\frac{dy}{dx}$ . 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:

$\frac{d}{dx} (3x^3 + 2xy + 4y^2 = 0)$

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:

$3x^3 \rightarrow 3(3x^3^-^1) \rightarrow 9x^2$ Power Rule

$2xy \rightarrow (2x \times \frac{dy}{dx} + y \times 2)$ Product Rule

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

Let’s combine these steps into one comprehensive operation:

$9x^2 + (2x \times \frac{dy}{dx} + y \times 2) + 8y \times \frac{dy}{dx} = 0$

Simplify.

$9x^2 + 2x\frac{dy}{dx} + 2y + 8y \frac{dy}{dx} = 0$

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

Subtract 9x^2 and from both sides of the equation.

$2x\frac{dy}{dx} +8y \frac{dy}{dx} = -9x^2-2y$

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

$\frac{dy}{dx}(2x+8y) = -9x^2-2y$

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

$\frac{dy}{dx}= \frac{-9x^2-2y}{2x+8y}$

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

$\frac{dy}{dx}= \frac{-9x^2-2y}{2(x+4y)}$ $\frac{dy}{dx}= -\frac{9x^2+2y}{2(x+4y)}$

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