Math Problem Statement

Differentiate the equation 4y^2 + y + 2x^2 = 0 implicitly to find dy/dx in terms of x and y.

Solution

Let's solve the given problem.

We are tasked with differentiating the equation 4y2+y+2x2=04y^2 + y + 2x^2 = 0 implicitly to find dydx\frac{dy}{dx} in terms of xx and yy.

Step 1: Differentiate both sides

The equation is: 4y2+y+2x2=04y^2 + y + 2x^2 = 0

Differentiating term by term with respect to xx, we apply the chain rule for yy, as yy is a function of xx:

  1. The derivative of 4y24y^2 is: ddx(4y2)=8ydydx\frac{d}{dx}(4y^2) = 8y \frac{dy}{dx}

  2. The derivative of yy is: ddx(y)=dydx\frac{d}{dx}(y) = \frac{dy}{dx}

  3. The derivative of 2x22x^2 is: ddx(2x2)=4x\frac{d}{dx}(2x^2) = 4x

  4. The derivative of 00 is simply: 00

Step 2: Combine the derivatives

Now substitute these derivatives into the equation: 8ydydx+dydx+4x=08y \frac{dy}{dx} + \frac{dy}{dx} + 4x = 0

Step 3: Factorize terms involving dydx\frac{dy}{dx}

Group terms with dydx\frac{dy}{dx} on one side: dydx(8y+1)+4x=0\frac{dy}{dx} (8y + 1) + 4x = 0

Solve for dydx\frac{dy}{dx}: dydx(8y+1)=4x\frac{dy}{dx} (8y + 1) = -4x

dydx=4x8y+1\frac{dy}{dx} = \frac{-4x}{8y + 1}

Final Answer:

dydx=4x8y+1\frac{dy}{dx} = \frac{-4x}{8y + 1}

Would you like a further breakdown or related examples?

Related Questions:

  1. How do we apply the chain rule in implicit differentiation?
  2. Why do we need to treat yy as a function of xx?
  3. How would the solution change if the equation was 4y2+y+2x3=04y^2 + y + 2x^3 = 0?
  4. Can implicit differentiation be used to find higher-order derivatives?
  5. What does dydx\frac{dy}{dx} represent geometrically in this case?

Tip:

Always group terms with dydx\frac{dy}{dx} together before isolating it—this simplifies solving!

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Chain Rule
Differentiation

Formulas

dy/dx = -4x / (8y + 1)

Theorems

Chain Rule for Implicit Differentiation

Suitable Grade Level

Grades 11-12