Math Problem Statement

Find π‘‘𝑦𝑑π‘₯οΏ½οΏ½οΏ½οΏ½ by implicit differentiation.

2π‘₯^3𝑦^2βˆ’4π‘₯^2𝑦=4

Solution

To find dydx\frac{dy}{dx} by implicit differentiation, we'll differentiate both sides of the equation 2x3y2βˆ’4x2y=42x^3y^2 - 4x^2y = 4 with respect to xx, while treating yy as a function of xx (i.e., using the chain rule when differentiating terms involving yy).

Step 1: Differentiate both sides

We differentiate each term:

ddx(2x3y2)βˆ’ddx(4x2y)=ddx(4)\frac{d}{dx}(2x^3y^2) - \frac{d}{dx}(4x^2y) = \frac{d}{dx}(4)

First term: 2x3y22x^3y^2

Using the product rule for differentiation: ddx(uv)=uβ€²v+uvβ€²\frac{d}{dx}(uv) = u'v + uv', where u=2x3u = 2x^3 and v=y2v = y^2, we get:

ddx(2x3y2)=6x2y2+2x3β‹…2yβ‹…dydx\frac{d}{dx}(2x^3y^2) = 6x^2y^2 + 2x^3 \cdot 2y \cdot \frac{dy}{dx}

Simplified: 6x2y2+4x3ydydx6x^2y^2 + 4x^3y \frac{dy}{dx}

Second term: βˆ’4x2y-4x^2y

Again, apply the product rule: u=βˆ’4x2u = -4x^2 and v=yv = y, so:

ddx(βˆ’4x2y)=βˆ’8xyβˆ’4x2β‹…dydx\frac{d}{dx}(-4x^2y) = -8xy - 4x^2 \cdot \frac{dy}{dx}

Right-hand side: 44

ddx(4)=0\frac{d}{dx}(4) = 0

Step 2: Combine results

Putting everything together, we have: 6x2y2+4x3ydydxβˆ’8xyβˆ’4x2dydx=06x^2y^2 + 4x^3y \frac{dy}{dx} - 8xy - 4x^2 \frac{dy}{dx} = 0

Step 3: Solve for dydx\frac{dy}{dx}

Group the terms involving dydx\frac{dy}{dx} on one side: 4x3ydydxβˆ’4x2dydx=βˆ’6x2y2+8xy4x^3y \frac{dy}{dx} - 4x^2 \frac{dy}{dx} = -6x^2y^2 + 8xy

Factor out dydx\frac{dy}{dx} on the left side: dydx(4x3yβˆ’4x2)=8xyβˆ’6x2y2\frac{dy}{dx}(4x^3y - 4x^2) = 8xy - 6x^2y^2

Finally, solve for dydx\frac{dy}{dx}: dydx=8xyβˆ’6x2y24x3yβˆ’4x2\frac{dy}{dx} = \frac{8xy - 6x^2y^2}{4x^3y - 4x^2}

Final Answer:

dydx=8xyβˆ’6x2y24x2(xyβˆ’1)\frac{dy}{dx} = \frac{8xy - 6x^2y^2}{4x^2(xy - 1)}

Would you like more details on any step, or have any questions?

Here are 5 related questions you might be interested in:

  1. How does implicit differentiation work?
  2. What is the product rule in differentiation?
  3. How does the chain rule apply when differentiating implicitly?
  4. Can implicit differentiation be applied to trigonometric functions?
  5. How do you solve implicit equations after differentiation?

Tip: When using implicit differentiation, always remember to apply the product rule and chain rule carefully to terms involving both xx and yy.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Product Rule
Chain Rule

Formulas

Product Rule: (uv)' = u'v + uv'
Chain Rule: dy/dx = dy/du * du/dx

Theorems

Implicit Differentiation Theorem

Suitable Grade Level

Grades 11-12 (Advanced High School Calculus)