Math Problem Statement

4x³+y-6xy=0 find the y prime

Solution

To find the derivative of yy with respect to xx (denoted as dydx\frac{dy}{dx} or yy'), we need to implicitly differentiate the given equation.

The equation is:

4x3+y6xy=04x^3 + y - 6xy = 0

Step 1: Differentiate both sides with respect to xx

We will differentiate each term individually with respect to xx, keeping in mind that yy is a function of xx, so we must apply the chain rule where necessary.

  1. Differentiate 4x34x^3: ddx(4x3)=12x2\frac{d}{dx}(4x^3) = 12x^2

  2. Differentiate yy with respect to xx: ddx(y)=y\frac{d}{dx}(y) = y'

  3. Differentiate 6xy-6xy with respect to xx. This requires the product rule:

    = -6 \left( x \cdot y' + y \cdot 1 \right) = -6xy' - 6y$$

Step 2: Substitute the derivatives back into the equation

Substitute the differentiated terms back into the original equation:

12x2+y6xy6y=012x^2 + y' - 6xy' - 6y = 0

Step 3: Solve for yy'

To solve for yy', first combine like terms:

y6xy=6y12x2y' - 6xy' = 6y - 12x^2

Factor out yy' on the left side:

y(16x)=6y12x2y'(1 - 6x) = 6y - 12x^2

Finally, solve for yy':

y=6y12x216xy' = \frac{6y - 12x^2}{1 - 6x}

So the derivative of yy with respect to xx is:

y=6y12x216xy' = \frac{6y - 12x^2}{1 - 6x}

Would you like more details or have any questions? Here are 5 questions related to this problem:

  1. How does the product rule apply when differentiating 6xy-6xy?
  2. What is the significance of implicit differentiation in this problem?
  3. Can you find the second derivative yy'' from this result?
  4. How does the chain rule help in differentiating terms like yy with respect to xx?
  5. What are the possible critical points for the function y(x)y(x) based on this derivative?

Tip: Always remember to apply the chain rule when differentiating a function that depends on another variable.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Product Rule
Chain Rule

Formulas

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Theorems

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Suitable Grade Level

Advanced High School