Math Problem Statement

get y'

Find y' for the equation x^2y - 3x + 2y^2 = 0 at the point (0, 1).

Solution

To find yy' for the given equation x2y3x+2y2=0x^2y - 3x + 2y^2 = 0 at the point (0,1)(0,1), we'll differentiate the equation implicitly with respect to xx.

Step-by-Step Solution:

  1. Given Equation: x2y3x+2y2=0x^2y - 3x + 2y^2 = 0

  2. Differentiate Implicitly with respect to xx: Using the product rule for x2yx^2y and the chain rule for y2y^2: ddx(x2y)ddx(3x)+ddx(2y2)=0\frac{d}{dx}(x^2y) - \frac{d}{dx}(3x) + \frac{d}{dx}(2y^2) = 0

    The derivatives are:

    • ddx(x2y)=2xy+x2dydx=2xy+x2y\frac{d}{dx}(x^2y) = 2xy + x^2\frac{dy}{dx} = 2xy + x^2y',
    • ddx(3x)=3\frac{d}{dx}(-3x) = -3,
    • ddx(2y2)=4ydydx=4yy\frac{d}{dx}(2y^2) = 4y\frac{dy}{dx} = 4yy'.

    Substituting these, we get: 2xy+x2y3+4yy=02xy + x^2y' - 3 + 4yy' = 0

  3. Simplify and Solve for yy': Combine all terms involving yy': x2y+4yy=32xyx^2y' + 4yy' = 3 - 2xy

    Factor out yy': y(x2+4y)=32xyy'(x^2 + 4y) = 3 - 2xy

    Solve for yy': y=32xyx2+4yy' = \frac{3 - 2xy}{x^2 + 4y}

  4. Substitute (x,y)=(0,1)(x, y) = (0, 1): At (0,1)(0,1), substitute x=0x = 0 and y=1y = 1: y=32(0)(1)02+4(1)y' = \frac{3 - 2(0)(1)}{0^2 + 4(1)}

    Simplify: y=34y' = \frac{3}{4}

Final Answer:

y=34y' = \frac{3}{4}

Would you like me to explain any step in more detail?


Related Questions:

  1. What is implicit differentiation, and why is it useful?
  2. Can this method be applied to equations without explicit yy terms?
  3. What would happen if the point given were not on the curve?
  4. How does the product rule work for terms like x2yx^2y?
  5. What is the geometrical interpretation of yy' at a specific point?

Tip:

When solving for yy' in implicit differentiation, always isolate the terms with yy' carefully to avoid errors in factoring.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Derivatives
Product Rule
Chain Rule

Formulas

Implicit differentiation: d/dx[f(x, y)] = 0
Product Rule: d/dx[uv] = u'v + uv'
Chain Rule: d/dx[g(y)] = g'(y) * dy/dx

Theorems

Rules of Differentiation
Implicit Differentiation Theorem

Suitable Grade Level

Grades 11-12, College Level Calculus