Math Problem Statement

find y' and y'' at the point (-1,1) on graph 3x^2y+2y^3=5x^2

Solution

To find the first and second derivatives, yy' and yy'', of the implicit function given by 3x2y+2y3=5x23x^2y + 2y^3 = 5x^2 at the point (1,1)(-1, 1), we need to follow these steps:

Step 1: Differentiate Implicitly to Find yy'

  1. Differentiate both sides of the equation with respect to xx: ddx(3x2y+2y3)=ddx(5x2)\frac{d}{dx}(3x^2y + 2y^3) = \frac{d}{dx}(5x^2)

  2. Apply the product rule to the term 3x2y3x^2y and the chain rule to 2y32y^3: 3(2x)y+3x2dydx+6y2dydx=10x3(2x)y + 3x^2 \frac{dy}{dx} + 6y^2 \frac{dy}{dx} = 10x

  3. Substitute dydx=y\frac{dy}{dx} = y' and solve for yy': 6xy+(3x2+6y2)y=10x6xy + (3x^2 + 6y^2)y' = 10x y=10x6xy3x2+6y2y' = \frac{10x - 6xy}{3x^2 + 6y^2}

  4. Substitute x=1x = -1 and y=1y = 1 to find yy' at (1,1)(-1, 1): y=10(1)6(1)(1)3(1)2+6(1)2=10+63+6=49y' = \frac{10(-1) - 6(-1)(1)}{3(-1)^2 + 6(1)^2} = \frac{-10 + 6}{3 + 6} = \frac{-4}{9}

Therefore, y=49y' = -\frac{4}{9} at the point (1,1)(-1, 1).

Step 2: Differentiate Again to Find yy''

  1. Differentiate yy' with respect to xx to obtain yy'': y=10x6xy3x2+6y2y' = \frac{10x - 6xy}{3x^2 + 6y^2}

  2. Using the quotient rule ddx(uv)=uvuvv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}:

    • Let u=10x6xyu = 10x - 6xy and v=3x2+6y2v = 3x^2 + 6y^2.

    • Differentiate uu and vv with respect to xx: u=106y6xyu' = 10 - 6y - 6xy' v=6x+12yyv' = 6x + 12yy'

    • Substitute into the quotient rule: y=(106y6xy)(3x2+6y2)(10x6xy)(6x+12yy)(3x2+6y2)2y'' = \frac{(10 - 6y - 6xy')(3x^2 + 6y^2) - (10x - 6xy)(6x + 12yy')}{(3x^2 + 6y^2)^2}

  3. Substitute x=1x = -1, y=1y = 1, and y=49y' = -\frac{4}{9} into the expression to find yy'':

    • Substitute values into uu' and vv': u=106(1)6(1)(49)=483=43u' = 10 - 6(1) - 6(-1)\left(-\frac{4}{9}\right) = 4 - \frac{8}{3} = \frac{4}{3} v=6(1)+12(1)(49)=6163=343v' = 6(-1) + 12(1)\left(-\frac{4}{9}\right) = -6 - \frac{16}{3} = -\frac{34}{3}

    • Calculate yy'': y=(43)(9)(4)(343)81=12136981=2827y'' = \frac{\left(\frac{4}{3}\right)(9) - (-4)(-\frac{34}{3})}{81} = \frac{12 - \frac{136}{9}}{81} = -\frac{28}{27}

So, y=2827y'' = -\frac{28}{27} at the point (1,1)(-1, 1).

To summarize:

  • y(1,1)=49y'(-1,1) = -\frac{4}{9}
  • y(1,1)=2827y''(-1,1) = -\frac{28}{27}

Would you like further details or have any questions?

Follow-Up Questions:

  1. How does implicit differentiation work and why do we use it?
  2. What other methods are there for finding derivatives if the function is not given implicitly?
  3. Why do we need to apply the quotient rule in finding the second derivative here?
  4. Can this problem be solved without finding yy' first?
  5. How can one verify that these derivatives are correct?

Tip:

When differentiating implicitly, always keep track of both xx and yy, and remember to apply the chain rule when differentiating terms with yy.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Quotient Rule
Chain Rule

Formulas

Implicit differentiation: d/dx[f(x, y)] = ∂f/∂x + (∂f/∂y)(dy/dx)
Quotient rule: d/dx(u/v) = (u'v - uv')/v^2
Product rule: d/dx(uv) = u'v + uv'

Theorems

Implicit Differentiation Theorem
Chain Rule

Suitable Grade Level

Grades 11-12, College Calculus