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

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

### Step 1: Differentiate Implicitly to Find \( y' \)
1. Differentiate both sides of the equation with respect to \( x \):
   \[
   \frac{d}{dx}(3x^2y + 2y^3) = \frac{d}{dx}(5x^2)
   \]
2. Apply the product rule to the term \( 3x^2y \) and the chain rule to \( 2y^3 \):
   \[
   3(2x)y + 3x^2 \frac{dy}{dx} + 6y^2 \frac{dy}{dx} = 10x
   \]
3. Substitute \( \frac{dy}{dx} = y' \) and solve for \( y' \):
   \[
   6xy + (3x^2 + 6y^2)y' = 10x
   \]
   \[
   y' = \frac{10x - 6xy}{3x^2 + 6y^2}
   \]

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

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

### Step 2: Differentiate Again to Find \( y'' \)
1. Differentiate \( y' \) with respect to \( x \) to obtain \( y'' \):
   \[
   y' = \frac{10x - 6xy}{3x^2 + 6y^2}
   \]

2. Using the quotient rule \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \):
   - Let \( u = 10x - 6xy \) and \( v = 3x^2 + 6y^2 \).
   - Differentiate \( u \) and \( v \) with respect to \( x \):
     \[
     u' = 10 - 6y - 6xy'
     \]
     \[
     v' = 6x + 12yy'
     \]

   - Substitute into the quotient rule:
     \[
     y'' = \frac{(10 - 6y - 6xy')(3x^2 + 6y^2) - (10x - 6xy)(6x + 12yy')}{(3x^2 + 6y^2)^2}
     \]

3. Substitute \( x = -1 \), \( y = 1 \), and \( y' = -\frac{4}{9} \) into the expression to find \( y'' \):
   - Substitute values into \( u' \) and \( v' \):
     \[
     u' = 10 - 6(1) - 6(-1)\left(-\frac{4}{9}\right) = 4 - \frac{8}{3} = \frac{4}{3}
     \]
     \[
     v' = 6(-1) + 12(1)\left(-\frac{4}{9}\right) = -6 - \frac{16}{3} = -\frac{34}{3}
     \]

   - Calculate \( y'' \):
     \[
     y'' = \frac{\left(\frac{4}{3}\right)(9) - (-4)(-\frac{34}{3})}{81} = \frac{12 - \frac{136}{9}}{81} = -\frac{28}{27}
     \]

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

To summarize:
- \( y'(-1,1) = -\frac{4}{9} \)
- \( y''(-1,1) = -\frac{28}{27} \)

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 \( y' \) first?
5. How can one verify that these derivatives are correct?

#### Tip:
When differentiating implicitly, always keep track of both \( x \) and \( y \), and remember to apply the chain rule when differentiating terms with \( y \).

Discover how to compute the first and second derivatives of the equation 3x^2y + 2y^3 = 5x^2 at the point (-1, 1) using implicit differentiation, the quotient rule, and the chain rule. This solution includes step-by-step differentiation and is suitable for advanced high school and college calculus students.

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