Let's solve the given problem step-by-step.
Given:
[tex]\[ 3xy + 4y^2 = 4 \][/tex]
We need to find the second derivative [tex]\(\frac{d^2 y}{dx^2}\)[/tex] at the point [tex]\((0, -1)\)[/tex].
### Step 1: Implicit Differentiation
First, let's start by differentiating the given equation implicitly with respect to [tex]\(x\)[/tex].
[tex]\[ \frac{d}{dx} (3xy + 4y^2) = \frac{d}{dx} (4) \][/tex]
Using the product rule on the term [tex]\(3xy\)[/tex] and the chain rule on the term [tex]\(4y^2\)[/tex], we get:
[tex]\[ 3y + 3x \frac{dy}{dx} + 8y \frac{dy}{dx} = 0 \][/tex]
Now, combine like terms:
[tex]\[ 3y + (3x + 8y) \frac{dy}{dx} = 0 \][/tex]
### Step 2: Solve for [tex]\(\frac{dy}{dx}\)[/tex]
Isolate [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ (3x + 8y) \frac{dy}{dx} = -3y \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{-3y}{3x + 8y} \][/tex]
### Step 3: Find the First Derivative at the Point [tex]\((0, -1)\)[/tex]
Substitute [tex]\(x = 0\)[/tex] and [tex]\(y = -1\)[/tex]:
[tex]\[ \frac{dy}{dx} \Big|_{(0, -1)} = \frac{-3(-1)}{3(0) + 8(-1)} = \frac{3}{-8} = -\frac{3}{8} \][/tex]
### Step 4: Differentiate the First Derivative to Find the Second Derivative
We need to find [tex]\(\frac{d^2 y}{dx^2}\)[/tex]. Using the expression for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} \left( \frac{-3y}{3x + 8y} \right) \][/tex]
Use implicit differentiation again:
[tex]\[ \frac{d}{dx} \left( \frac{-3y}{3x + 8y} \right) \][/tex]
Apply the quotient rule:
[tex]\[ \frac{d}{dx} \left( \frac{-3y}{3x + 8y} \right) = \frac{( (3x + 8y) \frac{d}{dx}(-3y) - (-3y) \frac{d}{dx}(3x + 8y) )}{(3x + 8y)^2} \][/tex]
[tex]\[ \frac{d}{dx} \left( -3y \right) = -3 \frac{dy}{dx} \][/tex]
[tex]\[ \frac{d}{dx} \left(3x + 8y\right) = 3 + 8 \frac{dy}{dx} \][/tex]
Now substitute back:
[tex]\[ -3(3x + 8y)\frac{dy}{dx} - (-3y)(3 + 8 \frac{dy}{dx}) \][/tex]
[tex]\[ \frac{( -3(3x + 8y)\frac{dy}{dx} + 3y (3 + 8 \frac{dy}{dx} ) )}{(3x + 8y)^2} \][/tex]
Simplifying:
[tex]\[ \frac{( -9xy - 24y^2) (\frac{dy}{dx}) + (-9y -24y \frac{dy}{dx)}}}{( 3x + 8y)
^2} \][/tex]
Now evaluate at [tex]\((0, -1)\)[/tex] :
[tex]\[3 \cdot 0 + 8 \cdot (-1) = -8 \][/tex]
[tex]\(\((3x + 8y)=0+( -8 )=(-8 )
^2 \)[/tex])=64\\
\[ d^2y dx^2fw
which is \(\(120) \approx 9 x 3 -5
\[
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