To find [tex]\(\frac{dy}{dx}\)[/tex] for the curve defined by the equation [tex]\(3xy + 2(xy)^2 + xy^3 = 1\)[/tex], we will employ implicit differentiation. Here's a step-by-step solution:
1. Start with the given equation:
[tex]\[
3xy + 2(xy)^2 + xy^3 = 1
\][/tex]
2. Differentiate both sides of the equation implicitly with respect to [tex]\(x\)[/tex]:
For the left-hand side, use the product rule and chain rule as needed:
- Differentiate [tex]\(3xy\)[/tex]:
[tex]\[
\frac{d}{dx}(3xy) = 3 \left( x \frac{dy}{dx} + y \right)
\][/tex]
- Differentiate [tex]\(2(xy)^2\)[/tex]:
Let [tex]\(u = xy\)[/tex]. Thus, the term becomes [tex]\(2u^2\)[/tex].
[tex]\[
\frac{d}{dx}(2u^2) = 4u \frac{du}{dx} \quad \text{where } u = xy
\][/tex]
Now, differentiate [tex]\(u = xy\)[/tex] using the product rule:
[tex]\[
\frac{du}{dx} = y + x \frac{dy}{dx}
\][/tex]
So,
[tex]\[
\frac{d}{dx}(2(xy)^2) = 4(xy) \left( y + x \frac{dy}{dx} \right)
\][/tex]
- Differentiate [tex]\(xy^3\)[/tex]:
[tex]\[
\frac{d}{dx}(xy^3) = x \cdot 3y^2 \frac{dy}{dx} + y^3
\][/tex]
Putting these differentiations together, the left-hand side becomes:
[tex]\[
3 \left( x \frac{dy}{dx} + y \right) + 4(xy) \left( y + x \frac{dy}{dx} \right) + x \cdot 3y^2 \frac{dy}{dx} + y^3
\][/tex]
3. Differentiate the right-hand side of the equation (which is 1):
[tex]\[
\frac{d}{dx}(1) = 0
\][/tex]
4. Combine the results to form an equation:
[tex]\[
3 \left( x \frac{dy}{dx} + y \right) + 4(xy) \left( y + x \frac{dy}{dx} \right) + 3xy^2 \frac{dy}{dx} + y^3 = 0
\][/tex]
5. Simplify the equation:
Distribute and combine like terms:
[tex]\[
3x \frac{dy}{dx} + 3y + 4xy^2 + 4x^2y \frac{dy}{dx} + 3xy^2 \frac{dy}{dx} + y^3 = 0
\][/tex]
Group the terms involving [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
\left( 3x + 4x^2y + 3xy^2 \right) \frac{dy}{dx} + \left( 3y + 4xy^2 + y^3 \right) = 0
\][/tex]
6. Isolate [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
\left( 3x + 4x^2y + 3xy^2 \right) \frac{dy}{dx} = - \left( 3y + 4xy^2 + y^3 \right)
\][/tex]
Solve for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
\frac{dy}{dx} = \frac{- \left( 3y + 4xy^2 + y^3 \right)}{3x + 4x^2y + 3xy^2}
\][/tex]
Simplify the expression:
[tex]\[
\frac{dy}{dx} = \frac{y(-4xy - y^2 - 3)}{x(4xy + 3y^2 + 3)}
\][/tex]
Thus, the derivative of [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex], given the equation [tex]\(3xy + 2(xy)^2 + xy^3 = 1\)[/tex], is:
[tex]\[
\frac{dy}{dx} = \frac{y(-4xy - y^2 - 3)}{x(4xy + 3y^2 + 3)}
\][/tex]
