Question Statement

Find $\frac{dy}{dx}$ given the implicit equation:

$x^3{y}^2=2x^2+y^2$

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Background and Explanation

This problem requires implicit differentiation, which is used when $y$ is defined implicitly as a function of $x$ rather than explicitly as $y=f(x)$. Since both sides of the equation contain terms involving $x$ and $y$, we differentiate both sides with respect to $x$ and apply the chain rule whenever we differentiate a $y$-term.

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Solution

We start with the given equation: $x^3{y}^2=2x^2+y^2$

Differentiate both sides with respect to $x$:

$\frac{d}{dx}\left(x^3{y}^2\right)=2\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(y^2\right)$

Apply the product rule to the left side $\frac{d}{dx}(x^3{y}^2)$. The product rule states that $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$. Here, treat $y^2$ as a function of $x$ (since $y$ depends on $x$), so we need the chain rule for the $y^2$ term:

$y^2\frac{d}{dx}\left(x^3\right)+x^3\frac{d}{dx}\left(y^2\right)=2(2x)+2y\frac{dy}{dx}$

Compute the derivatives:

• $\frac{d}{dx}(x^3)=3x^2$
• $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$ (by the chain rule)

Substituting these in:

$y^2\left(3x^2\right)+x^3\left(2y\frac{dy}{dx}\right)=4x+2y\frac{dy}{dx}$

Collect all terms containing $\frac{dy}{dx}$ on the left side and move all other terms to the right side:

$2x^{3}y\frac{dy}{dx}-2y\frac{dy}{dx}=4x-3x^2{y}^2$

Factor out $\frac{dy}{dx}$ on the left side:

$\left(2x^{3}y-2y\right)\frac{dy}{dx}=4x-3x^2{y}^2$

Solve for $\frac{dy}{dx}$ by dividing both sides by $(2x^{3}y-2y)$:

$\frac{dy}{dx}=\frac{4x-3x^2{y}^2}{2x^{3}y-2y}$

Factor the numerator and denominator to simplify:

• Numerator: $4x-3x^2{y}^2=x(4-3x{y}^2)$
• Denominator: $2x^{3}y-2y=2y(x^3-1)$

Thus, the final answer is:

$\frac{dy}{dx}=\frac{x\left(4-3x{y}^2\right)}{2y\left(x^3-1\right)}$

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Key Formulas or Methods Used

• Implicit Differentiation: Differentiating both sides of an equation with respect to $x$ when $y$ is defined implicitly as a function of $x$
• Product Rule: $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$ (used for the $x^3{y}^2$ term)
• Chain Rule: $\frac{d}{dx}(f(y))=f^{\prime }(y)\cdot \frac{dy}{dx}$ (used when differentiating $y^2$ with respect to $x$)
• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$

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Summary of Steps

1. Differentiate both sides of the equation $x^3{y}^2=2x^2+y^2$ with respect to $x$.
2. Apply the product rule to $x^3{y}^2$ and the chain rule to $y^2$ terms.
3. Expand and simplify the derivatives: $3x^2{y}^2+2x^{3}y\frac{dy}{dx}=4x+2y\frac{dy}{dx}$.
4. Collect $\frac{dy}{dx}$ terms on one side and remaining terms on the other: $2x^{3}y\frac{dy}{dx}-2y\frac{dy}{dx}=4x-3x^2{y}^2$.
5. Factor out $\frac{dy}{dx}$: $(2x^{3}y-2y)\frac{dy}{dx}=4x-3x^2{y}^2$.
6. Solve for $\frac{dy}{dx}$ and simplify by factoring: $\frac{dy}{dx}=\frac{x(4-3x{y}^2)}{2y(x^3-1)}$.