Using implicit differentiation, find $\frac{d y}{d x}$ . $\newline$ $-5 y^{3}-2 x^{4}+3 x+3 y+x y=2$

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Simplify expressions using trigonometric identities

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Q. Using implicit differentiation, find

\frac{d y}{d x}

\newline

-5 y^{3}-2 x^{4}+3 x+3 y+x y=2

Apply Differentiation Rules: We will differentiate each term of the equation with respect to $x$ , remembering to use the product rule for the term $xy$ and the chain rule for terms involving $y$ , since $y$ is a function of $x$ .
Differentiate $-5y^3$ : Differentiate $-5y^3$ with respect to $x$ . Since $y$ is a function of $x$ , we get $-5 \times 3y^2 \times (\frac{dy}{dx}) = -15y^2 \times (\frac{dy}{dx})$ .
Differentiate $-2x^4$ : Differentiate $-2x^4$ with respect to $x$ . This is a straightforward differentiation, yielding $-2 \times 4x^3 = -8x^3$ .
Differentiate $3x$ : Differentiate $3x$ with respect to $x$ . The derivative of $3x$ with respect to $x$ is $3$ .
Differentiate $3y$ : Differentiate $3y$ with respect to $x$ . Since $y$ is a function of $x$ , we get $3 \times \left(\frac{dy}{dx}\right)$ .
Differentiate $xy$ : Differentiate $xy$ with respect to $x$ using the product rule. The product rule states that $\frac{d(uv)}{dx} = u\left(\frac{dv}{dx}\right) + v\left(\frac{du}{dx}\right)$ , where $u = x$ and $v = y$ . So we get $y \cdot 1 + x \cdot \left(\frac{dy}{dx}\right) = y + x\left(\frac{dy}{dx}\right)$ .
Differentiate Constant: Differentiate the constant $2$ with respect to $x$ . The derivative of a constant is $0$ .
Combine Differentiated Terms: Combine all the differentiated terms to form the new equation: $-15y^2 \cdot \frac{dy}{dx} - 8x^3 + 3 + 3\frac{dy}{dx} + y + x\frac{dy}{dx} = 0$ .
Collect Terms on Each Side: Now, we collect all terms involving $\frac{dy}{dx}$ on one side and the rest on the other side. This gives us $-15y^2 \cdot \frac{dy}{dx} + 3\frac{dy}{dx} + x\frac{dy}{dx} = 8x^3 - 3 - y$ .
Factor out $\frac{dy}{dx}$ : Factor out $\frac{dy}{dx}$ from the terms on the left side of the equation to get $\frac{dy}{dx}(-15y^2 + 3 + x) = 8x^3 - 3 - y$ .
Solve for $(\frac{dy}{dx}):</b> Solve for \$(\frac{dy}{dx})$ by dividing both sides of the equation by $(-15y^2 + 3 + x)$ . This gives us $(\frac{dy}{dx}) = \frac{8x^3 - 3 - y}{-15y^2 + 3 + x}$ .

Using implicit differentiation, find dydx \frac{d y}{d x} dxdy​.\newline−5y3−2x4+3x+3y+xy=2 -5 y^{3}-2 x^{4}+3 x+3 y+x y=2 −5y3−2x4+3x+3y+xy=2

Using implicit differentiation, find $\frac{d y}{d x}$ . $\newline$ $-5 y^{3}-2 x^{4}+3 x+3 y+x y=2$