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

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Q. Using implicit differentiation, find

\frac{d y}{d x}

\newline

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

Apply Product Rule and Chain Rule: We need to differentiate both sides of the equation with respect to $x$ , using the product rule for the terms involving both $x$ and $y$ , and the chain rule for the terms involving $y$ , since $y$ is a function of $x$ .
Differentiate $-5x^{3}y^{4}$ : Differentiate the first term $-5x^{3}y^{4}$ with respect to $x$ . Using the product rule, we get $-5(3x^{2}y^{4} + x^{3}(4y^{3}\frac{dy}{dx}))$ .
Differentiate $xy^{4}$ : Differentiate the second term $xy^{4}$ with respect to $x$ . Using the product rule, we get $y^{4} + x(4y^{3}\frac{dy}{dx})$ .
Differentiate right side: Differentiate the right side of the equation, $3x + 5$ , with respect to $x$ . The derivative of $3x$ is $3$ , and the derivative of $5$ is $0$ , since it's a constant.
Combine differentiated terms: Combine the differentiated terms to form the new equation: $-5(3x^{2}y^{4} + x^{3}(4y^{3}\frac{dy}{dx})) + y^{4} + x(4y^{3}\frac{dy}{dx}) = 3$ .
Simplify equation: Simplify the equation by distributing the $-5$ and combining like terms: $-15x^{2}y^{4} - 20x^{3}y^{3}\frac{dy}{dx} + y^{4} + 4xy^{3}\frac{dy}{dx} = 3$ .
Group terms: Group the $\frac{dy}{dx}$ terms on one side and the remaining terms on the other side: $-20x^{3}y^{3}\frac{dy}{dx} + 4xy^{3}\frac{dy}{dx} = 3 + 15x^{2}y^{4} - y^{4}$ .
Factor out $\frac{dy}{dx}$ : Factor out $\frac{dy}{dx}$ from the left side of the equation: $\frac{dy}{dx}(-20x^{3}y^{3} + 4xy^{3}) = 3 + 15x^{2}y^{4} - y^{4}$ .
Simplify right side: Simplify the right side of the equation: $\frac{dy}{dx}(-20x^{3}y^{3} + 4xy^{3}) = 3 + 14x^{2}y^{4}$ .
Solve for dy/dx: Solve for $\frac{dy}{dx}$ by dividing both sides by $(-20x^{3}y^{3} + 4xy^{3})$ : $\frac{dy}{dx} = \frac{3 + 14x^{2}y^{4}}{-20x^{3}y^{3} + 4xy^{3}}$ .
Factor out $y^{3}$ : Factor out $y^{3}$ from the denominator to simplify the expression: $\frac{dy}{dx} = \frac{3 + 14x^{2}y^{4}}{y^{3}(-20x^{3} + 4x)}$ .
Further simplify: Factor out $4x$ from the denominator to simplify further: $\frac{dy}{dx} = \frac{3 + 14x^{2}y^{4}}{4xy^{3}(-5x^{2} + 1)}$ .