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

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Simplify expressions using trigonometric identities

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Q. Using implicit differentiation, find

\frac{d y}{d x}

\newline

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

Apply Product Rule and Chain Rule: We need to differentiate both sides of the equation with respect to $x$ . Remember that $y$ is a function of $x$ , so we will use the product rule and the chain rule for differentiation.
Differentiate $-x^2y^4$ : Differentiate the left side of the equation $-x^2y^4 - 4x^2y^3$ with respect to $x$ . For the first term $-x^2y^4$ , we apply the product rule: $\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ , where $u = -x^2$ and $v = y^4$ .
Differentiate $-4x^2y^3$ : Differentiate $u = -x^2$ with respect to $x$ to get $\frac{du}{dx} = -2x$ .
Combine Left Side Terms: Differentiate $v = y^4$ with respect to $x$ using the chain rule to get $\frac{dv}{dx} = 4y^3\frac{dy}{dx}$ .
Differentiate Right Side: Now apply the product rule to the first term: $-x^2y^4$ becomes $-2xy^4 - 4x^2y^3\left(\frac{dy}{dx}\right)$ .
Combine Terms and Factor: Repeat the process for the second term $-4x^2y^3$ . Differentiate $u = -4x^2$ to get $\frac{du}{dx} = -8x$ .
Isolate $\frac{dy}{dx}$ : Differentiate $v = y^3$ with respect to $x$ using the chain rule to get $\frac{dv}{dx} = 3y^2\left(\frac{dy}{dx}\right)$ .
Simplify $\frac{dy}{dx}$ : Now apply the product rule to the second term: $-4x^2y^3$ becomes $-8xy^3 - 12x^2y^2\left(\frac{dy}{dx}\right)$ .
Simplify $\frac{dy}{dx}$ : Now apply the product rule to the second term: $-4x^2y^3$ becomes $-8xy^3 - 12x^2y^2\frac{dy}{dx}$ .Combine the differentiated terms for the left side of the equation: $-2xy^4 - 4x^2y^3\frac{dy}{dx} - 8xy^3 - 12x^2y^2\frac{dy}{dx}$ .
Simplify $\frac{dy}{dx}$ : Now apply the product rule to the second term: $-4x^2y^3$ becomes $-8xy^3 - 12x^2y^2\frac{dy}{dx}$ .Combine the differentiated terms for the left side of the equation: $-2xy^4 - 4x^2y^3\frac{dy}{dx} - 8xy^3 - 12x^2y^2\frac{dy}{dx}$ .Differentiate the right side of the equation $2x + 5$ with respect to $x$ to get $2$ .
Simplify $\frac{dy}{dx}$ : Now apply the product rule to the second term: $-4x^2y^3$ becomes $-8xy^3 - 12x^2y^2\frac{dy}{dx}$ .Combine the differentiated terms for the left side of the equation: $-2xy^4 - 4x^2y^3\frac{dy}{dx} - 8xy^3 - 12x^2y^2\frac{dy}{dx}$ .Differentiate the right side of the equation $2x + 5$ with respect to $x$ to get $2$ .Now we have the equation $-2xy^4 - 4x^2y^3\frac{dy}{dx} - 8xy^3 - 12x^2y^2\frac{dy}{dx} = 2$ .
Simplify $\frac{dy}{dx}$ : Now apply the product rule to the second term: $-4x^2y^3$ becomes $-8xy^3 - 12x^2y^2\frac{dy}{dx}$ .Combine the differentiated terms for the left side of the equation: $-2xy^4 - 4x^2y^3\frac{dy}{dx} - 8xy^3 - 12x^2y^2\frac{dy}{dx}$ .Differentiate the right side of the equation $2x + 5$ with respect to $x$ to get $2$ .Now we have the equation $-2xy^4 - 4x^2y^3\frac{dy}{dx} - 8xy^3 - 12x^2y^2\frac{dy}{dx} = 2$ .Combine like terms and factor out $\frac{dy}{dx}$ : $-2xy^4 - 8xy^3 = 2 + \frac{dy}{dx}(4x^2y^3 + 12x^2y^2)$ .
Simplify $\frac{dy}{dx}$ : Now apply the product rule to the second term: $-4x^2y^3$ becomes $-8xy^3 - 12x^2y^2\frac{dy}{dx}$ .Combine the differentiated terms for the left side of the equation: $-2xy^4 - 4x^2y^3\frac{dy}{dx} - 8xy^3 - 12x^2y^2\frac{dy}{dx}$ .Differentiate the right side of the equation $2x + 5$ with respect to $x$ to get $2$ .Now we have the equation $-2xy^4 - 4x^2y^3\frac{dy}{dx} - 8xy^3 - 12x^2y^2\frac{dy}{dx} = 2$ .Combine like terms and factor out $\frac{dy}{dx}$ : $-2xy^4 - 8xy^3 = 2 + \frac{dy}{dx}(4x^2y^3 + 12x^2y^2)$ .Solve for $\frac{dy}{dx}$ by isolating it on one side of the equation: $-4x^2y^3$ $1$ .
Simplify $\frac{dy}{dx}$ : Now apply the product rule to the second term: $-4x^2y^3$ becomes $-8xy^3 - 12x^2y^2\frac{dy}{dx}$ .Combine the differentiated terms for the left side of the equation: $-2xy^4 - 4x^2y^3\frac{dy}{dx} - 8xy^3 - 12x^2y^2\frac{dy}{dx}$ .Differentiate the right side of the equation $2x + 5$ with respect to $x$ to get $2$ .Now we have the equation $-2xy^4 - 4x^2y^3\frac{dy}{dx} - 8xy^3 - 12x^2y^2\frac{dy}{dx} = 2$ .Combine like terms and factor out $\frac{dy}{dx}$ : $-2xy^4 - 8xy^3 = 2 + \frac{dy}{dx}(4x^2y^3 + 12x^2y^2)$ .Solve for $\frac{dy}{dx}$ by isolating it on one side of the equation: $-4x^2y^3$ $1$ .Simplify the expression for $\frac{dy}{dx}$ if possible. In this case, there is no further simplification.