Using implicit differentiation, find $\frac{d y}{d x}$ . $\newline$ $\left(-x^{3}-4 y^{2}\right)^{3}=-3 x y$

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Simplify expressions using trigonometric identities

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Q. Using implicit differentiation, find

\frac{d y}{d x}

\newline

\left(-x^{3}-4 y^{2}\right)^{3}=-3 x y

Differentiate with chain and product rules: Differentiate both sides of the equation with respect to $x$ . We will use the chain rule to differentiate the left side and the product rule to differentiate the right side. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
Apply chain rule to left side: Apply the chain rule to the left side of the equation. $\newline$ Let $u = -x^3 - 4y^2$ , then the left side becomes $u^3$ . The derivative of $u^3$ with respect to $u$ is $3u^2$ . Now we need to find $\frac{du}{dx}$ , which is the derivative of $u$ with respect to $x$ .
Differentiate $u$ with respect to $x$ : Differentiate $u = -x^3 - 4y^2$ with respect to $x$ . The derivative of $-x^3$ with respect to $x$ is $-3x^2$ . Since $y$ is a function of $x$ , we need to use implicit differentiation for $-4y^2$ , which gives us $x$ $0$ . So, $x$ $1$ .
Substitute $\frac{du}{dx}$ into left side: Substitute $\frac{du}{dx}$ back into the derivative of the left side. $\newline$ The derivative of the left side is $3u^2 \cdot \frac{du}{dx}$ , which becomes $3(-x^3 - 4y^2)^2 \cdot (-3x^2 - 8y\frac{dy}{dx})$ .
Apply product rule to right side: Apply the product rule to the right side of the equation. $\newline$ The derivative of $-3xy$ with respect to $x$ is $-3y - 3x\left(\frac{dy}{dx}\right)$ because the derivative of $-3x$ with respect to $x$ is $-3$ and the derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$ .
Set derivatives equal to each other: Set the derivatives of both sides equal to each other. $\newline$ $3(-x^3 - 4y^2)^2 \cdot (-3x^2 - 8y\frac{dy}{dx}) = -3y - 3x\frac{dy}{dx}$ .
Solve for $\frac{dy}{dx}$ : Solve for $\frac{dy}{dx}$ . $\newline$ To solve for $\frac{dy}{dx}$ , we need to collect all terms containing $\frac{dy}{dx}$ on one side and the rest on the other side. $\newline$ $3(-x^3 - 4y^2)^2 * (-3x^2) - 3y = 3(-x^3 - 4y^2)^2 * (-8y(\frac{dy}{dx})) + 3x(\frac{dy}{dx})$ .
Factor out $\frac{dy}{dx}$ : Factor out $\frac{dy}{dx}$ on the right side of the equation. $3(-x^3 - 4y^2)^2 * (-3x^2) - 3y = \frac{dy}{dx} * (3(-x^3 - 4y^2)^2 * (-8y) + 3x)$ .
Isolate $\frac{dy}{dx}$ : Isolate $\frac{dy}{dx}$ . $\frac{dy}{dx} = \frac{3(-x^3 - 4y^2)^2 * (-3x^2) - 3y}{3(-x^3 - 4y^2)^2 * (-8y) + 3x}.$