If -2x^(3)-y^(3)-4x^(2)-4-3x=0 then find (dy)/(dx) in terms of x and y. Answer: (dy)/(dx)=

Differentiate with respect to x: We are given the equation -2x^3 - y^3 - 4x^2 - 4 - 3x = 0. To find dy/dx, we need to differentiate both sides of the equation with respect to x, treating y as a function of x (implicit differentiation).Combine derivatives: Differentiate each term with respect to x. For terms involving only x, we use the power rule. For the term involving y, we use the chain rule, remembering to multiply by dy/dx because y is a function of x. d/dx(-2x^3) = -6x^2 d/dx(-y^3) = -3y^2 * (dy/dx) d/dx(-4x^2) = -8x d/dx(-4) = 0 d/dx(-3x) = -3Isolate term with dy/dx: Combine the derivatives to get the differentiated equation: -6x^2 - 3y^2 * (dy/dx) - 8x - 3 = 0Solve for dy/dx: Now, we need to solve for dy/dx. To do this, we isolate the term involving dy/dx on one side of the equation: -3y^2 * (dy/dx) = 6x^2 + 8x + 3Divide by -3y^2: Divide both sides by -3y^2 to solve for dy/dx: (dy/dx) = (6x^2 + 8x + 3) / (-3y^2)Simplify expression: Simplify the expression for dy/dx: (dy/dx) = -2x^2 / y^2 - 8x / (3y^2) - 1 / y^2

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If
-2x^(3)-y^(3)-4x^(2)-4-3x=0 then find
(dy)/(dx) in terms of
x and
y.
Answer:
(dy)/(dx)=

If $-2 x^{3}-y^{3}-4 x^{2}-4-3 x=0$ then find $\frac{d y}{d x}$ in terms of $x$ and $y$ . $\newline$ Answer: $\frac{d y}{d x}=$

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Algebra 1

Transformations of quadratic functions

Full solution

Q. If

-2 x^{3}-y^{3}-4 x^{2}-4-3 x=0

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Differentiate with respect to $x$ : We are given the equation $-2x^3 - y^3 - 4x^2 - 4 - 3x = 0$ . To find $\frac{dy}{dx}$ , we need to differentiate both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ (implicit differentiation).
Combine derivatives: Differentiate each term with respect to $x$ . For terms involving only $x$ , we use the power rule. For the term involving $y$ , we use the chain rule, remembering to multiply by $\frac{dy}{dx}$ because $y$ is a function of $x$ .
$\frac{d}{dx}(-2x^3) = -6x^2$
$\frac{d}{dx}(-y^3) = -3y^2 \cdot \left(\frac{dy}{dx}\right)$
$\frac{d}{dx}(-4x^2) = -8x$
$\frac{d}{dx}(-4) = 0$
$x$ $0$
Isolate term with dy/dx: Combine the derivatives to get the differentiated equation: $\newline$ $-6x^2 - 3y^2 * \left(\frac{dy}{dx}\right) - 8x - 3 = 0$
Solve for $\frac{dy}{dx}$ : Now, we need to solve for $\frac{dy}{dx}$ . To do this, we isolate the term involving $\frac{dy}{dx}$ on one side of the equation: $\newline$ \(-3y^ $2$ \cdot \left(\frac{dy}{dx}\right) = $6$ x^ $2$ + $8$ x + $3$
Divide by $-3y^2$ : Divide both sides by $-3y^2$ to solve for $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{6x^2 + 8x + 3}{-3y^2}$
Simplify expression: Simplify the expression for $\frac{dy}{dx}$ : $\frac{dy}{dx} = -\frac{2x^2}{y^2} - \frac{8x}{3y^2} - \frac{1}{y^2}$