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

Differentiate Equation: Differentiate both sides of the equation with respect to x. We will use the power rule, product rule, and the fact that the derivative of a constant is zero. Differentiate y^3 with respect to x using the chain rule, which gives us 3y^2(dy/dx). The derivative of -2 with respect to x is 0. The derivative of -3x with respect to x is -3. The derivative of 2xy with respect to x using the product rule is 2y + 2x(dy/dx). So, differentiating both sides of the equation y^3 - 2 - 3x = 2xy with respect to x gives us: 3y^2(dy/dx) - 0 - 3 = 2y + 2x(dy/dx).Rearrange for dy/dx: Rearrange the equation to solve for (dy/dx). We need to collect all the terms with (dy/dx) on one side and the rest on the other side. 3y^2(dy/dx) - 3 = 2y + 2x(dy/dx). Subtract 2x(dy/dx) from both sides to get: 3y^2(dy/dx) - 2x(dy/dx) = 2y + 3. Factor out (dy/dx) from the left side: (dy/dx)(3y^2 - 2x) = 2y + 3.Solve for dy/dx: Solve for (dy/dx). Divide both sides by (3y^2 - 2x) to isolate (dy/dx): (dy/dx) = (2y + 3) / (3y^2 - 2x). This gives us the derivative of y with respect to x in terms of x and y.

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

If $y^{3}-2-3 x=2 x y$ 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

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Q. If

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

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Differentiate Equation: Differentiate both sides of the equation with respect to $x$ . We will use the power rule, product rule, and the fact that the derivative of a constant is zero. Differentiate $y^3$ with respect to $x$ using the chain rule, which gives us $3y^2\frac{dy}{dx}$ . The derivative of $-2$ with respect to $x$ is $0$ . The derivative of $-3x$ with respect to $x$ is $-3$ . The derivative of $y^3$ $0$ with respect to $x$ using the product rule is $y^3$ $2$ . So, differentiating both sides of the equation $y^3$ $3$ with respect to $x$ gives us: $y^3$ $5$ .
Rearrange for $\frac{dy}{dx}$ : Rearrange the equation to solve for $\frac{dy}{dx}$ . We need to collect all the terms with $\frac{dy}{dx}$ on one side and the rest on the other side. $3y^2\frac{dy}{dx} - 3 = 2y + 2x\frac{dy}{dx}$ . Subtract $2x\frac{dy}{dx}$ from both sides to get: $3y^2\frac{dy}{dx} - 2x\frac{dy}{dx} = 2y + 3$ . Factor out $\frac{dy}{dx}$ from the left side: $\frac{dy}{dx}(3y^2 - 2x) = 2y + 3$ .
Solve for $\frac{dy}{dx}$ : Solve for $\frac{dy}{dx}$ . $\newline$ Divide both sides by $(3y^2 - 2x)$ to isolate $\frac{dy}{dx}$ : $\newline$ $\frac{dy}{dx} = \frac{2y + 3}{3y^2 - 2x}$ . $\newline$ This gives us the derivative of $y$ with respect to $x$ in terms of $x$ and $y$ .