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

Identify Given Equation: Identify the given equation and the requirement to differentiate with respect to x. Given equation: -y - y^2 + y^3 = 3x^2 We need to find (dy)/(dx).Differentiate with Respect: Differentiate both sides of the equation with respect to x, using implicit differentiation. d/dx(-y - y^2 + y^3) = d/dx(3x^2)Apply Chain Rule: Apply the chain rule to differentiate the terms involving y with respect to x. d/dx(-y) = -dy/dx d/dx(-y^2) = -2y * dy/dx d/dx(y^3) = 3y^2 * dy/dx d/dx(3x^2) = 6x So, -dy/dx - 2y(dy/dx) + 3y^2(dy/dx) = 6xCombine Like Terms: Combine like terms and solve for dy/dx. -dy/dx - 2y(dy/dx) + 3y^2(dy/dx) = 6x (dy/dx)(-1 - 2y + 3y^2) = 6xIsolate dy/dx: Isolate dy/dx on one side of the equation. dy/dx = 6x / (-1 - 2y + 3y^2)

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

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

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Math Problems

Algebra 2

Evaluate exponential functions

Full solution

Q. If

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

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Identify Given Equation: Identify the given equation and the requirement to differentiate with respect to $x$ . $\newline$ Given equation: $-y - y^2 + y^3 = 3x^2$ $\newline$ We need to find $\frac{dy}{dx}$ .
Differentiate with Respect: Differentiate both sides of the equation with respect to $x$ , using implicit differentiation. $\newline$ $\frac{d}{dx}(-y - y^2 + y^3) = \frac{d}{dx}(3x^2)$
Apply Chain Rule: Apply the chain rule to differentiate the terms involving $y$ with respect to $x$ .
$\frac{d}{dx}(-y) = -\frac{dy}{dx}$
$\frac{d}{dx}(-y^2) = -2y \cdot \frac{dy}{dx}$
$\frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}$
$\frac{d}{dx}(3x^2) = 6x$
So, $-\frac{dy}{dx} - 2y(\frac{dy}{dx}) + 3y^2(\frac{dy}{dx}) = 6x$
Combine Like Terms: Combine like terms and solve for $\frac{dy}{dx}$ . $\frac{-dy}{dx} - 2y\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 6x$ $\frac{dy}{dx}(-1 - 2y + 3y^2) = 6x$
Isolate $\frac{dy}{dx}$ : Isolate $\frac{dy}{dx}$ on one side of the equation. $\newline$ $\frac{dy}{dx} = \frac{6x}{-1 - 2y + 3y^2}$