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

Differentiate both sides: First, we need to differentiate both sides of the equation with respect to x. The left side of the equation is in terms of y, which is a function of x, so we will use the chain rule to differentiate it. The right side is already in terms of x, so we can differentiate it directly. Differentiate the left side: (d/dx)(3y^2 - 3y^3) = (d/dx)(3y^2) - (d/dx)(3y^3) Using the chain rule, we get: 6y(dy/dx) - 9y^2(dy/dx) Differentiate the right side: (d/dx)(-y - x^3) = -(dy/dx) - 3x^2 Now we have the differentiated equation: 6y(dy/dx) - 9y^2(dy/dx) = -(dy/dx) - 3x^2Solve for dy/dx: Next, we need to solve for (dy/dx). To do this, we will collect all the terms involving (dy/dx) on one side of the equation and the remaining terms on the other side. 6y(dy/dx) - 9y^2(dy/dx) + (dy/dx) = -3x^2 Combine like terms: (6y - 9y^2 + 1)(dy/dx) = -3x^2Isolate dy/dx: Now, we isolate (dy/dx) by dividing both sides of the equation by (6y - 9y^2 + 1): (dy/dx) = -3x^2 / (6y - 9y^2 + 1) This gives us the derivative of y with respect to x in terms of x and y.

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

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

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

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Differentiate both sides: First, we need to differentiate both sides of the equation with respect to $x$ . The left side of the equation is in terms of $y$ , which is a function of $x$ , so we will use the chain rule to differentiate it. The right side is already in terms of $x$ , so we can differentiate it directly. $\newline$ Differentiate the left side: $\newline$ $(\frac{d}{dx})(3y^2 - 3y^3) = (\frac{d}{dx})(3y^2) - (\frac{d}{dx})(3y^3)$ $\newline$ Using the chain rule, we get: $\newline$ $6y(\frac{dy}{dx}) - 9y^2(\frac{dy}{dx})$ $\newline$ Differentiate the right side: $\newline$ $(\frac{d}{dx})(-y - x^3) = -(\frac{dy}{dx}) - 3x^2$ $\newline$ Now we have the differentiated equation: $\newline$ $6y(\frac{dy}{dx}) - 9y^2(\frac{dy}{dx}) = -(\frac{dy}{dx}) - 3x^2$
Solve for $\frac{dy}{dx}$ : Next, we need to solve for $\frac{dy}{dx}$ . To do this, we will collect all the terms involving $\frac{dy}{dx}$ on one side of the equation and the remaining terms on the other side. $\newline$ $6y\frac{dy}{dx} - 9y^2\frac{dy}{dx} + \frac{dy}{dx} = -3x^2$ $\newline$ Combine like terms: $\newline$ $(6y - 9y^2 + 1)\frac{dy}{dx} = -3x^2$
Isolate $\frac{dy}{dx}$ : Now, we isolate $\frac{dy}{dx}$ by dividing both sides of the equation by $(6y - 9y^2 + 1)$ : $\frac{dy}{dx} = \frac{-3x^2}{(6y - 9y^2 + 1)}$ This gives us the derivative of $y$ with respect to $x$ in terms of $x$ and $y$ .