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

Differentiate with respect to x: Differentiate both sides of the equation with respect to x. The equation is xy^2 = -y^3 - x^3. We will use the product rule for the left side and the chain rule for the right side. Using the product rule: d(uv)/dx = u(dv/dx) + v(du/dx), where u = x and v = y^2. Using the chain rule: d(u^n)/dx = n*u^(n-1)*(du/dx), where u = y or x and n = 3.Apply product rule: Apply the product rule to the left side of the equation. Differentiating xy^2 with respect to x gives us: d(xy^2)/dx = x(d(y^2)/dx) + y^2(d(x)/dx) Since d(x)/dx = 1, we have: d(xy^2)/dx = x(2y(dy/dx)) + y^2Apply chain rule: Apply the chain rule to the right side of the equation. Differentiating -y^3 with respect to x gives us: d(-y^3)/dx = -3y^2(dy/dx) Differentiating -x^3 with respect to x gives us: d(-x^3)/dx = -3x^2Combine derivatives: Combine the derivatives from both sides of the equation. We have: x(2y(dy/dx)) + y^2 = -3y^2(dy/dx) - 3x^2Isolate terms: Isolate terms involving (dy/dx) on one side of the equation. Move all terms involving (dy/dx) to the left side and the rest to the right side: x(2y(dy/dx)) + 3y^2(dy/dx) = -y^2 - 3x^2Factor out (dy/dx): Factor out (dy/dx) from the left side of the equation. (dy/dx)(2xy + 3y^2) = -y^2 - 3x^2Solve for (dy/dx): Solve for (dy/dx). Divide both sides by (2xy + 3y^2) to isolate (dy/dx): (dy/dx) = (-y^2 - 3x^2) / (2xy + 3y^2)

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

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

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Algebra 1

Transformations of quadratic functions

Full solution

Q. If

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

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Differentiate with respect to $x$ : Differentiate both sides of the equation with respect to $x$ . The equation is $xy^2 = -y^3 - x^3$ . We will use the product rule for the left side and the chain rule for the right side. Using the product rule: $\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ , where $u = x$ and $v = y^2$ . Using the chain rule: $\frac{d(u^n)}{dx} = n\cdot u^{(n-1)}\cdot\frac{du}{dx}$ , where $u = y$ or $x$ and $n = 3$ .
Apply product rule: Apply the product rule to the left side of the equation. $\newline$ Differentiating $xy^2$ with respect to $x$ gives us: $\newline$ $\frac{d(xy^2)}{dx} = x\left(\frac{d(y^2)}{dx}\right) + y^2\left(\frac{d(x)}{dx}\right)$ $\newline$ Since $\frac{d(x)}{dx} = 1$ , we have: $\newline$ $\frac{d(xy^2)}{dx} = x(2y\frac{dy}{dx}) + y^2$
Apply chain rule: Apply the chain rule to the right side of the equation. $\newline$ Differentiating $-y^3$ with respect to $x$ gives us: $\newline$ $\frac{d(-y^3)}{dx} = -3y^2\frac{dy}{dx}$ $\newline$ Differentiating $-x^3$ with respect to $x$ gives us: $\newline$ $\frac{d(-x^3)}{dx} = -3x^2$
Combine derivatives: Combine the derivatives from both sides of the equation. $\newline$ We have: $\newline$ $x(2y\frac{dy}{dx}) + y^2 = -3y^2\frac{dy}{dx} - 3x^2$
Isolate terms: Isolate terms involving $\frac{dy}{dx}$ on one side of the equation. $\newline$ Move all terms involving $\frac{dy}{dx}$ to the left side and the rest to the right side: $\newline$ $x(2y\frac{dy}{dx}) + 3y^2\frac{dy}{dx} = -y^2 - 3x^2$
Factor out $\frac{dy}{dx}$ : Factor out $\frac{dy}{dx}$ from the left side of the equation. $\newline$ $\frac{dy}{dx}(2xy + 3y^2) = -y^2 - 3x^2$
Solve for (\frac{dy}{dx}): Solve for $(\frac{dy}{dx}).\(\newlineDivide both sides by (\(2xy + $3$ y^ $2$ ) to isolate (\frac{dy}{dx}):\(\newline\((\frac{dy}{dx}) = \frac{-y^\(2$ - $3$x^$2$}{$2$xy + $3$y^$2$}