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

Differentiate terms: To find the derivative (dy)/(dx), we need to differentiate the given equation with respect to x, treating y as a function of x (implicit differentiation). The given equation is: -xy^(3) + y - x^(2) = 0. Differentiate each term with respect to x.Product rule: Differentiate -xy^(3) with respect to x using the product rule (d/dx)[uv] = u'(v) + v'(u), where u = x and v = y^(3). The derivative of x with respect to x is 1, and the derivative of y^(3) with respect to x is 3y^(2)(dy/dx) by the chain rule. So, the derivative of -xy^(3) is -y^(3) - 3xy^(2)(dy/dx).Differentiate y: Differentiate y with respect to x. Since y is a function of x, its derivative is (dy/dx). So, the derivative of y is (dy/dx).Differentiate -x^2: Differentiate -x^(2) with respect to x. The derivative of -x^(2) with respect to x is -2x.Combine derivatives: Combine the derivatives of each term to get the derivative of the entire left side of the equation. The combined derivative is: -y^(3) - 3xy^(2)(dy/dx) + (dy/dx) - 2x = 0.Solve for (dy/dx): Now, we need to solve for (dy/dx). Group the terms containing (dy/dx) on one side and the rest on the other side. 3xy^(2)(dy/dx) - (dy/dx) = y^(3) + 2x.Factor out (dy/dx): Factor out (dy/dx) from the left side of the equation. (dy/dx)(3xy^(2) - 1) = y^(3) + 2x.Isolate (dy/dx): Divide both sides of the equation by (3xy^(2) - 1) to isolate (dy/dx). (dy/dx) = (y^(3) + 2x) / (3xy^(2) - 1).

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

If $-x y^{3}+y-x^{2}=0$ 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 2

Evaluate exponential functions

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Q. If

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

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Differentiate terms: To find the derivative $\frac{dy}{dx}$ , we need to differentiate the given equation with respect to $x$ , treating $y$ as a function of $x$ (implicit differentiation). $\newline$ The given equation is: $-xy^{3} + y - x^{2} = 0$ . $\newline$ Differentiate each term with respect to $x$ .
Product rule: Differentiate $-xy^{3}$ with respect to $x$ using the product rule $\frac{d}{dx}[uv] = u'(v) + v'(u)$ , where $u = x$ and $v = y^{3}$ . The derivative of $x$ with respect to $x$ is $1$ , and the derivative of $y^{3}$ with respect to $x$ is $x$ $0$ by the chain rule. So, the derivative of $-xy^{3}$ is $x$ $2$ .
Differentiate $y$ : Differentiate $y$ with respect to $x$ . $\newline$ Since $y$ is a function of $x$ , its derivative is $\frac{dy}{dx}$ . $\newline$ So, the derivative of $y$ is $\frac{dy}{dx}$ .
Differentiate $-x^2$ : Differentiate $-x^{2}$ with respect to $x$ . The derivative of $-x^{2}$ with respect to $x$ is $-2x$ .
Combine derivatives: Combine the derivatives of each term to get the derivative of the entire left side of the equation. $\newline$ The combined derivative is: $-y^{3} - 3xy^{2}\left(\frac{dy}{dx}\right) + \left(\frac{dy}{dx}\right) - 2x = 0$ .
Solve for (\frac{dy}{dx}): Now, we need to solve for \((\frac{dy}{dx}). Group the terms containing \((\frac{dy}{dx}) on one side and the rest on the other side. \(\(3xy^{ $2$ }(\frac{dy}{dx}) - (\frac{dy}{dx}) = y^{ $3$ } + $2$ x.
Factor out $\frac{dy}{dx}$ : Factor out $\frac{dy}{dx}$ from the left side of the equation. $\newline$ \left(\frac{dy}{dx}\right)(\(3xy^{ $2$ } - $1$ ) = y^{ $3$ } + $2$ x.
Isolate $(\frac{dy}{dx}):</b> Divide both sides of the equation by \$(3xy^{2} - 1)$ to isolate $(\frac{dy}{dx})$ . $\newline$ $(\frac{dy}{dx}) = \frac{y^{3} + 2x}{3xy^{2} - 1}$ .

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If −xy3+y−x2=0 -x y^{3}+y-x^{2}=0 −xy3+y−x2=0 then find dydx \frac{d y}{d x} dxdy​ in terms of x x x and y y y.\newlineAnswer: dydx= \frac{d y}{d x}= dxdy​=

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