If -y^(3)+4x^(2)-5xy=1 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 both sides of the equation with respect to x, treating y as a function of x (implicit differentiation).Combine Derivatives: Differentiate each term on the left side of the equation with respect to x. The derivative of -y^3 with respect to x is -3y^2(dy/dx) because y is a function of x. The derivative of 4x^2 with respect to x is 8x. The derivative of -5xy with respect to x is -5y - 5x(dy/dx) because we apply the product rule: the derivative of the first function (x) times the second function (y) plus the first function (x) times the derivative of the second function (y).Group Terms: Combine the derivatives to rewrite the equation. -3y^2(dy/dx) + 8x - 5y - 5x(dy/dx) = 0Factor Out: Group the terms with (dy/dx) on one side and the rest on the other side. -3y^2(dy/dx) - 5x(dy/dx) = 5y - 8xSolve for (dy/dx): Factor out (dy/dx) from the left side of the equation. (dy/dx)(-3y^2 - 5x) = 5y - 8xSolve for (dy/dx) by dividing both sides of the equation by (-3y^2 - 5x). (dy/dx) = (5y - 8x) / (-3y^2 - 5x)

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If
-y^(3)+4x^(2)-5xy=1 then find
(dy)/(dx) in terms of
x and
y.
Answer:
(dy)/(dx)=

If $-y^{3}+4 x^{2}-5 x y=1$ 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}+4 x^{2}-5 x y=1

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 both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ (implicit differentiation).
Combine Derivatives: Differentiate each term on the left side of the equation with respect to $x$ . The derivative of $-y^3$ with respect to $x$ is $-3y^2(\frac{dy}{dx})$ because $y$ is a function of $x$ . The derivative of $4x^2$ with respect to $x$ is $8x$ . The derivative of $-5xy$ with respect to $x$ is $-y^3$ $1$ because we apply the product rule: the derivative of the first function ( $x$ ) times the second function ( $y$ ) plus the first function ( $x$ ) times the derivative of the second function ( $y$ ).
Group Terms: Combine the derivatives to rewrite the equation. $\newline$ $-3y^2\frac{dy}{dx} + 8x - 5y - 5x\frac{dy}{dx} = 0$
Factor Out: Group the terms with $\frac{dy}{dx}$ on one side and the rest on the other side. $-3y^2\frac{dy}{dx} - 5x\frac{dy}{dx} = 5y - 8x$
Solve for $\frac{dy}{dx}$ : Factor out $\frac{dy}{dx}$ from the left side of the equation. $\newline$ $\frac{dy}{dx}(-3y^2 - 5x) = 5y - 8x$
Solve for $(\frac{dy}{dx}):</b> Factor out \$(\frac{dy}{dx})$ from the left side of the equation. $\newline$ $(\frac{dy}{dx})(-3y^2 - 5x) = 5y - 8x$ Solve for $(\frac{dy}{dx})$ by dividing both sides of the equation by $(-3y^2 - 5x)$ . $\newline$ $(\frac{dy}{dx}) = \frac{5y - 8x}{-3y^2 - 5x}$

If $-y^{3}+4 x^{2}-5 x y=1$ then find $\frac{d y}{d x}$ in terms of $x$ and $y$ . $\newline$ Answer: $\frac{d y}{d x}=$

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If −y3+4x2−5xy=1 -y^{3}+4 x^{2}-5 x y=1 −y3+4x2−5xy=1 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 $-y^{3}+4 x^{2}-5 x y=1$ then find $\frac{d y}{d x}$ in terms of $x$ and $y$ . $\newline$ Answer: $\frac{d y}{d x}=$