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

Apply Product Rule: We need to differentiate both sides of the equation with respect to x to find (dy)/(dx). Differentiate the left side using the product rule: d(uv)/dx = u(dv/dx) + v(du/dx), where u = 3y and v = x. Differentiate the right side with respect to x normally.Differentiate Left Side: Differentiate the left side: d(3xy)/dx = 3(d(xy)/dx) - d(1)/dx. Using the product rule: d(xy)/dx = x(dy/dx) + y(dx/dx). Since dx/dx = 1, we have d(xy)/dx = x(dy/dx) + y. So, d(3xy)/dx = 3(x(dy/dx) + y).Differentiate Right Side: Differentiate the right side: d(x^3 - 2y)/dx = d(x^3)/dx - d(2y)/dx. The derivative of x^3 with respect to x is 3x^2. The derivative of -2y with respect to x is -2(dy/dx). So, d(x^3 - 2y)/dx = 3x^2 - 2(dy/dx).Equate Derivatives: Now we equate the derivatives from both sides: 3(x(dy/dx) + y) - 0 = 3x^2 - 2(dy/dx). Simplify the equation: 3x(dy/dx) + 3y = 3x^2 - 2(dy/dx).Group Terms: Group all the terms containing (dy/dx) on one side and the rest on the other side: 3x(dy/dx) + 2(dy/dx) = 3x^2 - 3y. Factor out (dy/dx): (dy/dx)(3x + 2) = 3x^2 - 3y.Solve for dy/dx: Solve for (dy/dx): (dy/dx) = (3x^2 - 3y) / (3x + 2). This is the derivative of y with respect to x in terms of x and y.

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

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

3 x y-1=x^{3}-2 y

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Apply Product Rule: We need to differentiate both sides of the equation with respect to $x$ to find $\frac{dy}{dx}$ . Differentiate the left side using the product rule: $\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ , where $u = 3y$ and $v = x$ . Differentiate the right side with respect to $x$ normally.
Differentiate Left Side: Differentiate the left side: $d(3xy)/dx = 3(d(xy)/dx) - d(1)/dx$ . Using the product rule: $d(xy)/dx = x(dy/dx) + y(dx/dx)$ . Since $dx/dx = 1$ , we have $d(xy)/dx = x(dy/dx) + y$ . So, $d(3xy)/dx = 3(x(dy/dx) + y)$ .
Differentiate Right Side: Differentiate the right side: $\frac{d(x^3 - 2y)}{dx} = \frac{d(x^3)}{dx} - \frac{d(2y)}{dx}$ . The derivative of $x^3$ with respect to $x$ is $3x^2$ . The derivative of $-2y$ with respect to $x$ is $-2\frac{dy}{dx}$ . So, $\frac{d(x^3 - 2y)}{dx} = 3x^2 - 2\frac{dy}{dx}$ .
Equate Derivatives: Now we equate the derivatives from both sides: $\newline$ $3(x\frac{dy}{dx} + y) - 0 = 3x^2 - 2\frac{dy}{dx}$ . $\newline$ Simplify the equation: $\newline$ $3x\frac{dy}{dx} + 3y = 3x^2 - 2\frac{dy}{dx}$ .
Group Terms: Group all the terms containing $\frac{dy}{dx}$ on one side and the rest on the other side: $\newline$ $3x\frac{dy}{dx} + 2\frac{dy}{dx} = 3x^2 - 3y.$ $\newline$ Factor out $\frac{dy}{dx}$ : $\newline$ $\frac{dy}{dx}(3x + 2) = 3x^2 - 3y.$
Solve for $\frac{dy}{dx}$ : Solve for $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{3x^2 - 3y}{3x + 2}$ .This is the derivative of $y$ with respect to $x$ in terms of $x$ and $y$ .