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

Write Given Equation: Write down the given equation. We have the equation 2 + y^3 = xy^3.Differentiate with Respect: Differentiate both sides of the equation with respect to x. The left side becomes the derivative of 2 with respect to x, which is 0, plus the derivative of y^3 with respect to x, which is 3y^2(dy/dx) using the chain rule. The right side becomes the derivative of xy^3 with respect to x, which is y^3 plus x times the derivative of y^3 with respect to x, which is 3y^2(dy/dx), using the product rule. So, we have 0 + 3y^2(dy/dx) = y^3 + x*3y^2(dy/dx).Simplify and Solve: Simplify the equation and solve for (dy/dx). Subtract y^3 from both sides to get 3y^2(dy/dx) - x*3y^2(dy/dx) = y^3 - y^3. This simplifies to 3y^2(dy/dx)(1 - x) = 0.Factor Out Common Factor: Factor out 3y^2(dy/dx). We have 3y^2(dy/dx)(1 - x) = 0. Since 3y^2(dy/dx) is a common factor, we can divide both sides by 3y^2, assuming y is not zero (since y=0 would make the original equation 2=0, which is not true). This gives us (dy/dx)(1 - x) = 0.Isolate and Solve: Solve for (dy/dx). Divide both sides by (1 - x) to isolate (dy/dx), which gives us (dy/dx) = 0 / (1 - x). Since 0 divided by anything is 0, we have (dy/dx) = 0.

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

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

Find trigonometric ratios using reference angles

Full solution

Q. If

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

then find

\frac{d y}{d x}

in terms of

x

and

y

\newline

Answer:

\frac{d y}{d x}=

Write Given Equation: Write down the given equation. $\newline$ We have the equation $2 + y^3 = xy^3$ .
Differentiate with Respect: Differentiate both sides of the equation with respect to $x$ . The left side becomes the derivative of $2$ with respect to $x$ , which is $0$ , plus the derivative of $y^3$ with respect to $x$ , which is $3y^2\frac{dy}{dx}$ using the chain rule. The right side becomes the derivative of $xy^3$ with respect to $x$ , which is $y^3$ plus $x$ times the derivative of $y^3$ with respect to $x$ , which is $3y^2\frac{dy}{dx}$ , using the product rule. So, we have $2$ $4$ .
Simplify and Solve: Simplify the equation and solve for $\frac{dy}{dx}$ . Subtract $y^3$ from both sides to get $3y^2\frac{dy}{dx} - x\cdot3y^2\frac{dy}{dx} = y^3 - y^3$ . This simplifies to $3y^2\frac{dy}{dx}(1 - x) = 0$ .
Factor Out Common Factor: Factor out $3y^2\frac{dy}{dx}$ . We have $3y^2\frac{dy}{dx}(1 - x) = 0$ . Since $3y^2\frac{dy}{dx}$ is a common factor, we can divide both sides by $3y^2$ , assuming $y$ is not zero (since $y=0$ would make the original equation $2=0$ , which is not true). This gives us $\frac{dy}{dx}(1 - x) = 0$ .
Isolate and Solve: Solve for $(\frac{dy}{dx})$ . $\newline$ Divide both sides by $(1 - x)$ to isolate $(\frac{dy}{dx})$ , which gives us $(\frac{dy}{dx}) = \frac{0}{(1 - x)}$ . $\newline$ Since $0$ divided by anything is $0$ , we have $(\frac{dy}{dx}) = 0$ .