If x^(3)+3xy+2y^(3)=17, then in terms of x and y,(dy)/(dx)= -((x^(2)+y)/(x+2y^(2))) -((x^(2)+y)/(x+2y)) -((x^(2)+y)/(2y^(2))) -((x^(2)+y)/(x+y^(2)))

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If  x^(3)+3xy+2y^(3)=17, then in terms of  x and  y,(dy)/(dx)=  -((x^(2)+y)/(x+2y^(2)))  -((x^(2)+y)/(x+2y))  -((x^(2)+y)/(2y^(2)))  -((x^(2)+y)/(x+y^(2)))

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Question Prompt:  Question Prompt: If $x^3 + 3xy + 2y^3 = 17$, find $\frac{dy}{dx}$ in terms of $x$ and $y$. Implicit Differentiation:  Use implicit differentiation on both sides of the equation with respect to $x$. Differentiate each term separately: - Derivative of $x^3$ is $3x^2$. - Derivative of $3xy$ using the product rule is $3x \frac{dy}{dx} + 3y$. - Derivative of $2y^3$ using the chain rule is $6y^2 \frac{dy}{dx}$. - The derivative of the constant $17$ is $0$. Combine Derivatives:  Combine the derivatives: $$ 3x^2 + 3x \frac{dy}{dx} + 3y + 6y^2 \frac{dy}{dx} = 0 $$ Solve for dy/dx:  Rearrange to solve for $\frac{dy}{dx}$: $$ 3x \frac{dy}{dx} + 6y^2 \frac{dy}{dx} = -3x^2 - 3y $$ $$ (3x + 6y^2) \frac{dy}{dx} = -3x^2 - 3y $$ $$ \frac{dy}{dx} = \frac{-3x^2 - 3y}{3x + 6y^2} $$ $$ \frac{dy}{dx} = \frac{-(x^2 + y)}{x + 2y^2} $$

If $x^{3}+3xy+2y^{3}=17$ , then in terms of $x$ and $y$ , $(dy)/(dx) =$ $\newline$ a) $((x^{2}+y)/(x+2y^{2}))$ $\newline$ b) $((x^{2}+y)/(x+2y))$ $\newline$ c) $((x^{2}+y)/(2y^{2}))$ $\newline$ d) $((x^{2}+y)/(x+y^{2}))$

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