If 1-4y^(2)+2y=-x^(3)-y^(3) then find (dy)/(dx) at the point (3,-2). Answer: (dy)/(dx)|_((3,-2))=

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If  1-4y^(2)+2y=-x^(3)-y^(3) then find  (dy)/(dx) at the point  (3,-2). Answer:  (dy)/(dx)|_((3,-2))=

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Differentiate with respect to x: First, we need to differentiate both sides of the equation with respect to x to find dy/dx. The equation is 1 - 4y^2 + 2y = -x^3 - y^3. We will apply the chain rule to differentiate terms involving y with respect to x.Left side differentiation: Differentiate the left side of the equation with respect to x: d/dx(1) - d/dx(4y^2) + d/dx(2y). The derivative of a constant is 0, so d/dx(1) = 0. For the other terms, we need to use the chain rule because they are functions of y, which is a function of x. So, d/dx(4y^2) = 8y(dy/dx) and d/dx(2y) = 2(dy/dx).Right side differentiation: Differentiate the right side of the equation with respect to x: d/dx(-x^3) - d/dx(y^3). The derivative of -x^3 with respect to x is -3x^2. For the term -d/dx(y^3), we again use the chain rule, giving us -3y^2(dy/dx).Combine derivatives: Now we combine the derivatives from both sides to form the equation: 0 - 8y(dy/dx) + 2(dy/dx) = -3x^2 - 3y^2(dy/dx). Simplifying, we get -8y(dy/dx) + 2(dy/dx) + 3y^2(dy/dx) = -3x^2.Factor out dy/dx: We can factor out dy/dx on the left side of the equation: dy/dx(-8y + 2 + 3y^2) = -3x^2. This gives us dy/dx = -3x^2 / (-8y + 2 + 3y^2).Substitute point: Now we substitute the point (3, -2) into the equation to find the value of dy/dx at that point. Plugging in x = 3 and y = -2, we get dy/dx = -3(3)^2 / (-8(-2) + 2 + 3(-2)^2).Calculate value: Calculate the value: dy/dx = -3(9) / (16 + 2 + 3(4)). This simplifies to dy/dx = -27 / (16 + 2 + 12).Simplify denominator: Further simplifying the denominator: dy/dx = -27 / (30). This reduces to dy/dx = -27/30, which can be simplified to dy/dx = -9/10.

If $1-4 y^{2}+2 y=-x^{3}-y^{3}$ then find $\frac{d y}{d x}$ at the point $(3,-2)$ . $\newline$ Answer: $\left.\frac{d y}{d x}\right|_{(3,-2)}=$

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If 1−4y2+2y=−x3−y3 1-4 y^{2}+2 y=-x^{3}-y^{3} 1−4y2+2y=−x3−y3 then find dydx \frac{d y}{d x} dxdy​ at the point (3,−2) (3,-2) (3,−2).\newlineAnswer: dydx∣(3,−2)= \left.\frac{d y}{d x}\right|_{(3,-2)}= dxdy​∣∣​(3,−2)​=

If $1-4 y^{2}+2 y=-x^{3}-y^{3}$ then find $\frac{d y}{d x}$ at the point $(3,-2)$ . $\newline$ Answer: $\left.\frac{d y}{d x}\right|_{(3,-2)}=$