If 4-5x^(2)-2y^(2)=y^(3)-x then find (dy)/(dx) at the point (1,-2). Answer: (dy)/(dx)|_((1,-2))=

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If  4-5x^(2)-2y^(2)=y^(3)-x then find  (dy)/(dx) at the point  (1,-2). Answer:  (dy)/(dx)|_((1,-2))=

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Differentiate with Chain Rule: First, we need to find the derivative of both sides of the equation with respect to x. This will involve using the chain rule for the terms involving y, since y is a function of x. Differentiate both sides of the equation with respect to x: d/dx(4 - 5x^2 - 2y^2) = d/dx(y^3 - x) This gives us: 0 - 10x - 4y(dy/dx) = 3y^2(dy/dx) - 1Rearrange to Solve: Now, we rearrange the terms to solve for dy/dx: -4y(dy/dx) - 3y^2(dy/dx) = -10x + 1 Combine like terms: dy/dx(-4y - 3y^2) = -10x + 1Isolate dy/dx: Next, we isolate dy/dx by dividing both sides by (-4y - 3y^2): dy/dx = (-10x + 1) / (-4y - 3y^2)Substitute Point: Now we substitute the point (1, -2) into the equation to find the value of dy/dx at that point: dy/dx|_(1,-2) = (-10(1) + 1) / (-4(-2) - 3(-2)^2)Perform Calculations: Perform the calculations: dy/dx|_(1,-2) = (-10 + 1) / (8 - 3(4)) dy/dx|_(1,-2) = (-9) / (8 - 12) dy/dx|_(1,-2) = (-9) / (-4)Simplify Fraction: Finally, simplify the fraction: dy/dx|_(1,-2) = 9/4

If $4-5 x^{2}-2 y^{2}=y^{3}-x$ then find $\frac{d y}{d x}$ at the point $(1,-2)$ . $\newline$ Answer: $\left.\frac{d y}{d x}\right|_{(1,-2)}=$

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If 4−5x2−2y2=y3−x 4-5 x^{2}-2 y^{2}=y^{3}-x 4−5x2−2y2=y3−x then find dydx \frac{d y}{d x} dxdy​ at the point (1,−2) (1,-2) (1,−2).\newlineAnswer: dydx∣(1,−2)= \left.\frac{d y}{d x}\right|_{(1,-2)}= dxdy​∣∣​(1,−2)​=

If $4-5 x^{2}-2 y^{2}=y^{3}-x$ then find $\frac{d y}{d x}$ at the point $(1,-2)$ . $\newline$ Answer: $\left.\frac{d y}{d x}\right|_{(1,-2)}=$