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

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

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Q. If −4−y+x2=y2+3y3 -4-y+x^{2}=y^{2}+3 y^{3} then find dydx \frac{d y}{d x} at the point (−3,1) (-3,1) .\newlineAnswer: dydx∣(−3,1)= \left.\frac{d y}{d x}\right|_{(-3,1)}=
  1. Differentiate with respect to xx: First, we need to differentiate both sides of the equation with respect to xx to find dydx\frac{dy}{dx}. The equation is −4−y+x2=y2+3y3-4 - y + x^2 = y^2 + 3y^3. Differentiating both sides with respect to xx gives us: ddx(−4−y+x2)=ddx(y2+3y3)\frac{d}{dx}(-4 - y + x^2) = \frac{d}{dx}(y^2 + 3y^3).
  2. Apply chain rule: On the left side, the derivative of −4-4 with respect to xx is 00, the derivative of −y-y with respect to xx is −dydx-\frac{dy}{dx} (since yy is a function of xx), and the derivative of x2x^2 with respect to xx is xx00. So, we have: xx11.
  3. Collect terms and solve: On the right side, we use the chain rule to differentiate y2y^2 and 3y33y^3 with respect to xx. The derivative of y2y^2 with respect to xx is 2y(dydx)2y(\frac{dy}{dx}), and the derivative of 3y33y^3 with respect to xx is 9y2(dydx)9y^2(\frac{dy}{dx}). So, we have: −dydx+2x=2y(dydx)+9y2(dydx)-\frac{dy}{dx} + 2x = 2y(\frac{dy}{dx}) + 9y^2(\frac{dy}{dx}).
  4. Factor out dy/dx: Now, we collect all the terms involving dy/dxdy/dx on one side of the equation to solve for dy/dxdy/dx. This gives us: −dy/dx−2y(dy/dx)−9y2(dy/dx)=2x- dy/dx - 2y(dy/dx) - 9y^2(dy/dx) = 2x.
  5. Solve for dydx\frac{dy}{dx}: Factor out dydx\frac{dy}{dx} from the left side of the equation:\newlinedydx(−1−2y−9y2)=2x\frac{dy}{dx}(-1 - 2y - 9y^2) = 2x.
  6. Evaluate at (−3,1)(-3,1): Now, we can solve for dydx\frac{dy}{dx} by dividing both sides by (−1−2y−9y2)(-1 - 2y - 9y^2):\newlinedydx=2x−1−2y−9y2.\frac{dy}{dx} = \frac{2x}{-1 - 2y - 9y^2}.
  7. Perform calculations: We need to evaluate dydx\frac{dy}{dx} at the point (−3,1)(-3,1). Substitute x=−3x = -3 and y=1y = 1 into the equation:\newlinedydx∣(−3,1)=2(−3)(−1−2(1)−9(1)2)\frac{dy}{dx}\bigg|_{(-3,1)} = \frac{2(-3)}{(-1 - 2(1) - 9(1)^2)}.
  8. Simplify the denominator: Now, perform the calculations:\newlinedydx∣(−3,1)=−6(−1−2−9).\frac{dy}{dx}\bigg|_{(-3,1)} = \frac{-6}{(-1 - 2 - 9)}.
  9. Divide to find value: Simplify the denominator: dydx∣(−3,1)=−6−12\frac{dy}{dx}\bigg|_{(-3,1)} = \frac{-6}{-12}.
  10. Divide to find value: Simplify the denominator: dydx∣(−3,1)=−6−12\frac{dy}{dx}\bigg|_{(-3,1)} = \frac{-6}{-12}.Finally, divide −6-6 by −12-12 to get the value of dydx\frac{dy}{dx} at the point (−3,1)(-3,1): dydx∣(−3,1)=12\frac{dy}{dx}\bigg|_{(-3,1)} = \frac{1}{2}.

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