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

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

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Q. If −3−3x2−5y=−y3−y2 -3-3 x^{2}-5 y=-y^{3}-y^{2} then find dydx \frac{d y}{d x} at the point (−1,−2) (-1,-2) .\newlineAnswer: dydx∣(−1,−2)= \left.\frac{d y}{d x}\right|_{(-1,-2)}=
  1. Rewrite equation: First, we need to rewrite the given equation to make it easier to differentiate with respect to xx. The given equation is: −3−3x2−5y=−y3−y2-3 - 3x^2 - 5y = -y^3 - y^2 We can rewrite this as: 3x2+5y+3=y3+y23x^2 + 5y + 3 = y^3 + y^2
  2. Differentiate with respect to xx: Next, we differentiate both sides of the equation with respect to xx, keeping in mind that yy is a function of xx, so we will use the chain rule for terms involving yy.
    ddx(3x2+5y+3)=ddx(y3+y2)\frac{d}{dx}(3x^2 + 5y + 3) = \frac{d}{dx}(y^3 + y^2)
    This gives us: 6x+5dydx=3y2dydx+2ydydx6x + 5\frac{dy}{dx} = 3y^2\frac{dy}{dx} + 2y\frac{dy}{dx}
  3. Solve for dydx\frac{dy}{dx}: Now, we need to solve for dydx\frac{dy}{dx}. To do this, we collect all the terms involving dydx\frac{dy}{dx} on one side of the equation.\newline5(dydx)−3y2(dydx)−2y(dydx)=−6x5\left(\frac{dy}{dx}\right) - 3y^2\left(\frac{dy}{dx}\right) - 2y\left(\frac{dy}{dx}\right) = -6x\newlineThis simplifies to: (5−3y2−2y)(dydx)=−6x(5 - 3y^2 - 2y)\left(\frac{dy}{dx}\right) = -6x
  4. Isolate dydx\frac{dy}{dx}: We can now isolate dydx\frac{dy}{dx} by dividing both sides of the equation by (5−3y2−2y)(5 - 3y^2 - 2y).dydx=−6x(5−3y2−2y)\frac{dy}{dx} = \frac{-6x}{(5 - 3y^2 - 2y)}
  5. Substitute point: Finally, we substitute the point (−1,−2)(-1, -2) into the equation to find the value of dydx\frac{dy}{dx} at that point.\newline(dydx)∣(−1,−2)=−6(−1)(5−3(−2)2−2(−2))\left(\frac{dy}{dx}\right)|_{(-1,-2)} = \frac{-6(-1)}{(5 - 3(-2)^2 - 2(-2))}\newlineThis simplifies to: (dydx)∣(−1,−2)=6(5−3(4)+4)\left(\frac{dy}{dx}\right)|_{(-1,-2)} = \frac{6}{(5 - 3(4) + 4)}
  6. Calculate denominator: Now we calculate the denominator: 5−3(4)+4=5−12+4=−7+4=−35 - 3(4) + 4 = 5 - 12 + 4 = -7 + 4 = -3\newlineSo, (dy/dx)∣(−1,−2)=6/−3(dy/dx)|_{(-1,-2)} = 6 / -3
  7. Find dydx\frac{dy}{dx}: Finally, we find the value of dydx\frac{dy}{dx} at the point (−1,−2)(-1, -2).(dydx)∣(−1,−2)=6−3=−2\left(\frac{dy}{dx}\right)|_{(-1,-2)} = \frac{6}{-3} = -2

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