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

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

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Evaluate a nonlinear functionright chevron icon

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Q. If −y3−y+2+y2=x2 -y^{3}-y+2+y^{2}=x^{2} then find dydx \frac{d y}{d x} at the point (4,−2) (4,-2) .\newlineAnswer: dydx∣(4,−2)= \left.\frac{d y}{d x}\right|_{(4,-2)}=
  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}.\newlineGiven equation: −y3−y+2+y2=x2-y^{3} - y + 2 + y^{2} = x^{2}\newlineDifferentiating both sides with respect to xx, we get:\newlineddx(−y3−y+2+y2)=ddx(x2)\frac{d}{dx}(-y^{3} - y + 2 + y^{2}) = \frac{d}{dx}(x^{2})\newlineSince yy is a function of xx, we will use the chain rule for differentiating terms involving yy.
  2. Left side differentiation: Differentiating the left side of the equation with respect to xx:
    ddx(−y3)=−3y2⋅dydx\frac{d}{dx}(-y^{3}) = -3y^{2} \cdot \frac{dy}{dx}
    ddx(−y)=−1⋅dydx\frac{d}{dx}(-y) = -1 \cdot \frac{dy}{dx}
    ddx(2)=0\frac{d}{dx}(2) = 0 (since the derivative of a constant is 00)
    ddx(y2)=2y⋅dydx\frac{d}{dx}(y^{2}) = 2y \cdot \frac{dy}{dx}
    Adding these up, we get:
    −3y2⋅dydx−dydx+0+2y⋅dydx-3y^{2} \cdot \frac{dy}{dx} - \frac{dy}{dx} + 0 + 2y \cdot \frac{dy}{dx}
    Simplifying, we have:
    (−3y2−1+2y)⋅dydx(-3y^{2} - 1 + 2y) \cdot \frac{dy}{dx}
  3. Right side differentiation: Differentiating the right side of the equation with respect to xx:ddx(x2)=2x\frac{d}{dx}(x^{2}) = 2x
  4. Solve for dydx\frac{dy}{dx}: Now we equate the derivatives from both sides:\newline(−3y2−1+2y)⋅dydx=2x(-3y^{2} - 1 + 2y) \cdot \frac{dy}{dx} = 2x\newlineTo find dydx\frac{dy}{dx}, we need to solve for it:\newlinedydx=2x−3y2−1+2y\frac{dy}{dx} = \frac{2x}{-3y^{2} - 1 + 2y}
  5. Substitute point (4,−2)(4,-2): We substitute the point (4,−2)(4,-2) into the equation to find the value of dydx\frac{dy}{dx} at that point:\newline(dydx)∣(4,−2)=2⋅4−3⋅(−2)2−1+2⋅(−2)\left(\frac{dy}{dx}\right)|_{(4,-2)} = \frac{2\cdot 4}{-3\cdot (-2)^{2} - 1 + 2\cdot (-2)}
  6. Perform calculations: Now we perform the calculations:\newline(dydx)∣(4,−2)=8(−3⋅4−1−4)(\frac{dy}{dx})|_{(4,-2)} = \frac{8}{(-3\cdot 4 - 1 - 4)}\newline(dydx)∣(4,−2)=8(−12−1−4)(\frac{dy}{dx})|_{(4,-2)} = \frac{8}{(-12 - 1 - 4)}\newline(dydx)∣(4,−2)=8(−17)(\frac{dy}{dx})|_{(4,-2)} = \frac{8}{(-17)}
  7. Final answer: Simplifying the fraction, we get the final answer: (dydx)∣(4,−2)=−817(\frac{dy}{dx})|_{(4,-2)} = \frac{-8}{17}

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