Solve the implicit derivative

Mathematics

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Answer

y′′=0y'' = 0

Step-by-step solution

5 steps

To find y′′y'' for the equation x3+y3+1=0x^3 + y^3 + 1 = 0 at the point (−1,1)(-1, 1), we will use implicit differentiation.

  1. Differentiate the equation x3+y3+1=0x^3 + y^3 + 1 = 0 with respect to xx.

    ddx(x3)+ddx(y3)+ddx(1)=ddx(0)\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) + \frac{d}{dx}(1) = \frac{d}{dx}(0) 3x2+3y2dydx+0=03x^2 + 3y^2 \frac{dy}{dx} + 0 = 0 3x2+3y2y′=03x^2 + 3y^2 y' = 0

  2. Solve for y′y'.

    3y2y′=−3x23y^2 y' = -3x^2 y′=−3x23y2y' = -\frac{3x^2}{3y^2} y′=−x2y2y' = -\frac{x^2}{y^2}

  3. Evaluate y′y' at the point (−1,1)(-1, 1).

    Substitute x=−1x = -1 and y=1y = 1 into the expression for y′y'.

    y′=−(−1)2(1)2=−11=−1y' = -\frac{(-1)^2}{(1)^2} = -\frac{1}{1} = -1

  4. Differentiate y′=−x2y2y' = -\frac{x^2}{y^2} with respect to xx to find y′′y''.

    We will use the quotient rule: ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}. Here u=−x2u = -x^2 and v=y2v = y^2.

    u′=ddx(−x2)=−2xu' = \frac{d}{dx}(-x^2) = -2x v′=ddx(y2)=2ydydx=2yy′v' = \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} = 2y y'

    Now substitute these into the quotient rule:

    y′′=(−2x)(y2)−(−x2)(2yy′)(y2)2y'' = \frac{(-2x)(y^2) - (-x^2)(2y y')}{(y^2)^2} y′′=−2xy2+2x2yy′y4y'' = \frac{-2xy^2 + 2x^2y y'}{y^4}

  5. Substitute the values of x=−1x = -1, y=1y = 1, and y′=−1y' = -1 into the expression for y′′y''.

    y′′=−2(−1)(1)2+2(−1)2(1)(−1)(1)4y'' = \frac{-2(-1)(1)^2 + 2(-1)^2(1)(-1)}{(1)^4} y′′=2(1)+2(1)(1)(−1)1y'' = \frac{2(1) + 2(1)(1)(-1)}{1} y′′=2−21y'' = \frac{2 - 2}{1} y′′=0y'' = 0
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