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Part (a)

Product Rule u’v + uv’. Factor out a (dy/dx). (dy/dx) (8y - 3x) = 3y – 2x. (dy/dx) =. 3y – 2x. 8y – 3x. Part (a). I chose to use implicit differentiation. x 2 + 4y 2 = 7 + 3x y. 2x + 8y(dy/dx) = 3y + 3x(dy/dx). 8y(dy/dx) - 3x(dy/dx) = 3y – 2x. From Part (a). = 0. 3y – 2x.

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Part (a)

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  1. Product Ruleu’v + uv’ Factor out a (dy/dx) (dy/dx) (8y - 3x) = 3y – 2x (dy/dx) = 3y – 2x 8y – 3x Part (a) I chose to use implicit differentiation. x2 + 4y2 = 7 + 3x y 2x + 8y(dy/dx) = 3y + 3x(dy/dx) 8y(dy/dx) - 3x(dy/dx) = 3y – 2x

  2. From Part (a) = 0 3y – 2x Cross multiply 8y – 3x (dy/dx) = 3y – 2x y = 2 8y – 3x Part (b) They’re telling us to set the derivative equal to zero. 3y – 2x = 0 3y – 6 = 0 There’s a horizontal tangent at (3,2)

  3. Part (b) x2 + 4y2 = 7 + 3xy 32 + (4)(2)2 = 7 + 3(3)(2) 9 + 16 = 7 + 18 25 = 25 Yup! It would be a good idea to make sure that (3,2) is actually on our curve!!! There’s a horizontal tangent at (3,2)

  4. d2y/dx2 = [3(dy/dx) – 2](8y – 3x) - (3y – 2x) [8(dy/dx) – 3] (8y – 3x)2 (dy/dx) = 3y – 2x 8y – 3x Part (c) First, we’re being asked to find the value of the 2nd derivative at (3,2). Quotient Rule(u’v - uv’)/v2 u’ = 3(dy/dx) – 2v’ = 8(dy/dx) - 3

  5. d2y/dx2 = [3(dy/dx) – 2](8y – 3x) - (3y – 2x) [8(dy/dx) – 3] (8y – 3x)2 d2y/dx2 = [3(0) – 2](8y – 3x) - (3y – 2x) [8(0) – 3] (8y – 3x)2 Part (c) First, we’re asked to find the value of the 2nd derivative at (3,2). Keep in mind that the value of the 1st derivative at (3,2) is zero.

  6. d2y/dx2 = = (–2)(7) – (0)(-3) -2 (7)2 7 d2y/dx2 = (–2)[(8)(2) – 3(3)] – [(3)(2) – (2)(3)](–3) [(8)(2)– (3)(3)]2 @ (3,2) d2y/dx2 = [3(0) – 2](8y – 3x) - (3y – 2x) [8(0) – 3] (8y – 3x)2 Part (c)

  7. d2y/dx2 = = (–2)(7) – (0)(-3) -2 (7)2 7 Part (c) The 1st derivative is zero at (3,2) while the 2nd derivative is negative there. Therefore, according to the 2nd derivative test, the curve has a local maximum at (3,2).

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