Calculus, implicit diff help.........
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Calculus, implicit diff help.........

[From: ] [author: ] [Date: 11-09-18] [Hit: ]
55001705y = 28.6 (1 decimal place)dy/dx = 3x^2 - 2x - 20d^2 y / dx^2 = 6x - 2Setting d^2 y / dx^2 to zero, we have:6x - 2 = 06x = 2x = 0.333333333333.........

y = (x + 4) * (x^2 - 5x)
y = x^3 + 4x^2 - 5x^2 - 20x
y = x^3 - x^2 - 20x

dy/dx = 3x^2 - 2x - 20

Setting dy/dx to zero, we have:
3x^2 - 2x - 20 = 0
Using the quadratic formula, we will get:
x = [2 +/- sqrt((-2)^2 - 4(3)(-20))] / [(2)(3)]
x = [2 +/- sqrt(244)] / 6
x = 2.936749892 or x = -2.270083225
x = 2.9 or x = -2.3 [1 decimal place]

When x = 2.936749892,
y = (2.936749892)^3 - (2.936749892)^2 - 20(2.936749892)
y = -42.03149854
y = -42.0 (1 decimal place)

When x = -2.270083225,
y = (-2.270083225)^3 - (-2.270083225)^2 - 20(-2.270083225)
y = 28.55001705
y = 28.6 (1 decimal place)

dy/dx = 3x^2 - 2x - 20
d^2 y / dx^2 = 6x - 2

Setting d^2 y / dx^2 to zero, we have:
6x - 2 = 0
6x = 2
x = 0.333333333333...
x = 0.3 (1 decimal place)

When x = 0.333333..., we have:
y = (1/3)^3 - (1/3)^2 - 20(1/3)
y = -6.74074074....
y = -6.7 (1 decimal place)

Hence, we have:
p(2.9 , -42.0), p(-2.3 , 28.6), p(0.3 , -6.7)

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1.      y⁴ + x + sin(y) = 7            ← differentiate implicitly.

4y³y' + 1 + cos(y)·y' = 0
      4y³y' + cos(y)·y' = -1
      [4y³ + cos(y)]·y' = -1
                     ➊    y' = -1/[4y³ + cos(y)]            ← now, use the Quotient Rule


           [4y³ + cos(y)]•0 - (-1)•[12y²y' - sin(y)·y']
y'' = ———————————————————
                                 [4y³ + cos(y)]²


                      12y²y' - sin(y)·y'
y'' = ————————————
                        [4y³ + cos(y)]²


                [12y² - sin(y)]·y'
y'' = ———————————            ← ANSWER
                  [4y³ + cos(y)]²                              OR
                                                    If you need y'' in terms of y only,
                                            you can substitute ➊ in for y' and simplify


           - [12y² - sin(y)]
y'' = ——————————      ← substituted ➊ in for y' and simplified
              [4y³ + cos(y)]³


               sin(y) - 12y²
y'' = —————————            ← ANSWER
              [4y³ + cos(y)]³
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