How do you use implicit differentiation to find d 2 y d x 2 of x 3 + y 3 = 1

Question

How do you use implicit differentiation to find                               d          2                y                              d          x                2             of             x      3        +          y      3        =    1

yermarvg · Accepted Answer

By implicitly differentiating twice, we can find                              d          2                y                                          d            x                    2                      =    -                  2        x                    y        5            First, let us find                               d          y                                      d          x                                x      3        +          y      3        =    1  by differentiating with respect to x      ⇒    3          x      2        +    3          y      2                                d          y                                      d          x                      =    0  by subtracting       3          x      2            ⇒    3          y      2                                d          y                                      d          x                      =    -    3          x      2      by dividing by       3          y      2            ⇒                            d          y                                      d          x                      =    -                            x          2                                      y          2                    Now, let us find                               d          2                y                                          d            x                    2                    by differentiating with respect to x      ⇒                            d          2                y                                          d            x                    2                      =    -                  2        x        ⋅                  y          2                -                  x          2                ⋅        2        y                                            d              y                                                          d              x                                                                        (                          y              2                        )                    2                      =    -                  2        x                  (                      y            2                    -          x          y                                                    d                y                                                                    d                x                                              )                                      y          4                    by plugging in                               d          y                                      d          x                      =    -                            x          2                                      y          2                          ⇒                            d          2                y                                          d            x                    2                      =    -                  2        x                  [                      y            2                    -          x          y                      (            -                                          x                2                                            y                2                                      )                    ]                            y        4              =    -                  2        x                  (                      y            2                    +                                    x              3                        y                    )                            y        4            by multiplying the numerator and the denominator by y      ⇒                            d          2                y                                          d            x                    2                      =    -                  2        x                  (                      y            3                    +                      x            3                    )                            y        5            by plugging in             y      3        +          x      3        =    1        ⇒                            d          2                y                                          d            x                    2                      =    -                  2        x                    y        5

How do you use implicit differentiation to find (d^2y)/dx^2 of x^3+y^3=1

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