find d y d x of x 3 + y 3 + 8 using partial derivativeSo what I need to do is ∂ ∂ x and ∂ ∂ y which are: 3 x 2 and 3 y 2 and than to divide to get ∂ ∂ x ∂ ∂ y = ∂ y ∂ x = 1?

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find             d      y              d      x       of       x    3    +      y    3    +  8 using partial derivativeSo what I need to do is       ∂          ∂      x       and       ∂          ∂      y       which are:  3      x    2   and   3      y    2   and than to divide to get                     ∂                  ∂          x                            ∂                  ∂          y                      =            ∂      y              ∂      x        =  1?

gortepap6yb · Accepted Answer

Yes, but sign should be also correct, see how the negative sign comes in.Let  u  =      x    3    +      y    3    +  8By two variable implicit differentiation at constant   u,   d  u  =  0  d  u  =            ∂      u              ∂      x        d  x  +            ∂      u              ∂      y        d  y  =  0            d      y              d      x        =  −                                          ∂            u                                ∂            x                                                            ∂            u                                ∂            y                                =  −                    3                  x          2                            3                  y          2                      =  −                    x        2                    y        2              .

pighead73283r · Accepted Answer

I suppose that this is for   f  =      x    3    +      y    3    +  8  =  0 and not for   f  =      x    3    +      y    3    +  8.So       f    x    ′    =  3      x    2  ,       f    y    ′    =  3      y    2   and, by the implicit function theorem,            d      y              d      x        =  −            f      x      ′              f      y      ′        =  −            x      2              y      2

find dy/dx of x^3+y^3+8 using partial derivative

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