Using implicit differentiation, find d y d x x 3 y 6 = ( x + y ) 9 My Attempt: x 3 y 6 = ( x + y ) 9 Differentiating both sides with respect to x, we get: d d x ( x 3 y 6 ) = d d x ( ( x + y ) 9 ) x 3 . d d x y 6 + y 6 . d d x x 3 = 9 ( x + y ) 8 d d x ( x + y ) 6 x 3 y 5 . d y d x + 3 x 2 y 6 = 9 ( x + y ) 8 ( 1 + d y d x ) d y d x = 9 ( x + y ) 8 − 3 x 2 y 6 x 3 y 5 − 9 ( x + y ) 8 But the answer is d y d x = y x

Question

Using implicit differentiation, find                     d        y                    d        x                  x    3        y    6    =  (  x  +  y      )    9  My Attempt:      x    3        y    6    =  (  x  +  y      )    9  Differentiating both sides with respect to   x, we get:            d              d        x              (      x    3        y    6    )  =            d              d        x              (  (  x  +  y      )    9    )      x    3    .            d              d        x                  y    6    +      y    6    .            d              d        x                  x    3    =  9  (  x  +  y      )    8              d              d        x              (  x  +  y  )  6      x    3        y    5    .                    d        y                    d        x              +  3      x    2        y    6    =  9  (  x  +  y      )    8    (  1  +                    d        y                    d        x              )                    d        y                    d        x              =                    9        (        x        +        y                  )          8                −        3                  x          2                y                    6                  x          3                          y          5                −        9        (        x        +        y                  )          8                    But the answer is                    d        y                    d        x              =            y      x

Houston Ellis · Accepted Answer

I fixed a typo in your question, and with that, your implicit differentiation is perfectly correct. However, it looks difficult (as in, I cannot see how) to blackuce it from that to the expected simple answer.However, if you&#039;re dealing only with the real domain it makes things a lot easier to take cube roots right at the outset. This simplifies things greatly.      x    3        y    6    =  (  x  +  y      )    9  For   x  ,  y  ∈      R  ,  x      y    2    =  (  x  +  y      )    3  Differentiating implictly wrt   x,      y    2    +  2  y      y    ′    x  =  3  (  x  +  y      )    2    (  1  +      y    ′    )      y    ′    (  2  y  x  −  3  (  x  +  y      )    2    )  =  3  (  x  +  y      )    2    −      y    2              y      ′        =                  3        (        x        +        y                  )          2                −                  y          2                            2        y        x        −        3        (        x        +        y                  )          2                    Use the initial relationship given on the RHS,            y      ′        =                                        3            (            x            +            y                          )              3                                            x            +            y                          −                  y          2                            2        y        x        −                              3            (            x            +            y                          )              3                                            x            +            y                                          y      ′        =                                        3            x                          y              2                                            x            +            y                          −                  y          2                            2        y        x        −                              3            x                          y              2                                            x            +            y                              The rest is just algebra,            y      ′        =                  3        x                  y          2                −                  y          2                (        x        +        y        )                    2        y        x        (        x        +        y        )        −        3        x                  y          2                                y      ′        =                  3        x                  y          2                −        x                  y          2                −                  y          3                            2        y                  x          2                +        2                  y          2                x        −        3        x                  y          2                                y      ′        =                  2        x                  y          2                −                  y          3                            2        y                  x          2                −        x                  y          2                                y      ′        =                            y          2                (        2        x        −        y        )                    y        x        (        2        x        −        y        )                        y      ′        =          y      x      as requiblack, et voilà!EDIT: And just to flesh out hint involving logarithmic differentiation:  3  log  ⁡  x  +  6  log  ⁡  y  =  9  log  ⁡  (  x  +  y  )            3      x        +          6      y              y      ′        =          9              x        +        y              (    1    +          y      ′        )              y      ′        (          6      y        −          9              x        +        y              )    =          9              x        +        y              −          3      x                  y      ′        (                  6        x        −        3        y                    y        (        x        +        y        )              )    =                  6        x        −        3        y                    x        (        x        +        y        )                        y      ′        =          y      x      A decidedly much quicker route to the answer.

spatularificw2 · Accepted Answer

For  (  x  +  y      )          m      +      n        =      x    m        y    n  Take logarithm wrt base   e  ,  (  m  +  n  )  ln  ⁡  (  x  +  y  )  =  m  ln  ⁡  x  +  n  ln  ⁡  yNow differentiate wrt   x

Using implicit differentiation, find (dy)/(dx) x^3y^6=(x+y)^9

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Answer & Explanation

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