I'm struggling somewhat to understand how to use implicit differentiation to solve the following equation: cos ⁡ cos ⁡ ( x 3 y 2 ) − x cot ⁡ y = − 2 yI figublack that the calculation requires the chain rule to differentiate the composite function, but I'm not sure how to 'remove' the y with respect to x from inside the composite function. My calculations are: d y d x [ cos ⁡ cos ⁡ ( x 3 y 2 ) − x cot ⁡ y ] = d y d x [ − 2 y ] d y d x [ cos ⁡ cos ⁡ ( x 3 y 2 ) ] = sin ⁡ cos ⁡ ( x 3 y 2 ⋅ y ′ ( x ) ) ⋅ sin ⁡ ( x 3 y 2 ⋅ y ′ ( x ) ) ⋅ 6 x 2 y ⋅ y ′ ( x )This seems a bit long and convoluted. I'm also not sure how this will allow me to solve for y ′ ( x ). Carrying on... d y d x [ x cot ⁡ y ] = − csc 2 ⁡ y ⋅ y ′ ( x ) d y d x [ − 2 y ] = − 2Is my calculation correct so far? This seems to be a very complex derivative. Any comments or feedback would be appreciated.

Question

I&#039;m struggling somewhat to understand how to use implicit differentiation to solve the following equation:  cos  ⁡  cos  ⁡  (      x    3        y    2    )  −  x  cot  ⁡  y  =  −  2  yI figublack that the calculation requires the chain rule to differentiate the composite function, but I&#039;m not sure how to &#039;remove&#039; the y with respect to x from inside the composite function. My calculations are:            d      y              d      x        [  cos  ⁡  cos  ⁡  (      x    3        y    2    )  −  x  cot  ⁡  y  ]  =            d      y              d      x        [  −  2  y  ]            d      y              d      x        [  cos  ⁡  cos  ⁡  (      x    3        y    2    )  ]  =  sin  ⁡  cos  ⁡  (      x    3        y    2    ⋅      y    ′    (  x  )  )  ⋅  sin  ⁡  (      x    3        y    2    ⋅      y    ′    (  x  )  )  ⋅  6      x    2    y  ⋅      y    ′    (  x  )This seems a bit long and convoluted. I&#039;m also not sure how this will allow me to solve for       y    ′    (  x  ). Carrying on...            d      y              d      x        [  x  cot  ⁡  y  ]  =  −      csc    2    ⁡  y  ⋅      y    ′    (  x  )            d      y              d      x        [  −  2  y  ]  =  −  2Is my calculation correct so far? This seems to be a very complex derivative. Any comments or feedback would be appreciated.

Melanie Wong · Accepted Answer

First, you should be writing             d              d      x      , not             d      y              d      x      .             d      y              d      x       refers to the derivative of y with respect to   x, while here you are taking the derivative of some complicated function with respect to   x. After that, this is just an application of the chain rule. On the right-hand side,      d          d      x        (  −  2  y  )  =  −  2            d      y              d      x        =  −  2      y    ′    (  x  )  .On the left-hand side,                              d                      d            x                          [        cos        ⁡        cos        ⁡        (                  x          3                          y          2                )        −        x        cot        ⁡        y        ]                            =                  d                      d            x                          (        cos        ⁡        cos        ⁡        (                  x          3                          y          2                )        )        −                  d                      d            x                          (        x        cot        ⁡        y        )                                          =        −        sin        ⁡        cos        ⁡        (                  x          3                          y          2                )        ⋅                  d                      d            x                          (        cos        ⁡        (                  x          3                          y          2                )        )        −        cot        ⁡        y        −        x                  d                      d            x                          (        cot        ⁡        y        )                                          =        −        sin        ⁡        cos        ⁡        (                  x          3                          y          2                )                  (          −          sin          ⁡          (                      x            3                                y            2                    )          )                ⋅                  d                      d            x                          (                  x          3                          y          2                )        −        cot        ⁡        y        +        x                  csc          2                ⁡        y        ⋅                              d            y                                d            x                                                            =        sin        ⁡        cos        ⁡        (                  x          3                          y          2                )        sin        ⁡        (                  x          3                          y          2                )                  (          3                      x            2                                y            2                    +          2                      x            3                    y                                    d              y                                      d              x                                )                −        cot        ⁡        y        +        x                  csc          2                ⁡        y        ⋅                              d            y                                d            x                                                            =        sin        ⁡        cos        ⁡        (                  x          3                          y          2                )        sin        ⁡        (                  x          3                          y          2                )        (        3                  x          2                          y          2                +        2                  x          3                y                  y          ′                )        −        cot        ⁡        y        +        x                  y          ′                          csc          2                ⁡        y        .            Set those two equal and solve for       y    ′  .