I have the following expression which I need to implicitly differentiate: x y 2 + x 2 + y + sin ⁡ ( x 2 y ) = 0I'm a little confused as I'm not entirely sure what to do with the trig function. Here is my work so far: d y d x [ x y 2 + x 2 + y + sin ⁡ ( x 2 y ) ] = d y d x 0 d y 2 d x + 2 x + d y d x + cos ⁡ ( x 2 y ) ( 2 x d y d x ) = 0How should I proceed?

Question

I have the following expression which I need to implicitly differentiate:  x      y    2    +      x    2    +  y  +  sin  ⁡  (      x    2    y  )  =  0I&#039;m a little confused as I&#039;m not entirely sure what to do with the trig function. Here is my work so far:            d      y              d      x        [  x      y    2    +      x    2    +  y  +  sin  ⁡  (      x    2    y  )  ]  =            d      y              d      x        0            d              y        2                    d      x        +  2  x  +            d      y              d      x        +  cos  ⁡  (      x    2    y  )  (  2  x            d      y              d      x        )  =  0How should I proceed?

Simeon Hester · Accepted Answer

d          d      x        [  x      y    2    +      x    2    +  y  +  sin  ⁡  (      x    2    y  )  ]  =      d          d      x        (  0  )      ⟹        y    2    +  2  x  y            d      y              d      x        +  2  x  +            d      y              d      x        +  cos  ⁡  (      x    2    y  )  (  2  x  y  +      x    2              d      y              d      x        )  =  0We use the product rule and chain rule here, and also the operator for differentiation is      d          d      x

zaiskaladu · Accepted Answer

First, you should take the derivative of both sides of the equation (apply       d          d      x       instead of             d      y              d      x      ). Then make use of the chain and product rules. It helps to think of   y as a function of   x (i.e.   y  =  y  (  x  )).                              d                      d            x                          [        x                  y          2                +                  x          2                +        y        +        s        i        n        (                  x          2                y        )        ]                            =                  d                      d            x                          0                                      y          2                +        x        (        2        y        )                              d            y                                d            x                          +        2        x        +                              d            y                                d            x                          +        cos        ⁡        (                  x          2                y        )                  [                      d                          d              x                                (                      x            2                    y          )          ]                                    =        0                                      y          2                +        x        (        2        y        )                              d            y                                d            x                          +        2        x        +                              d            y                                d            x                          +        cos        ⁡        (                  x          2                y        )        (        2        x        y        +                  x          2                                      d            y                                d            x                          )                            =        0                                                  d            y                                d            x                                    (          2          x          y          +          1          +                      x            2                    cos          ⁡          (                      x            2                    y          )          )                                    =        −        (                  y          2                +        2        x        )                                                  d            y                                d            x                                              =        −                              (                          y              2                        +            2            x            )                                (            2            x            y            +            1            +                          x              2                        cos            ⁡            (                          x              2                        y            )            )

I have the following expression which I need to implicitly differentiate: xy^2+x^2+y+sin(x^2y)=0

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