I'm learning implicit differentiation and I've hit a snag with the following equation. f ( x , y ) = x + x y + y = 2 D x ( x ) + D x ( x y ) + D x ( y ) = D x ( 2 ) 1 + x y ′ + y + y ′ = 0 x y ′ + y ′ = − 1 − y y ′ ( x + 1 ) = 1 + y y ′ = ( 1 + y ) ( x + 1 ) y ″ = ( x + 1 ) y ′ − ( 1 + y ) ( x + 1 ) 2 y ″ = ( x + 1 ) ( 1 + y ) ( x + 1 ) − ( 1 + y ) ( x + 1 ) 2 Ok now what according to this y'' = 0 which is wrong.

Question

I&#039;m learning implicit differentiation and I&#039;ve hit a snag with the following equation.  f  (  x  ,  y  )  =  x  +  x  y  +  y  =  2  D  x  (  x  )  +  D  x  (  x  y  )  +  D  x  (  y  )  =  D  x  (  2  )  1  +  x      y    ′    +  y  +      y    ′    =  0  x      y    ′    +      y    ′    =  −  1  −  y      y    ′    (  x  +  1  )  =  1  +  y      y    ′    =                    (        1        +        y        )                    (        x        +        1        )                  y    ″    =                    (        x        +        1        )                  y          ′                −        (        1        +        y        )                    (        x        +        1                  )          2                          y    ″    =                    (        x        +        1        )                                            (              1              +              y              )                                      (              x              +              1              )                                      −        (        1        +        y        )                    (        x        +        1                  )          2                    Ok now what according to this y&#039;&#039; = 0 which is wrong.

Dwayne Hood · Accepted Answer

Your mistake seems to originate when moving from this:  x      y    ′    +      y    ′    =  −  1  −  y...to this, where you &quot;lost the sign&quot;:      y    ′    (  x  +  1  )  =  1  +  yWe need      y    ′    (  x  +  1  )  =  −  (  1  +  y  )Let&#039;s back up:                    1        +        x                  y          ′                +        y        +                  y          ′                                    =        0                                          x                  y          ′                +                  y          ′                                    =        −        1        −        y                                                        =        −        (        1        +        y        )            Then we factor out       y    ′   on the left-hand side, giving us:      y    ′    (  x  +  1  )  =  −  (  1  +  y  )Fixing for that, then, we get:                              y          ′                                    =                                            −              (              1              +              y              )                                      (              x              +              1              )                                                                                  y          ″                                    =                              (            x            +            1            )            (            −                          y              ′                        )            −            [            −            (            1            +            y            )            ]                                (            x            +            1                          )              2                                                                                      =                              −            (            x            +            1            )                                                            −                  (                  1                  +                  y                  )                                                  (                  x                  +                  1                  )                                                      +            (            1            +            y            )                                (            x            +            1                          )              2                                                                                      =                              (            1            +            y            )            +            (            1            +            y            )                                (            x            +            1                          )              2                                                                                      =                              2            (            y            +            1            )                                (            x            +            1                          )              2

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New Questions in Calculus 2