How to use Implicit differentiation find x 2 + 2 x y - y 2 + x = 2 and to find an equation of the tangent line to the curve, at the point (1,2)?

Question

How to use Implicit differentiation find             x      2        +    2    x    y    -          y      2        +    x    =    2   and to find an equation of the tangent line to the curve, at the point (1,2)?

Marin Valencia · Accepted Answer

x      2        +    2    x    y    -          y      2        +    x    =    2    Distinguish the two sides based on       x  .            d                        d          x                            (              x        2            +      2      x      y      -              y        2            +      x      )        =          d                        d          x                            (      2      )        (Write this every time)            d                        d          x                            (              x        2            )        +          d                        d          x                            (      +      2      x      y      )        -          d                        d          x                            (              y        2            )        +          d                        d          x                            (      x      )        =          d                        d          x                            (      2      )       Remember that y is the name of some function(s) of       x   that I haven&#039;t found expressions for.So       2    x    y   is really       2    x    ⋅          (              some f of x            )        We&#039;ll need the product and chain rules.(I use the product rule written as             (      F      S      )        ′    =    F    ′    S    +    F    S    ′  .)Similarly             y      2       is really                       (                  some f of x                )            2       so we&#039;ll need the power and chain rules.  (Implicit differentiation is using the chain rule.)      2    x                            d          x                                      d          x                      +          [              d                              d            x                                      (        +        2        x        )            ⋅      y      +              (        +        2        x        )                    d                              d            x                                      (        y        )            ]        -    2    y                  d        y                    d        x              +    1    =    0        2    x    +    2    y    +    2    x                            d          y                                      d          x                      -    2    y                  d        y                    d        x              +    1    =    0  We could solve for                               d          y                                      d          x                      =                  -                  (          2          x          +          2          y          +          1          )                            2        x        -        2        y            , but if we don&#039;t have to, it&#039;s usually easier to substitute numbers now:At             (      1      ,      2      )      , we get:      2          (      1      )        +    2          (      2      )        +    2          (      1      )                                d          y                                      d          x                      -    2          (      2      )                      d        y                    d        x              +    1    =    0   so       2    +    4    +    2                            d          y                                      d          x                      -    4                  d        y                    d        x              +    1    =    0          7    -    2                            d          y                                      d          x                      =    0   and finally:                              d          y                                      d          x                      =          7      2      The tangent line contains the point             (      1      ,      2      )       and has slope       m    =          7      2       so its equation is:      y    =          7      2        x    -          3      2      With experience, your solution will look more like:            x      2        +    2    x    y    -          y      2        +    x    =    2               d                        d          x                            (              x        2            +      2      x      y      -              y        2            +      x      )        =          d                        d          x                            (      2      )             2    x    +    2    y    +    2    x                            d          y                                      d          x                      -    2    y                  d        y                    d        x              +    1    =    0  At             (      1      ,      2      )      :      2    +    4    +    2                            d          y                                      d          x                      -    4                  d        y                    d        x              +    1    =    0                                 d          y                                      d          x                      =          7      2

How to use Implicit differentiation find x^2+ 2xy- y^2 + x=2 and to find an equation of the tangent line to the curve, at the point (1,2)?

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