I'm supposed to implicitly differentiate the following and give the answer in terms of y ′ tan ⁡ ( x − y ) = y 1 + x 2 ( 1 + x 2 ) y ′ − 2 x y ( 1 + x 2 ) 2 How do I solve for y ′?Edit: After a ludicrous amount of algebra, i finally ended up at s e c 2 ( x − y ) ( 1 + x 2 ) 2 + 2 x y ( 1 + x 2 ) ( 1 + s e c 2 ( x − y ) ( 1 + x 2 ) ) which apparently is correct.

Question

I&#039;m supposed to implicitly differentiate the following and give the answer in terms of   y  ′  tan  ⁡  (  x  −  y  )  =            y              1        +                  x          2                                        (        1        +                  x          2                )        y        ′        −        2        x        y                    (        1        +                  x          2                          )          2                    How do I solve for   y  ′?Edit: After a ludicrous amount of algebra, i finally ended up at                    s        e                  c          2                (        x        −        y        )        (        1        +                  x          2                          )          2                +        2        x        y                    (        1        +                  x          2                )        (        1        +        s        e                  c          2                (        x        −        y        )        (        1        +                  x          2                )        )            which apparently is correct.

Regan Holloway · Accepted Answer

hint: You differentiate both sides of the equation. What you did was differentiate the right side. For the left side, it is by the Chain Rule that   tan  ⁡  (  x  −  y      )    ′    =  (  1  −      y    ′    )      sec    2    ⁡  (  x  −  y  ), and set it equal to the expression you got. Can you continue toward the solution?

kaltEvallwsr · Accepted Answer

Differentiating both sides from  tan  ⁡  (  x  −  y  (  x  )  )  =            y      (      x      )                                x          2                    +      1      gives:      1                                        cos                    2                    (      x      −      y      (      x      )      )        −                    y        ′            (      x      )                                            cos                    2                    (      x      −      y      (      x      )      )        =                    y        ′            (      x      )                                x          2                    +      1        −            2      x      y      (      x      )                                                          (                                                                    x                    2                                                  +                1                            )                                2                    which can be solved and simplified for             y      ′        (    x    )   like so:      y    ′    (  x  )  =            2      x      y      (      x      )                                x          2                    +      1        +            2              (                                            x              2                                −          2          x          y          (          x          )          +          1                )                    2                        x          2                    +      cos      ⁡      (      2      (      x      −      y      (      x      )      )      )      +      3

I'm supposed to implicitly differentiate the following and give the answer in terms of y′. tan(x−y)=y/(1+x^2) ((1+x^2)y′−2xy)/((1+x^2)^2) How do I solve for y′?

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