I'm given x 3 + y 3 = 6 x y. It's stated that y is a function of x and I'm tasked to differentiate with respect to x.The implicit differentiation is: 3 x 2 + 3 y 2 y ′ = 6 x y ′ + 6 yNow simplify and express in terms of y ′ . I'm going to number these steps.1. x 2 + y 2 y ′ = 2 x y ′ + 2 y2. y ′ = 2 x y ′ + 2 y − x 2 y 2 3. y ′ = 2 x y ′ y 2 + 2 y − x 2 y 2 4. y ′ = y ′ ∗ 2 x y 2 + 2 y − x 2 y 2 5. y ′ y ′ = 2 x y 2 + 2 y − x 2 y 2 . We know that y ′ y ′ = 1 and therefore I've made a mistake in my algebra. But I'm not sure what is wrong.Here is the correct simplification, taking it from step 1 again:1. x 2 + y 2 y ′ = 2 x y ′ + 2 y2. y 2 y ′ − 2 x y ′ = 2 y − x 2 3. y ′ ( y 2 − 2 x ) = 2 y − x 2 4. y ′ = 2 y − x 2 y 2 − 2 x This makes complete sense. What was wrong with my simplification? If it's not wrong, how can I arrive at the correct final expression of y ′ ?

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I&#039;m given       x    3    +      y    3    =  6  x  y. It&#039;s stated that   y is a function of   x and I&#039;m tasked to differentiate with respect to   x.The implicit differentiation is:  3      x    2    +  3      y    2        y    ′    =  6  x      y    ′    +  6  yNow simplify and express in terms of       y    ′  . I&#039;m going to number these steps.1.       x    2    +      y    2        y    ′    =  2  x      y    ′    +  2  y2.       y    ′    =            2      x              y        ′            +      2      y      −              x        2                    y      2      3.       y    ′    =            2      x              y        ′                    y      2        +            2      y      −              x        2                    y      2      4.       y    ′    =      y    ′    ∗            2      x              y      2        +            2      y      −              x        2                    y      2      5.             y      ′              y      ′        =            2      x              y      2        +            2      y      −              x        2                    y      2        . We know that             y      ′              y      ′        =  1 and therefore I&#039;ve made a mistake in my algebra. But I&#039;m not sure what is wrong.Here is the correct simplification, taking it from step 1 again:1.       x    2    +      y    2        y    ′    =  2  x      y    ′    +  2  y2.       y    2        y    ′    −  2  x      y    ′    =  2  y  −      x    2  3.       y    ′    (      y    2    −  2  x  )  =  2  y  −      x    2  4.       y    ′    =            2      y      −              x        2                            y        2            −      2      x      This makes complete sense. What was wrong with my simplification? If it&#039;s not wrong, how can I arrive at the correct final expression of       y    ′  ?

Marley Stone · Accepted Answer

Gather the       y    ′   terms together to get       y    ′    (  6  x  −  3      y    2    )  =  3      x    2    −  6  y.Then divide:       y    ′    =                    3                  x          2                −        6        y                    6        x        −        3                  y          2                    .

I'm given x^3+y^3=6xy. It's stated that y is a function of x and I'm tasked to differentiate with respect to x.

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