Find the derivative y x ′ (edit: y x ′ = d y d x ) from the equation y 3 + x 2 = x e y 2 − y sin ⁡ x

Question

Find the derivative       y    x    ′   (edit:       y    x    ′    =                    d        y                    d        x            ) from the equation      y    3    +      x    2    =  x      e                  y        2              −  y  sin  ⁡  x

levraijournalk1o · Accepted Answer

Let y(x) be the solution, then:  0  =      ∂          ∂      x        (      x    2    +  y  (  x      )    3    −  x      e          y      (      x              )        2              +  y  (  x  )  sin  ⁡  (  x  )  )  =  =  3  y  (  x      )    2        y    ′    (  x  )  −  2  x      e          y      (      x              )        2              y  (  x  )      y    ′    (  x  )  +  sin  ⁡  (  x  )      y    ′    (  x  )  −      e          y      (      x              )        2              +  y  (  x  )  cos  ⁡  (  x  )  +  2  xWith this we have:      y    ′    (  x  )  =            −              e                  y          (          x                      )            2                              +      y      (      x      )      cos      ⁡      (      x      )      +      2      x              −      3      y      (      x              )        2            +      2      x              e                  y          (          x                      )            2                              y      (      x      )      −      sin      ⁡      (      x      )        =  =                    x        2            −      y      (      x              )        3            −      y      (      x      )      sin      ⁡      (      x      )      +      x      y      (      x      )      cos      ⁡      (      x      )              x              (        2                  x          2                y        (        x        )        +        2        y        (        x                  )          4                −        3        y        (        x                  )          2                +        2        y        (        x                  )          2                sin        ⁡        (        x        )        −        sin        ⁡        (        x        )        )

kemecryncqe9 · Accepted Answer

Hint:Deriving with respect to x we have:      d          d      x            (          y      3        +          x      2        −    x          e                        y          2                      +    y    sin    ⁡    x    )    =      d          d      x            (          y      3        )    +      d          d      x            (          x      2        )    −      d          d      x            (    x          e                        y          2                      )    +      d          d      x            (    y    sin    ⁡    x    )    =  0now use the product rule and the chain rule to evaluate the derivatives:      d          d      x            (          y      3        )    =  2      y    2              d      y              d      x            d          d      x            (          x      2        )    =  2  x      d          d      x            (    x          e                        y          2                      )    =      e                  y        2              +  2  x  y      e                  y        2                        d      y              d      x            d          d      x            (    y    sin    ⁡    x    )    =  sin  ⁡  x            d      y              d      x        +  y  cos  ⁡  xnow put all together and find             d      y              d      x

Find the derivative y′_x (edit: y′_x=(dy)/(dx)) from the equation y^3+x^2=xe^(y^2)−y sin x

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