I am trying to find the derivative d y d x of the following function y e x y = sin ⁡ x using implicit differentiation.What I have is the following using product rule: d d x y e x y = d d x sin ⁡ x ⟹ y 2 e x y + d y d x e x y = cos ⁡ xHowever, I noticed in the solution manual it has it as the following and not sure why: d d x y e x y = d d x sin ⁡ x ⟹ y e x y ( x d y d x + y ) + d y d x e x y = cos ⁡ x

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I am trying to find the derivative             d      y              d      x       of the following function   y      e          x      y        =  sin  ⁡  x using implicit differentiation.What I have is the following using product rule:      d          d      x        y      e          x      y        =      d          d      x        sin  ⁡  x    ⟹        y    2        e          x      y        +            d      y              d      x            e          x      y        =  cos  ⁡  xHowever, I noticed in the solution manual it has it as the following and not sure why:      d          d      x        y      e          x      y        =      d          d      x        sin  ⁡  x    ⟹    y      e          x      y            (    x                  d        y                    d        x              +    y    )    +            d      y              d      x            e          x      y        =  cos  ⁡  x

matracavade · Accepted Answer

d          d      x        y      e          x      y        =      d          d      x        sin  ⁡  xThis line is not correct:    ⟹              y      2              e              x        y              +            d      y              d      x            e          x      y        =  cos  ⁡  xNote that you can do this:      d          d      x        y      e          x      y        =      e          x      y                  d      y              d      x        +  y            d              e                  x          y                            d      x      And apply the chain rule:            d              e                  x          y                            d      x        =            d              e                  x          y                            d      x      y                  d      x      y              d      x        =      e          x      y        (      y    ′    x  +  y  )Therefore:      d          d      x        y      e          x      y        =      e          x      y            y    ′    +  y      e          x      y        (      y    ′    x  +  y  )

epifizamvg · Accepted Answer

Your issue is that   y      d          d      x            e          x      y        =  y      e          x      y        [      d          d      x        x  y  ]  =  y      e          x      y        [  y  +  x      y    ′    ]  .

I am trying to find the derivative (dy)/(dx) of the following function ye^(xy)=sinx using implicit differentiation.

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