If y + sin ⁡ ( y ) = cos ⁡ ( x ), then d y d x = − sin ⁡ ( x ) 1 + cos ⁡ ( y ) ,so d y d x is not defined when cos ⁡ ( y ) = − 1.But cos ⁡ ( x ) is differentiable at every real number, so y + sin ⁡ ( y ) should also be differentiable, and so y should be differentiable at every point. But it is not differentiable when ? cos ⁡ ( y ) = − 1

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If       y    +  sin  ⁡  (  y  )  =  cos  ⁡  (  x  ), then            d      y              d      x          =    −            sin      ⁡      (      x      )              1      +      cos      ⁡      (      y      )        ,so             d      y              d      x       is not defined when   cos  ⁡  (  y  )  =  −  1.But   cos  ⁡  (  x  ) is differentiable at every real number, so       y    +  sin  ⁡  (  y  ) should also be differentiable, and so   y should be differentiable at every point. But it is not differentiable when ?  cos  ⁡  (  y  )  =  −  1

Mario Dorsey · Accepted Answer

As it comes, your implicit function is differentiable at all its points.Namely,       |    cos  ⁡  x      |    ≤  1 for all   x. On the other hand, if you assume   cos  ⁡  y  =  −  1 then   y is an odd multiple of   π, so   sin  ⁡  y  =  0 and so       |    y  +  sin  ⁡  y      |    =      |    y      |    ≥  π. Thus, the case   cos  ⁡  y  =  −  1 never happens on any of the points of the curve   y  +  sin  ⁡  y  =  cos  ⁡  x.The moral here is: the derivative is decided on each point on the curve, with the stress on the words on the curve.

If y+sin(y)=cos(x), then dy/dx=−(sin(x))/(1+cos(y)), so dy/dx is not defined when cos(y)=−1. But cos(x) is differentiable at every real number, so y+sin(y) should also be differentiable, and so y should be differentiable at every point. But it is not differentiable when cos(y)=−1?

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