When I search implicit differentiation for equation x 2 + y 2 = r 2 I find results of two versions: one using derivative and the other using differential.Version1: d d x ( x 2 + y 2 = r 2 ) ⇔ 2 x + 2 y d y d x = 0Version2: d ( x 2 + y 2 = r 2 ) ⇔ 2 x d x + 2 y d y = 0Using both methods, I can derive the result: d y d x = − x y However, I am confused, could you please provide some answers to:1. Which one (derivative /differential) is the "real" implicit differentiation?2. What are the differences between using these two methods?3. When should differential be used rather than derivative?

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When I search implicit differentiation for equation       x    2    +      y    2    =      r    2   I find results of two versions: one using derivative and the other using differential.Version1:       d          d      x        (      x    2    +      y    2    =      r    2    )  ⇔  2  x  +  2  y            d      y              d      x        =  0Version2:   d  (      x    2    +      y    2    =      r    2    )  ⇔  2  x  d  x  +  2  y  d  y  =  0Using both methods, I can derive the result:             d      y              d      x        =  −      x    y  However, I am confused, could you please provide some answers to:1. Which one (derivative /differential) is the &quot;real&quot; implicit differentiation?2. What are the differences between using these two methods?3. When should differential be used rather than derivative?

embraci4i · Accepted Answer

I think the geometric set-up is not really clear. In the first case you are viewing   y as a function of   x, so the equation       x    2    +      y    2    =      r    2   defines two semicircles in             R        2  , i.e.   y  =  ±  (      r    2    −      x    2        )          1              /            2      . Then it makes sense to write   d  y      /    d  x, and the resulting equation in version one will be an equation in the only variable   x.In the second case the dependence of   y on   x is not clear. In principle they could be independent variables. You would then have a vanishing function of two variables   x  ,  y which is   f  (  x  ,  y  )  =      x    2    +      y    2    −      r    2  , whose differential is zero:   d  f  =  2  x  d  x  +  2  y  d  y  =  0. Since   d  x  ,  d  y are independent this gives   x  =  y  =  0, meaning that   (  x  ,  y  )  =  (  0  ,  0  ) is a critical point of   f. This is clear geometrically as   f represents a cone in             R        3   whose vertex is   (  0  ,  0  ,  −      r    2    ).

Nathanael Perkins · Accepted Answer

Any two-variable relation can be summarized by the equation  f  (  x  ,  y  )  =  0In your example this would be      x    2    +      y    2    −      r    2    =  0The total derivative of   f with respect to an arbitrary variable   u is                    d            f                      d            u        =            ∂      f              ∂      x                          d            x                      d            u        +            ∂      f              ∂      y                          d            y                      d            u      We can plug in   u  =  x to obtain what you mention. Implicit differentiation is just another way to view total multivariable differentiation.

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