Determining divergence and gradient using chain ruleSuppose r → = x i ^ + y j ^ + z k ^ and r = | r | . Also, let f be a scalar function of r and let A → be a vector function of r. We wish to determine ∇ f and ∇ . A → .It seems to be obvious that the chain rule is the way to go here: ∇ f = ∂ f ∂ x i ^ + ∂ f ∂ y j ^ + ∂ f ∂ z k ^ = ∂ f ∂ r ∂ r ∂ x i ^ + ∂ f ∂ r ∂ r ∂ y j ^ + ∂ f ∂ r ∂ r ∂ z k ^ = ∂ f ∂ r ( ∂ r ∂ x i ^ + ∂ r ∂ y j ^ + ∂ r ∂ z k ^ ) = ∂ f ∂ r r ^ Which is correct. However, ∇ . A → = ∂ A → ∂ x . i ^ + ∂ A → ∂ y . j ^ + ∂ A → ∂ z . k ^ = ∂ A → ∂ r ∂ r ∂ x . i ^ + ∂ A → ∂ r ∂ r ∂ y . j ^ + ∂ A → ∂ r ∂ r ∂ z . k ^ = ∂ A → ∂ r . ( ∂ r ∂ x i ^ + ∂ r ∂ y j ^ + ∂ r ∂ z k ^ ) = ∂ A → ∂ r . r ^ This expressions turns out to be incorrect.The only scope for error seems to be in the usage of the chain rule, however if it(my usage of the chain rule) worked for the gradient, why did it fail here ?

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Determining divergence and gradient using chain ruleSuppose             r      →        =  x            i      ^        +  y            j      ^        +  z            k      ^       and   r  =      |    r      |  . Also, let   f  be a scalar function of r  and let             A      →       be a vector function of r. We wish to determine   ∇  f and   ∇  .            A      →      .It seems to be obvious that the chain rule is the way to go here:  ∇  f  =                    ∂        f                    ∂        x                        i      ^        +                    ∂        f                    ∂        y                        j      ^        +                    ∂        f                    ∂        z                        k      ^        =                    ∂        f                    ∂        r                                ∂        r                    ∂        x                        i      ^        +                    ∂        f                    ∂        r                                ∂        r                    ∂        y                        j      ^        +                    ∂        f                    ∂        r                                ∂        r                    ∂        z                        k      ^        =                    ∂        f                    ∂        r              (                    ∂        r                    ∂        x                        i      ^        +                    ∂        r                    ∂        y                        j      ^        +                    ∂        r                    ∂        z                        k      ^        )  =                    ∂        f                    ∂        r                        r      ^      Which is correct. However,  ∇  .            A      →        =                    ∂                              A            →                                      ∂        x              .            i      ^        +                    ∂                              A            →                                      ∂        y              .            j      ^        +                    ∂                              A            →                                      ∂        z              .            k      ^        =                    ∂                              A            →                                      ∂        r                                ∂        r                    ∂        x              .            i      ^        +                    ∂                              A            →                                      ∂        r                                ∂        r                    ∂        y              .            j      ^        +                    ∂                              A            →                                      ∂        r                                ∂        r                    ∂        z              .            k      ^        =                    ∂                              A            →                                      ∂        r              .  (                    ∂        r                    ∂        x                        i      ^        +                    ∂        r                    ∂        y                        j      ^        +                    ∂        r                    ∂        z                        k      ^        )  =                    ∂                              A            →                                      ∂        r              .            r      ^      This expressions turns out to be incorrect.The only scope for error seems to be in the usage of the chain rule, however if it(my usage of the chain rule) worked for the gradient, why did it fail here ?

Abril Orr · Accepted Answer

Ok, so here&#039;s the problem, you didn&#039;t account for the unit vectors changing.  ∇  ⋅            A      →        =      [                  x        ^              ⋅          ∂              ∂        x              +                  y        ^              ⋅          ∂              ∂        y              +                  z        ^              ⋅          ∂              ∂        z              ]              A      →      Now, let&#039;s consider the dot product with             x      ^      :                    (1)                                          x            ^                          ⋅                  ∂                      ∂            x                          (                              A            →                          )            We can write the differential operator as:      ∂          ∂      x        =            ∂      r              ∂      x            ∂          ∂      r        +            ∂      θ              ∂      x            ∂          ∂      θ      Hence from (1), we get:            x      ^        ⋅  (            ∂                        A          →                            ∂      r        +            ∂      θ              ∂      x                  ∂                        A          →                            ∂      θ        )Particularly speaking, the term of interest is             ∂                        A          →                            ∂      θ      , this we can write as(*);            ∂                        A          →                            ∂      θ        =            ∂              |            A      (      r      )              |                              r          ^                            ∂      θ        =      |    A      |              ∂                        r          ^                            ∂      θ        =      |    A      |              θ      ^      Even though we the function is not directly dependent on   θ or   ϕ the unit vectors are and hence contribute to derivative. Hope this helped!*: I assumed             A      →        =      |              A      →            |              r      ^

Suppose vec(r) =x hat(i)+y hatj+z hatk, and r=|r|. Also, let f be a scalar function of r and let vec(A) be a vector function of r. We wish to determine grad f and grad .vec(A).

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