Can the gradient exist for a function of n+1 variables?For a function of n + 1 variables f ( x 0 , x 1 , x 2 , . . . x n ) can a gradient exist?When I asked my professor this during class he said, "no, at most a gradient will exist for a function of three variables f(x,y,z) because there are only at most three standard basis vectors with which to represent a vector."This is a calculus 3 class so perhaps this answer was given to keep the concept of the gradient within the scope of the class, but I suspect this isn't the whole story and there is more to this than my professor is telling.Edit:The definition of the gradient for a function of two variables given during class was: Let z=f(x,y) be a function, then the gradient of f is defined as ∇ f = f x i → + f y j →

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Can the gradient exist for a function of n+1 variables?For a function of   n  +  1 variables   f  (      x    0    ,      x    1    ,      x    2    ,  .  .  .      x    n    ) can a gradient exist?When I asked my professor this during class he said, &quot;no, at most a gradient will exist for a function of three variables f(x,y,z) because there are only at most three standard basis vectors with which to represent a vector.&quot;This is a calculus 3 class so perhaps this answer was given to keep the concept of the gradient within the scope of the class, but I suspect this isn&#039;t the whole story and there is more to this than my professor is telling.Edit:The definition of the gradient for a function of two variables given during class was: Let z=f(x,y) be a function, then the gradient of f is defined as   ∇  f  =      f    x              i      →        +      f    y              j      →

Arturo Mays · Accepted Answer

I think the correction to your teacher is purely algebraic: the n-dimensional vector space             R        d   does have a standard basis formed by d vectors       e    1    ,  …      e    d  . Each basis vector is given by       e    i    =  (  0  …  1  …  0      )    ⊤   where the only non-zero component is at the i-th position.The extension of the definition of partial derivative is also straightforward. Given a function   f  :            R        d    →      R  , and a point   p  ∈            R        d  ,   p  =  (      p    1    ,  …      p    d    ), consider the function       g    i    :      R    →      R   given by   g  (  x  )  =  f  (      p    1    ,  …      p          i      −      1        ,  x  ,      p          i      +      1        ,  …      p    d    ). The partial derivative of f with respect to the i-th component at the point p is:                          f                              x            i                              |        p    =      g    i    ′    (      p    i    )Then the gradient of f at the point p is the vector       ∑    i                            f                              x            i                              |        p          e    i

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