, , .
if it exists, is denoted and called the directional derivative of at , in the direction .
This is obvious from the definitions.
One has
which shows the result whenever . The case when is trivially true.
The last assertion follows from the additivity of the function .
With the notation of Problem 2-4, part (a) of that problem says that exists for all . Now suppose . Then , But .
By Problem 1-26 (a), for all .
Show that is differentiable at 0 but is not continuous at 0.
Clearly, is differentiable at . At , one has since .
For , one has . The first term has limit 0 as approaches 0. But the second term takes on all values between -1 and 1 in every open neighborhood of . So, does not even exist.
Show that is differentiable at (0,0) but that is not continuous at .
The derivative at (0, 0) is the zero linear transformation because , just as in part (a).
However, for where is as in part (a). It follows from the differentiability of , that and are defined for . (The argument given above also shows that they are defined and 0 at .) Further the partials are equal to up to a sign, and so they cannot be continuous at 0.
Proceed as in the proof of Theorem 2-8 for all . In the case, it suffices to note that follows from the definition of . This is all that is needed in the rest of the proof.
Applying Theorem 2-9 to gives . On the other hand, and so . Substituting in these two formulas show the result.
Following the hint, let . Then . On the other hand, Theorem 2-9 gives . So, we have the result with .