[Math] Directional directive vs partial derivative

Question

Can someone clarify the relationship between directional derivative and partial derivatives for a function from $\mathbb{R}^n$ to $\mathbb{R}$?

To my understanding, if the function is continuously differentiable, then both directional and partial derivatives exist. Is that correct?

Consider this function:
\begin{align}
f(x,y) =
\begin{cases}
\sin( \frac{y^2}{x})\sqrt{x^2 + y^2}& x \ne 0 \\
0 & x = 0
\end{cases}
\end{align}
How would one verify this function has direction derivatives at $(0,0)$?

Answer 1

To be short:

• When defining partial derivatives of a function, one needs to choose bases for both the domain and the range of the function and thus the definition is not coordinate free. (Contrasting to defining the derivative of a function.) See the definition below in Rudin's book (your case is $m=1$).
• Defining directional derivatives, on the other hand, is coordinate free.

• One could say partial derivatives are special directional derivatives. Assuming existence, one can use partial derivatives to calculate directional derivativs.

• Suppose $E$ is an open subset of $\mathbb{R}^n$. If $f:E\to\mathbb{R}^m$ is continuously differentiable, then all the partial derivatives exist and are continuous on $E$. The converse is also true. See for instance Theorem 9.21 in Rudin's Principle of Mathematical Analysis.

• Regarding your example: \begin{align} f(x,y) = \begin{cases} \sin( \frac{y^2}{x})\sqrt{x^2 + y^2}& x \ne 0 \\ 0 & x = 0 \end{cases} \end{align} "How would one verify this function has direction derivatives at $(0,0)$?"

  One needs to specify the direction. For instance in the direction of $u=\frac{1}{\sqrt2}(1,1)$, one has $$ f(x,x)=\sqrt 2 x\sin x,\quad x\geq 0. $$ Thus the functional derivative at $(0,0), $in the direction $u$ exists. You can similarly check other directions by letting $y=mx$ and $x=ny$.

Rudin explains clearly in his book partial derivatives, directional derivatives and their relation: