# Directional directive vs partial derivative

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 $(0,0)$?

• You can think of partial derivatives as directional derivative in two directions $x$-direction and $y$-direction. – Jacky Chong Oct 17 '16 at 23:21
• I understand that, partial derivatives are just directional derivatives on the axis. But can the existence of partial directives imply the existence of directional derivatives in any direction? Since directional derivatives are composed of partial derivatives. – B.Li Oct 17 '16 at 23:23
• Nope. You can have the existence of partial derivatives but not all directional derivative exists. – Jacky Chong Oct 17 '16 at 23:24
• So how can the above equation be verified? – B.Li Oct 17 '16 at 23:24

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:

Let $f: \mathbb{R}^n \to \mathbb{R}$ and $x^1,...,x^n$ be the standard coordinate functions on $\mathbb{R}^n$. Suppose $f$ is also differentiable about $p$. Then we define the derivative of $f$ in the direction $\vec{v}$ at a point $p$ to be:

$$D_{\vec{v}} f(p) = \lim_{t \to 0} \frac{ f(p+t\vec{v})-f(p)}{t} = \nabla f(p) \cdot \vec{v} = \sum_j v^i\frac{\partial f}{\partial x^j}(p)$$

Observe that if $\vec{v} = \textbf{e}^i = \langle 0,...,x^i=1,...,0\rangle$ then the above definition becomes:

$$D_{\textbf{e}^i}f(p) = \frac{\partial f}{\partial x^i}(p)$$

i.e directional derivatives are a generalization of partial derivatives. If you wish to compute the partials at $(0,0)$ for your function, you will have to proceed by definition.

$$\frac{\partial f}{\partial x}(0,0) = \lim_{t = 0} \frac{f(t,0) - f(0,0)}{t} = \lim_{t \to 0} \frac{f(t,0)-0}{t} = 0$$

$$\hspace{-.4in} \frac{\partial f}{\partial y}(0,0) = \lim_{t = 0} \frac{f(0,t) - f(0,0)}{t} = \lim_{t \to 0} \frac{0-0}{t} = 0$$

• The second equality in the first equation is only true if the function is differentiable. – Jose27 Oct 17 '16 at 23:42
• @Jose27 Everyone knows that we are speaking only of functions which are differentiable. Of course it wouldn't make sense otherwise. – Faraad Armwood Oct 17 '16 at 23:43
• @FaraadArmwood Directionally differentiable is not the same as differentiable, in general. The question only asks about directional differentiability, so differentiability (which is a stronger condition) can't be assumed. – user137731 Oct 17 '16 at 23:44
• @Bye_World of course! differentiability speaks first on a neighborhood of $p$ i.e all directions about $p$. The OP is clearly in calc III. Adding all of the extra rigor is more confusing than helping at this point. – Faraad Armwood Oct 17 '16 at 23:46
• @FaraadArmwood Considering the function the OP gives is not differentiable at the point in question I think it would be misleading NOT to worry about this issue. – Jose27 Oct 18 '16 at 0:03

Hint: To check whether all directional derivatives at $(0,0)$ exist, all we need to is answer (i) Does $x\to f(x,mx)$ have a one-variable derivative at $x=0$ for any $m\in \mathbb R$? (ii) Does $y\to f(0,y)$ have a one-variable derivative at $y=0?$ These are not to hard to answer here; give it a shot.

Let me add to the discussion. Consider the following example \begin{align} f(x,y) = \begin{cases} \frac{x^2-y^2}{x^2+y^2} &\ \text{ if } \ \ (x,y) \neq (0, 0)\\ 0 & \ \text{ if } \ \ (x, y) = (0 , 0) \end{cases} \end{align} which is not continuous at $(0, 0)$.

It's doesn't have partial derivatives at $(0, 0)$ since \begin{align} \lim_{h\rightarrow 0} \frac{f(h, 0)-f(0, 0)}{h} = \lim_{h\rightarrow 0}\frac{1}{h} \end{align} which doesn't exists and likewise \begin{align} \lim_{k\rightarrow 0} \frac{f(0, k)-f(0, 0)}{k} = \lim_{k\rightarrow 0} \frac{-1}{k}. \end{align} Nevertheless, if we look at the direction $x=y$, we have \begin{align} \lim_{t\rightarrow 0} \frac{f(t, t)-f(0, 0)}{t}= 0 \end{align} that is $f$ has a directional derivative in the $x=y$ direction and the directional derivative is definitely not given by \begin{align} \nabla f(0,0)\cdot (1/\sqrt{2}, 1/\sqrt{2}). \end{align}

Here's a hint for your problem. Consider your given function, then \begin{align} \lim_{t\rightarrow 0} \frac{f(tv, tk)-f(0, 0)}{t} \end{align} where $\|(v, k)\|=1$ and $v\neq 0$. Explicitly we have \begin{align} \lim_{t\rightarrow 0} \frac{1}{t}\sin\left( \frac{t^2k^2}{tv}\right)\sqrt{(tv)^2+(tk)^2}. \end{align}