# How to find the directional derivative at every point?

Given a function $$f: \mathbb{R}^2 \to \mathbb{R}$$ defined by $$f(x,y) = \begin{cases} \left( x^2 + y^2 \right) \cos \left(\frac{1}{x^2 + y^2} \right) & \text { if } (x,y) \neq (0,0)\\ 0 & \text{ if } (x,y) = (0,0) \end{cases}$$ compute the directional derivatives of $$f$$ at every point.

My solution:

$$(x^2 + y^2)cos(\frac{1}{x^2 + y^2})$$ at $$(x,y) = (p,p)$$ in the direction $$\vec{v}=(\vec{v_1},\vec{v_2})$$.

The partial derivatives are:

$$\frac{(2p^2)cos(\frac{1}{2p^2})+sin(\frac{1}{2p^2})}{p}$$ and $$\frac{(2p^2)cos(\frac{1}{2p^2})+sin(\frac{1}{2p^2})}{p}$$

The $$\vec{v}$$ becomes: $$\frac{\vec{v_1}}{\sqrt{\vec{v_1^2} + \vec{v_2^2}}}$$ and $$\frac{\vec{v_2}}{\sqrt{\vec{v_1^2} + \vec{v_2^2}}}$$

Then the product of the partial derivatives and the vectors is:

$$D_\vec{v}f(p,p) = \frac{(v_1 + v_2)2p^2cos(\frac{1}{2p^2})+sin(\frac{1}{2p^2})}{p\sqrt{v_1^2 + v_2^2}}$$

Am I correct?

• $\vec{v_1}$ and $\vec{v_2}$ are components of a vector; they are scalars. It does not make sense to put arrows above the scalars.
– user1046533
Commented May 20, 2022 at 16:15
• @user1046533 Thank you!
– EUEU
Commented May 20, 2022 at 16:33

You should not set $$(x,y)=(p,p)$$: for an arbitrary point on the plane $$\mathbf{R}^2$$, its $$x$$ and $$y$$ coordinates are not necessarily the same.
For $$(x,y)\ne (0,0)$$, you simply apply the formula $$D_vf(x,y)=\nabla f(x,y)\cdot v=f_x(x,y)v_1+f_y(x,y)v_2.$$ where $$v=(v_1,v_2)$$ is a unit vector.
For $$(x,y)= (0,0)$$, work with the limit definition of partial derivatives.