# Directional Derivative of functions $f:S\rightarrow\mathbb{R}$

Suppose $S\subseteq\mathbb{R}^3$ is a regular surface and let $f:S\rightarrow \mathbb{R}$ is differentiable. I know that a vector $v\in T_pS$ is of the form $v=\alpha'(0)$, where $\alpha:]-\epsilon,\epsilon[\rightarrow S$ is such that $\alpha(0)=p$. Now we have defined the directional derivative of f with direction v at point p as $Df_p(v)=(f\circ\alpha)'(0)$.

My goal is to prove that $Df_p$ is linear. And this should be easy but I don't know how to it.

Let $v,w\in T_pS$ and choose $\alpha,\beta$ curves such that $\alpha'(0)=v$, $\beta'(0)=w$. I want to prove that $D_{v+w}f(p)=D_vf(p)+D_wf(p)$, right? Should I fix $\gamma$ such that $\gamma'(0)=v+w$? I don't seem to go anywhere.

Thank you very much!

## 1 Answer

Yes, this sort of abstract notation can be confusing sometimes. Let me write your expression for the directional derivative in coordinates, expanding it out using the chain rule: $$Df_p (v) = \frac{d(f \circ \alpha)} {dt} (0)= \frac{\partial f}{\partial x}(p) \frac{d \alpha_x}{dt}(0) + \frac{\partial f}{\partial y}(p) \frac{d \alpha_y}{dt}(0) + \frac{\partial f}{\partial z}(p) \frac{d \alpha_z}{dt}(0)$$ $$\ \ \ \ = \frac{\partial f}{\partial x}(p) v_x + \frac{\partial f}{\partial y}(p) v_y + \frac{\partial f}{\partial z}(p)v_z.$$ Similarly, replacing $v$ with $w$ and replacing $\alpha$ with $\beta$, we get $$Df_p (w) = \frac{\partial f}{\partial x}(p) w_x + \frac{\partial f}{\partial y}(p) w_y + \frac{\partial f}{\partial z}(p)w_z.$$ Finally, replacing $v$ with $v+ w$ and replacing $\alpha$ with $\gamma$, we get $$Df_p (v+w) = \frac{\partial f}{\partial x}(p) (v_x+ w_x) + \frac{\partial f}{\partial y}(p) (v_y + w_y) + \frac{\partial f}{\partial z}(p) (v_z + w_z).$$

Now that we have written all of this out, it should be clear that $$Df_p (v+w) = Df_p(v) + Df_p(w).$$ which is the same as saying that $Df_p$ acts linearly on tangent vectors in $T_p$.