Directional derivative and composite functions

Question

We have a function $f(p_0)$ with $p_0 \in \mathbb{R}^n$, and the vector $\vec{v}$ with $||\vec{v}|| = 1$. The derivative of $f(p)$ to $\lambda$ in the point $p = p_0 + \lambda \vec{v}$, calculated for $\lambda = 0$, is called directional derivative along $\vec{v}$ and is indicated with the symbol $$\left( \dfrac{\partial f}{\partial v} \right)_{p_{0}}$$

From the composite function derivation rules, must be

$$\left[ \dfrac{d}{d \lambda} f(p_0 + \lambda \vec{v})\right]_{\lambda = 0} = f'_{x_1}(p_0) \cdot v_1 +... f'_{x_n}(p_0) \cdot v_n$$

Although I have consulted several books, this passage is still not clear to me, especially how the function $f(p_0 + \lambda \vec{v})$ should be a composite function. Can you help me?

Thank you in advance

Answer 1

Well, $f(p)$ is a function of $p$ to begin with, but here $p=p(\lambda)=p_0 + \lambda \vec v$ is a function of $\lambda$. So you get a composite function which in the end depends on $\lambda$. And its derivative is found using the multivariable chain rule.

Answer 2

You can write $f(p)$ as a function of all the components of $p_0$:$$f(p)=f(p_0+\lambda\vec v)=f(p_{0,1}+\lambda v_1,p_{0,2}+\lambda v_2,...,p_{0,n}+\lambda v_n)$$ We now have $n$ components, so $$f'(p_0+\lambda\vec v)=\frac{d(p_{0,1}+\lambda v_1)}{d\lambda}\partial_{p_{0,1}+\lambda v_1}f+...+\frac{d(p_{0,n}+\lambda v_n)}{d\lambda}\partial_{p_{0,n}+\lambda v_n}f\\=f'_{p_{0,1}+\lambda v_1}(p_0+\lambda\vec v)v_1+...+f'_{p_{0,n}+\lambda v_n}(p_0+\lambda\vec v)v_n$$