# Why does $\frac{\partial F}{\partial x_i}(\pmb{a}) = \lim_{h\rightarrow 0} \frac{F(\pmb{a}+h\pmb{e}_i)-F(\pmb{a})}{h}$ gives a vector?

I am reading about partial derivatives in some notes on Multivariable Calculus. On page 2, it says:

Definition: Partial Derivatives

Let $$U \subset \mathbb R^m$$ be open, let $$F: U \to \mathbb R^m$$, and let $$\mathbf a \in U$$. If $$i \in \{1, \ldots , m\}$$, the partial derivative of $$F$$ with respect to $$x_i$$ at $$\mathbf a$$ is defined as follows: $$\frac{\partial F}{\partial x_i}(\mathbf a) = \lim_{h \to 0} \frac{F(\mathbf a + h\mathbf e_i) - F(\mathbf a)}{h}.$$ Here $$\mathbf e_i$$ denotes a unit vector in the xi direction.

Then, later it says:

As we have defined it, the partial derivative $$(\partial F)/(\partial x_i)(\mathbf a)$$ is a vector. Specifically, if $$F(\mathbf x) = (f_1(\mathbf x), \ldots, f_n(\mathbf x))$$, then $$\frac{\partial F}{\partial x_i}(\mathbf a) = \left(\frac{\partial f_1}{\partial x_i}(\mathbf a), \ldots, \frac{\partial f_n}{\partial x_i}(\mathbf a)\right)$$

I don't see how the limit expression can be a vector. Based on my observation, the numerator will be a $$n\times 1$$ and the denominator will be a $$m \times 1$$ vector. Therefore we have division by vectors which is not defined.

Can I have a step by step explanation of how the limit expression results the vector described in point 2? I've taken a course on multi variable calculus but have not encountered taking the limit of vectors.

• How could it not be a vector? The function $F$ maps to $\mathbb R^n$. The denominator is a scalar $h$. Nov 7 '14 at 13:33
• Yes, you are right. I did not realize that $h$ is a scalar. So the expression in the limit indeed results in a vector. But I still don't see how it will be equal to the vector described in point 2. Could you please show me how to do this algebraically? Nov 7 '14 at 13:50

As mentioned in the comments to the question, $\dfrac {\partial F}{\partial x_i}(\mathbf {a})\color{grey}{=\lim \limits_{h\to 0}\left[\dfrac{F(\mathbf a+h\mathbf e_i)-F(\mathbf a)}h\right]}$ is a vector simply because $F$ is a vector-valued function.
Presumably $F=(f_1, \ldots ,f_n)$ where $f_1, \ldots ,f_n$ are scalar functions defined in $U$.
Given $\mathbf {a}=(x_1, \ldots , x_m)$ one gets for all $i\in \{1, \ldots ,m\}$
\begin{align} \dfrac {\partial F}{\partial x_i}(\mathbf {a})&=\lim \limits_{h\to 0}\left[\dfrac{F(\mathbf a+h\mathbf e_i)-F(\mathbf a)}h\right]\\ &=\lim \limits_{h\to 0}\left[\dfrac{(f_1(\mathbf a+h\mathbf e_i), \ldots,f_n(\mathbf a+h\mathbf e_i))-(f_1(\mathbf a), \ldots ,f_n\left(\mathbf a)\right)}{h}\right]\\ &=\lim \limits_{h\to 0}\left[\dfrac {(f_1(\mathbf a+h\mathbf e_i)-f_1(\mathbf a), \ldots , f_n(\mathbf a+h\mathbf e_i)-f_n(\mathbf a))}h\right]\\ &=\lim \limits_{h\to 0}\left[\left(\dfrac{f_1(\mathbf a+h\mathbf e_i)}h, \ldots ,\dfrac{f_n(\mathbf a+h\mathbf e_i)}h\right)\right]\\ &=\left(\lim \limits_{h\to 0}\left[\dfrac{f_1(\mathbf a+h\mathbf e_i)}h\right], \ldots ,\lim \limits_{h\to 0}\left[\dfrac{f_n(\mathbf a+h\mathbf e_i)}h\right]\right)\\ &=\left(\dfrac{\partial f_1}{\partial x_i}(\mathbf a), \ldots, \dfrac{\partial f_n}{\partial x_i}(\mathbf a)\right)_. \end{align}
A vector space is invariant under both: $$(v_1, v_1)\to v_1 - v_2\\ (\alpha,v)\to \alpha v$$ for scalar value $\alpha$. So the increments $\frac{F(a+hv)-F(a)}{h}$ are vectors as well, when $F$ is vector-valued.