# Question on directional derivative for vector-valued functions

Let $\;f:\mathbb R^n\rightarrow \mathbb R^m\;$. The gradient matrix of $\;f\;$ is :

$\;\nabla f=\begin{pmatrix} \frac{\partial f_1}{\partial x_1} \dots \frac{\partial f_1}{\partial x_n}\\ \;\dots\;\\ \frac{\partial f_m}{\partial x_1} \dots \frac{\partial f_m}{\partial x_n}\\ \end{pmatrix}\;$

And $\;v=(v^1,\dots,v^n)\;$ is the unit normal vector field.

I know $\;\frac{\partial f}{\partial v}= \nabla f \cdot v\;$.

Question:

If I take the transpose matrix $\;(\nabla f)^T\;$, then would it be true to claim $\;\frac{\partial f}{\partial v}=(\nabla f)^T \cdot v\;$?

• Your first identity $\frac{\partial f}{\partial v} = v \cdot \nabla f$ doesn't make sense because $v$ is a $1 \times n$ vector while $\nabla f$ is a $m \times n$ matrix. Do you think of elements of $\mathbb{R}^n / \mathbb{R}^m$ as row vectors or column vectors? Is the gradient of a function $f \colon \mathbb{R}^n \rightarrow \mathbb{R}$ a row vector or a column vector for you? – levap Aug 24 '17 at 13:23
• Again, this doesn't make sense. If you want the result $\frac{\partial f}{\partial v}$ to be a row vector, the correct identity is $\frac{\partial f}{\partial v} = (\nabla f \cdot v^T)^T = v \cdot (\nabla f)^T.$ – levap Aug 24 '17 at 14:13
• @levap I'm sorry, multiplication of matrices with vectors is very confusing to me! I want $\;\frac{\partial f}{\partial v}\;$ to be a column vector... – kaithkolesidou Aug 24 '17 at 14:18
• Then the correct equation is $\nabla f \cdot v^T$. – levap Aug 24 '17 at 14:30