Question on directional derivative for vector-valued functions

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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 $\;\frac{\partial f}{\partial v}=(\nabla f)^T \cdot v\;$?

Thanks in advance!