Understanding Equation (3.13) from Bishop's Pattern Recognition and Machine Learning I'm having trouble deriving the mentioned equation from the text. Specifically, I'm having trouble obtaining the $\mathbf{\phi}(x_n)^T$ term. I checked the errata and while the lack of the $\beta$  parameter is mentioned there is nothing about this term.
Given that
$$
\ln\left(p(\mathbf{t}|\mathbf{w}, \beta)\right) 
= \frac{N}{2} \ln(\beta) 
- \frac{N}{2} \ln(2 \pi)
- \beta \frac{1}{2} \sum\limits_{i=1}^{N} 
\left(t_i - \mathbf{w}^T \mathbf{\phi}(\mathbf{x}_i) \right)^2.
$$
The author states that
$$
\nabla \ln(p(\mathbf{t}|\mathbf{w}, \beta) )
= \beta \sum\limits_{i=1}^{N} 
\left(t_i - \mathbf{w}^T \mathbf{\phi}(\mathbf{x}_i) \right)
\mathbf{\phi}(\mathbf{x}_i)^T,
$$
where $\nabla=(\partial w_0, \dots, \partial w_M)^T$,
$\mathbf{\phi}(\mathbf{x}_i) = (\phi_0(\mathbf{x}_i), \dots, \phi_M(\mathbf{x}_i))^T$ and $\mathbf{w}=(w_0, \dots, w_M)^T$.
However, when I
expand the inner product and take the gradient I obtain
$$
\begin{aligned}
\nabla \ln(p(\mathbf{t}|\mathbf{w}, \beta) 
&= -\beta \frac{1}{2} \sum\limits_{i=1}^{N} 
\nabla \left(t_i - \mathbf{w}^T \mathbf{\phi}(\mathbf{x}_i) \right)^2  
\\
&= -\beta \frac{1}{2} \sum\limits_{i=1}^{N} \nabla
\left(t_i - w_0 \phi_0 - \dots - w_M \phi_M \right)^2 
\\
&= -\beta \sum\limits_{i=1}^{N} 
\left(t_i - w_0 \phi_0 - \dots - w_M \phi_M \right) 
\left(-\phi_0, -\phi_1, \dots, -\phi_M \right)^T \\
&= \beta \sum\limits_{i=1}^{N} 
\left(t_i - \mathbf{w}^T \mathbf{\phi}(\mathbf{x}_i) \right) 
\mathbf{\phi}(\mathbf{x}_i) 
\end{aligned}
$$
Can somebody please illustrate where the error comes from?

If denominator layout is used while taking derivatives, then $\mathbf \phi(x)$ is correct; if numerator layout is used, $\phi(x)^T$ is correct. And, the author doesn't specify which one is used, but as it appears, it implicitly chooses the numerator layout. So, there is no right or wrong answer. Same follow-up equations can be derived by manipulating both expressions.
For example, the book continues with equating the gradient to zero and expanding the multiplication as follows:
$$0
= \sum\limits_{i=1}^{N} 
\left(t_i - \mathbf{w}^T \mathbf{\phi}(\mathbf{x}_i) \right)
\mathbf{\phi}(\mathbf{x}_i)^T \rightarrow0=\sum_i t_i\phi(\mathbf x_i)^T-\mathbf{w}^T\sum_i (\phi(\mathbf x_i)\phi(\mathbf x_i)^T)$$
For the other version, you can have
$$0
= \sum\limits_{i=1}^{N} 
\left(t_i - \mathbf{w}^T \mathbf{\phi}(\mathbf{x}_i) \right)
\mathbf{\phi}(\mathbf{x}_i)=\sum\limits_{i=1}^{N} 
\mathbf{\phi}(\mathbf{x}_i)\left(t_i -  \mathbf{\phi}(\mathbf{x}_i)^T\mathbf{w} \right)\\\rightarrow0=\sum_i t_i\phi(\mathbf x_i)-\left(\sum_i\phi(\mathbf x_i)\phi^T(\mathbf x_i)\right)\mathbf{w}$$
which is just the transpose of the first one.
