I was using This Wolfram Alpha Page to calculate the partial derivatives of the sum of two gaussians. It believes the answer to be: 
$$
e^{-(b-x)^2/2 c^2}
$$
but the working shows it as ending on:
$$
e^{-(x-b)^2/2 c^2}
$$
Which is what I had thought the answer was.

Is this a bug in Wolfram's engine or have I missed an obvious step?

The $\max$-central limit theorem (Fisher-Tippet-Gnedenko theorem) can be used to provide a decent approximation when $n$ is large. See this example at reference page for extreme value distribution in Mathematica.
The $\max$-central limit theorem states that $F_\max(x) = \left(\Phi(x)\right)^n \approx F_{\text{EV}}\left(\frac{x-\mu_n}{\sigma_n}\right)$, where $F_{EV} = \exp(-\exp(-x))$ is the cumulative distribution function for the extreme value distribution, and
$$
   \mu_n = \Phi^{-1}\left(1-\frac{1}{n} \right) \qquad \qquad
   \sigma_n = \Phi^{-1}\left(1-\frac{1}{n} \cdot \mathrm{e}^{-1}\right)- \Phi^{-1}\left(1-\frac{1}{n} \right)
$$
Here $\Phi^{-1}(q)$ denotes the inverse cdf of the standard normal distribution.
The mean of the maximum of the size $n$ normal sample, for large $n$, is well approximated by
$$ \begin{eqnarray}
  m_n &=&  \sqrt{2} \left((\gamma -1) \Phi^{-1}\left(2-\frac{2}{n}\right)-\gamma  \Phi^{-1}\left(2-\frac{2}{e n}\right)\right) \\ &=& \sqrt{\log \left(\frac{n^2}{2 \pi  \log \left(\frac{n^2}{2\pi} \right)}\right)} \cdot \left(1 + \frac{\gamma}{\log (n)} + \mathcal{o} \left(\frac{1}{\log (n)} \right) \right) 
\end{eqnarray}$$
where $\gamma$ is the Euler-Mascheroni constant.
