Partial derivative involving ln()

I'm looking at a paper involving a constrained optimisation problem using Lagrange multipliers, in which the following Lagrangian function appears:

$\gamma = \max \sum_{i=1}^nr_i(1-e^{-a_ix_i})+\lambda(1-\sum_{i=1}^{n}x_i)$

For the partial derivative of $\gamma$ with respect to $x_1$, the authors write:

$\frac{\partial\gamma}{\partial x_1} = a_1x_1+ \ln(\lambda)-\ln(r_1a_1)$

As a thoroughgoing newbie to all this, I thought that the partial derivative with respect to $x_1$ would be:

$\frac{\partial\gamma}{\partial x_1} = -a_1r_1e^{-a_1x_1} +\lambda$

...because I thought the partial derivative with respect to $x$ of $e^{kx}$ is $k\cdot e^{kx}$

Clearly, I'm missing something pretty fundamental (!), so I'd be really grateful if anyone could explain how the partial derivative above appears! Thanks in advance!

• You are totally correct (may be, just a sign error we don't care about since the expression will be set equal to zero). I really wonder how they made the loarithm appearing ! Typo's, once more I bet. – Claude Leibovici Dec 27 '15 at 13:14
• You are correct and the authors are not just wrong, but horribly wrong (you have the wrong signs, though). Assuming that you understood correctly their article, I suggest not reading it anymore. – Alex M. Dec 27 '15 at 13:16

But let $$F=\sum _{i=1}^n r_i \left(1-e^{-a_i x_i}\right)+\lambda \left(1-\sum _{i=1}^n x_i\right)$$ which gives $$\frac{dF}{dx_i}=a_i r_i e^{-a_i x_i}-\lambda$$ what you properly found. But they now set the derivative equal equal to $0$. Then, and only then, $$a_i r_i e^{-a_i x_i}=\lambda$$ Now, take the logarithms. But this does not represent the derivative of anything.