Gradient interpretation Let
$$k=Ae^{-\frac{x}{RT}}$$
It is always the case that $A, x, R, T > 0$. Now we can note that if $k_1 := k(T=T_1), k_2 := k(T=T_2)$, then
$$\frac{k_2}{k_1} = e^{\frac{x}{R}\big(\frac{1}{T_1}-\frac{1}{T_2}\big)}$$
This tells us that $k$ increases if $T$ increases (proof: $T_2>T_1 \rightarrow 1/{T_1}-1/{T_2}>0$, hence $\frac{x}{R} (1/{T_1}-1/{T_2}) > 0$, hence $k_2/k_1>1$). It also tells us that the amount by which $k$ increases, for a given increase in $T$, is greater as $x$ is greater.
My question is this: I want to prove the same by a perhaps more natural route, differentiation. First step works fine:
$$\frac{dk}{dT} = \frac{Ax}{RT^2}e^{-\frac{x}{RT}}$$
where we can see $\frac{dk}{dT}>0$, i.e. $k$ increases as $T$ increases. But if we differentiate now with respect to $x$ we get a different conclusion regarding the dependency on $x$:
$$\frac{d^2k}{dT dx} = \frac{A}{RT^2}e^{-\frac{x}{RT}} \Big(1-\frac{x}{RT} \Big)$$
according to which $\frac{dk}{dT}>0$ does not necessarily increase with increasing $x$ as expected from $k_2/k_1$. How should I interpret this differential $\frac{d^2k}{dT dx}$ and its signs?
 A: First of all it is probably a bit more rigorous to use partial derivatives for this problem. Keep in mind though that this is a second order derivative and not first order derivative. You are no longer tracking the rate of change of $k$ but instead tracking the rate of change of $\frac{\partial k}{\partial T}$. When $\frac{\partial^2 k}{\partial x \partial T} > 0$ then  $\frac{\partial k}{\partial T}$ is increasing. When $\frac{\partial^2 k}{\partial x \partial T} < 0$ then  $\frac{\partial k}{\partial T}$ is decreasing.
Also as a side note when we write mixed partials of a function the first derivative is written last so that
$$\frac{\partial^2 k}{\partial x \partial T} = \frac{\partial^2 }{\partial x \partial T} [k] = \frac{\partial }{\partial x} \bigg[\frac{\partial }{\partial T} [k] \bigg].$$
That being said, I am being slightly pedantic as we do have equality of mixed partials in this case so $$\frac{\partial^2 k}{\partial x \partial T}=\frac{\partial^2 k}{\partial T \partial x}. $$
This is not always the case however and you can read about that more here.
The factor by which $k$ increases when $T_1<T_2$ is fixed will increase with $x$ but that is not what the second derivative is tracking. It tracks the actual rate of this change (not the factor). Notice what happens as $x$ gets large. $k_2$ gets small but $k_1$ gets much smaller! Geometrically this would mean that the derivative doesn't have to be as large to let $k$ increase from $k_1$ to $k_2$ in a given interval $[T_1, T_2].$
Giving a tangible example, if $\Delta T = 1$ then an increase from $k_1=.0001$ to $k_2=.1$ yields ${k_2}/{k_1} =  1000$ whereas $\frac{dk}{dT} \approx .1$ (which is clearly much smaller). In fact, from this example it should be clear that as $x$ gets large, $\frac{dk}{dT}$ should get very small $\big($ in other words $\frac{\partial^2 k}{\partial x \partial T} < 0\big)$ even though ${k_2}/{k_1}$ increases.
The sign change of $\frac{\partial^2 k}{\partial x \partial T}$ gives conditions for which $\frac{dk}{dT}$ is increasing, decreasing, or neither. We can go as far as to say that $\frac{dk}{dT}$ switches from increasing to decreasing when $$\frac{x}{RT}=1.$$
What you want to show is actually that $\frac{d}{dx}\bigg[\frac{k_2}{k_1}\bigg] > 0.$ Showing this means that the factor by which $k_2$ is larger than $k_1$ increases as $x$ increases.
