derivative with respect to $\log(x)$
I have a dynamic equation, $$ \frac{\dot{k}}{k} = s k^{\alpha - 1} + \delta + n$$ Where $\dot{k}/k$ is the capital growth rate as a function of savings $s$, capital $k$, capital depreciation rate $\delta$, and population growth rate $n$.
I have been asked to find the change in the growth rate as $k$ increases. This is of course $$\frac{\partial \dot{k}/k}{\partial k} = (\alpha - 1) s k^{\alpha -2}$$ But what I want to find now is the change in growth rate as $k$ increases proportionately. This should be $$\frac{\partial \dot{k}/k}{\partial \ln(k)} = ?$$ How do you calculate the partial derivative with respect to the logarithm of a variable? I'm sure the answer is simple, but my analytical calculus is pretty rusty.
$\endgroup$3 Answers
$\begingroup$The simplest way is via chain rule: $$ \dfrac{\partial \dot{k}/k}{\partial k} = \dfrac{\partial \dot{k}/k}{\partial \ln{k}}\dfrac{\partial \ln{k}}{\partial k} $$ ...from which you can move things around to get your quantity of interest, e.g. $$\dfrac{\partial \dot{k}/k}{\partial \ln{k}} = \dfrac{\partial \dot{k}/k}{\partial k}k = (\alpha-1)sk^{\alpha-1}$$ (since $(\ln{x})' = 1/x$).
$\endgroup$ 1 $\begingroup$Notation for partial derivatives is inherently awkward. e.g. "growth rate as $k$ increases" doesn't actually make sense: there's implicit context "... while holding $s$, $\alpha$, $n$, and $\delta$ constant".
Differentials become cleaner when notation gets confusing. We have
$$ d \frac{\dot{k}}{k} = (\alpha - 1) s k^{\alpha - 2} dk + k^{\alpha - 1 } ds + s \log k k^{\alpha-1} d\alpha + d\delta + dn$$
Of course, I'm being a bit gratuitious here: we've already decided $ds=d\alpha=d\delta=dn=0$, so I could have just written
$$ d \frac{\dot{k}}{k} = (\alpha - 1) s k^{\alpha - 2} dk $$
Additionally, we have
$$ d \log k = \frac{1}{k} dk $$
What I'm guessing is that you're expected to find the ratio of these two differentials given that we've held all the other variables constant. Since both are multiples of $dk$, and thus should be multiples of each other, it makes sense to write
$$ \frac{d \frac{\dot{k}}{k}}{d \log k} = \frac{(\alpha - 1) s k^{\alpha - 2}}{\frac{1}{k}}$$
when everything is defined. (e.g. if we restrict to $k>0$)
Warning: sometimes people mean something different entirely by similar notation. e.g. if I saw
$$ y = x + \log x $$
and I saw someone asking for
$$ \frac{\partial y}{\partial \log x} $$
I'd put decent odds that they were looking for the answer $1$ -- more precisely, they were expecting you to substitute $\log x = z$ and write
$$ y = x + z $$
and pretend $x$ and $z$ are independent, then take the partial derivative of $y$ with respect to $z$, holding $x$ constant. (and the substitute $z = \log x$ back into the result)
$\endgroup$ 1 $\begingroup$The more formal/modern/natural way to answer this is via a change of variables.
Let $z \equiv \log(k)$ on $\mathbb R_{>0}$, so that pairs $(k, z) $ belong the graph of the function $\log$. Note that, for any differentiable real-valued function $ k \mapsto g(k)$ on $\mathbb R_{>0}$, the function $z \mapsto h(z) = (g \circ \exp)(z)$ satisfies:
$$ g(k) = h(z) \quad \text{for every $k$ in $\mathbb R_{>0}$}. $$Now note that, via the chain rule, the derivative $h'$ of $h$ with respect to $z$ is$$ h'(z) = g'(k) \cdot \exp'(z) = g'(k) \cdot (\exp \circ \log)(k) = g'(k) \cdot k .$$For $g'(k) = (\alpha - 1)\cdot s\cdot k^{\alpha - 2}$, we arrive at the result the OP was after.
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