Derivative notation with an expression instead of a plain dependent variable

Question

I've been trying to implement a set of differential equations from a research paper 1 and I'm stumped by this notation:

$$\dfrac{\mathrm{d}}{\mathrm{d}t}(\log P) = \log\left(\dfrac{S}{D}\right)$$

What does it mean? It's odd to see the dependent variable be an expression. How do I express $\frac{\mathrm{d}P}{\mathrm{d}t}$ using the above equation?

I'm sure that it's obvious to any mathematician, but alas I'm a programmer.

Thanks

Answer 1

On the assumption that $P$, $S$, and $D$ are functions of $t$, we get: $$ \begin{eqnarray} \ln\left( \dfrac{S(t)}{D(t)} \right) &=& \dfrac{\mathrm{d}\ln(P(t))}{\mathrm{d}t} \tag{given} \\ &=& \dfrac{\mathrm{d}\ln(P(t))}{\mathrm{d}P(t)} \cdot \dfrac{\mathrm{d}P(t)}{\mathrm{d}t} \tag{chain rule} \\ &=& \dfrac{1}{P(t)} \cdot \dfrac{\mathrm{d}P(t)}{\mathrm{d}t} \tag{derivative of $\log$} \;, \end{eqnarray} $$ and therefore, $$ \dfrac{\mathrm{d}P(t)}{\mathrm{d}t} = P(t) \cdot \ln\left( \dfrac{S(t)}{D(t)} \right) \;. $$ If the initial condition is given as $P_0 = P(t_0)$, this can be solved by separation of the variables: $$ P(t) = P_0 \cdot \exp\left( \int\limits_{t_0}^{t} \ln\left( \dfrac{S(\tau)}{D(\tau)} \right) \mathrm{d}\tau \right) \;. $$

Answer 2

Simply See Logarithmic Derivative

$$\frac {d}{dt}(\ln P)=(\ln P)'=\frac {P'}{P}=\frac 1 P  \frac {dP}{dt}$$

Answer 3

There is nothing unusual about it. It means the derivative of the function $\log P$ (wrt $t$) is equal to $\log S/D$.

In order to find $dP/dt$ in terms of the other symbols, we use the chain rule (a theorem in calculus that allows us to differentiate finite compositions of functions). Here, we use the chain rule because we have a nested sequence of just two functions, namely $P$ and $\log P$, so that we think of $\log P$ as first taking each $t$ to some unique $P(t)$, then taking each $P(t)$ to some unique $\log P(t)$. You should see where the 'chain' in the name of the method comes from now. I could go into why it works here, but you have not actually asked about that. Since you only want to use it, I shall just state it and apply it to your equation.

Chain Rule: Suppose we have two differentiable functions $f$ and $g$ of a variable $x$ and furthermore another function $k$ such that $k=g\circ f$ (that is, $f$ is composed with $g$), then we have that $${dk\over dx}={dg\over df}{df\over dx}.$$

In your equation, $k=\log P$, so we have that $${d\over dt}\log P={dk\over dP}{dP\over dt}.$$ If you recall that the derivative of $\log y$ is $1/y$, then I'm sure you know how to proceed from here.