Let $f(t)$ and $g(t)$ both be smooth functions of time and $dh(t)/dt = af(t) - bh(t)$. At the turning point in $h(t)$ we have

$$ af(t) - bh(t) = 0 $$

Can I re-arrange this and take the time derivative of both sides as follows?

$$ \frac{a}{b}\frac{df(t)}{dt} = \frac{dh(t)}{dt} = 0 $$

thus generating a condition for the turning point of $h$ to be

$$ \frac{df(t)}{dt} $$

  • $\begingroup$ That condition is only true for a critical (turning) point of $h$, so the derivative of the condition may not be $0$, $\endgroup$
    – Keshav
    May 3, 2020 at 15:39

1 Answer 1


There is an inconsistency in how you are treating your symbols between these two equations:

$$h'(t) = af(t) - bh(t)$$ $$0 = af(t) - bh(t)$$

Note that in the first, you establish that the image of a particular value $t$ under the function $h'$, is equal to some algebraic combination of the images of this particular $t$ under two different functions, $f$ and $h$.

In the second equation, you continue to treat $f$ and $h$ as functions. However, since you have asserted that $h'(t) = 0$, we are now saying that for some particular value of $t$ in the domain, call it $t_0$, that $$h'(t_0)= 0 = af(t_0) - bh(t_0)$$

Rewriting it this way we can see that the RHS is reduced to a value. It is the image of $t_0$ under the function $h'$.

In other words, $\frac{a}{b} \frac{d}{dt} f(t_0) = \frac{d}{dt} h(t_0) = 0$. This is not the same as the general conclusion $\frac{a}{b} \frac{d}{dt} f(t) = \frac{d}{dt} h(t) = 0$. In the former we are taking the derivative with respect to a value, in the latter it is the derivative with respect to a function.


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