# Can a time derivative be taken?

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

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

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

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