Differentiating both sides of a non-differential equation

Solution 1:

To answer the general question of "Why can't [one] differentiate each side [of an equation]?": Your original equation, of the form $f(t) = g(t)$, acts as a condition (i.e., is only true for some real $t$, in this case finitely many), not as an identity (true for all $t$ in some open interval).

When you differentiate a function $f$ at one point $a$, you implicitly use the values of $f$ in some neighborhood of $a$. Since your $f$ and $g$ are not equal in any open neighborhood, you can't expect differentiating to yield a new true condition.

In case an example clarifies, take $f(t) = t$ and $g(t) = 0$. The equation $t = 0$ certainly has a solution, but differentiating both sides gives $1 = 0$.

By contrast, it's safe to differentiate both sides of, e.g., $\cos(2t) = \cos^2 t - \sin^2 t$, since this equation is true for all real $t$.

Solution 2:

Let $f(t) = \ln t - 3 (1 - 1/t)$. You're saying that $f(1)= 0$ and $f'(3) = 0$; there is no problem here. The point is that when you try to find a value of $t$ that satisfies the first equation, you're looking for a root of $f$, while looking for a value of $t$ that satisfies the second equation is the same as finding a root of $f'$. In general, these are not the same and you have no reason to expect that they'll give you the same answer.

I'm not sure that there are easier ways to solve this than by inspection, though.