I'm working on solving for $t$ in the expression $$\ln t=3\left(1-\frac{1}{t}\right)$$ and although I can easily tell by inspection and by graphing that $t=1$, I'd like to prove it more rigorously.

I got stuck trying to solve this algebraically, so I tried to take the derivative of each side with respect to $t$ to get


However, this implies that $t=3$, which is incorrect.

Why can't I take the derivative of each side like this? What am I doing wrong or misunderstanding?

  • $\begingroup$ You have your answer to the question you asked, I have also done some analysis of how you would find the values for $t$ in this equation. Note that there is more than one solution for $t$. $\endgroup$
    – abiessu
    Oct 17, 2013 at 3:53

3 Answers 3


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 $t$.

  • 2
    $\begingroup$ Thanks! That's the kind of explanation I was looking for. $\endgroup$
    – wchargin
    Oct 16, 2013 at 23:40
  • $\begingroup$ What about doing differentiation on something like $x^2 + y^2 = 4$. Doesn't that act as an implicit condition since it only works for certain values of x,y $\endgroup$
    – 1110101001
    Mar 14, 2017 at 2:01
  • 1
    $\begingroup$ @1110101001: By "some real $t$", I meant at isolated points. If you treat the two-variable equation $x^{2} + y^{2} = 4$ as a condition defining $y$ as a function of $x$ with no further qualification, you're effectively assuming $y = f(x)$ for some differentiable function $f$ defined in the open interval $(-2, 2)$. $\endgroup$ Mar 14, 2017 at 13:00
  • $\begingroup$ For those who want to know the definition of identity equations and conditional equations, see this. $\endgroup$ Aug 2, 2020 at 12:01

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.

  • $\begingroup$ So the $t$s are not the same, in other words? $\endgroup$
    – wchargin
    Oct 16, 2013 at 22:01
  • 1
    $\begingroup$ $t$ is a variable, and finding a $t$ so that $f(t) = g(t)$ and finding a $t$ so that $f'(t) = g'(t)$ are different, and generally unrelated problems. $\endgroup$
    – user82196
    Oct 16, 2013 at 22:04

The derivative of a function measures the gradient of the function at various points, and this does not give you the solution for $t$ of the original. We can use the method of differentiation to analyze the equation, however:

$$\text{Let }f(t)=\ln t-3+\frac 3t$$

$$\text{Then }f'(t)=\frac 1t-\frac 3{t^2}$$

We can see that $f'(3)=0,\forall t\in (0,3),f(t)$ is decreasing, and $\forall t\in (3,+\infty),f(t)$ is increasing, so $t=3$ is a global minimum for the valid range of $t\in \mathbb R^+.$

Continuing the analysis:

$$\ln t=3\left(1-\frac 1t\right)\implies t={e^3\over e^{\frac 3t}}\implies e^{\frac 3t}={e^3\over t}$$

which is a nice near-symmetric equation. Observation now immediately shows that $t=1$ is a solution, but the other solution starts at the beginning again:

$$\ln t=3\left(1-\frac 1t\right)\implies \ln t^{\frac 13}=1-\frac 1t\le \ln e$$

$$\implies t^{\frac 13}\le e\implies t\le e^3\text{ and } \ln e^3\gt 3\left(1-\frac 1{e^3}\right)$$

$$\text{Also, }\ln (e^3-e^2)=2+\ln (e-1)\lt 3-\frac 3{e^3-e^2}$$

With this, we know that the other solution is for some $t\in (e^3-e^2,e^3).$ Graphing shows the actual value at $t\approx 16.8,$ and it is likely that further analysis could come up with a decent closed expression for this value.

  • $\begingroup$ Yes, I tried that method but to no avail. $\endgroup$
    – wchargin
    Oct 16, 2013 at 22:21
  • $\begingroup$ Are you allowed to use Newton's method ? If yes, you have all required elements (initial value : 16.8, function and derivative definitions). So,j you will solve in few iterations to very hich accuracy. $\endgroup$ Oct 17, 2013 at 5:49
  • $\begingroup$ @ClaudeLeibovici: Newton's method is a good suggestion for arriving at a good approximation. Since I'm doing this by hand for the benefit of the OP (and since my graphing program can give me lots of significant digits very quickly as well), I'm not so concerned about getting a high-accuracy value as I am about offering some avenues to narrow down where the solution might be. $\endgroup$
    – abiessu
    Oct 17, 2013 at 12:48
  • 1
    $\begingroup$ There is a Lambert W closed form for the $16.8…$ root. $\endgroup$ Dec 23, 2022 at 13:25

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