Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I found something strange when I try to solve this equatiin of $x$:

$\int_0^t \frac{1}{xW(\frac{1}{xf(\tau)})}d\tau=c_0t$,

where $t$ and $c_0$ are constants. $f(\tau)$ is a known polynomial function. $W(z)$ is the Lambert W function, i.e. $z=W(z)e^{W(z)}$.

If I take the derivative of the equation on both sides, I can get some solution of $x$. However, the result seems wrong since it require for any $x\in (0,t)$, $\frac{1}{xW(\frac{1}{xf(\tau)})}=c_0$ holds. And this is not what I want.

In brief, I have the following two questions:

1, What is wrong with taking the derivative of the equation?

2, Is there any other way to solve $x$?

share|cite|improve this question
To answer 1): If $t$ is a constant, then you can't differentiate and expect to get a meaningful result. For a much simpler example, imagine that you were looking at $\int_0^t xy dy = c_0 t$, with $c_0=1$ and $t=2$. Then the equation solves to $x=1$, but differentiating both sides gets you $xy=c_0$ which is obviously garbage. Differentiating only works if you expect the equation to hold as a function of $t$. – Steven Stadnicki Nov 7 '12 at 3:29
Great! Differentiating only works if you expect the equation to hold as a function of $t$. I think this is the key point! Thanks. – Severals-user45972 Nov 7 '12 at 5:55
Take $\int_0^t x ydy=t$ as an example. The solution can be derived as $x=\frac{2}{x}$. If I consider $x$ as $x(t)$ and take derivative on the equation as $x'(t)\frac{t^2}{2}+x(t)t=1$. Then I can get $x(t)=\frac{C_1}{t^2}+\frac{2}{t}$. The only difficult is to determine $C_1$. – Severals-user45972 Nov 7 '12 at 6:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.