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I made this problem:

$f(x)=e^{f^{\prime \prime}}$

I have just been taught the first derivative, and was thinking about what if the derivative depended upon it own derivative. I understand that $e^x$ is its "own" derivative, but the problem I made I was thinking that the first derviative is not logical, because to know the first derivative you then must know the 2nd or 3rd derviative, it seems self-referenecing.

Is the problem I made, a real problem or just some abstract idea?

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  • $\begingroup$ Do you mean to solve for $f$ since you have $f$ on both sides? $\endgroup$
    – JACKY88
    Oct 29, 2012 at 1:20
  • $\begingroup$ I just edited the question to be more clear. I hope its more clear now. $\endgroup$
    – yiyi
    Oct 29, 2012 at 1:22

2 Answers 2

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Such kind of problems usually ask for the function $f$ itself (of course, then its derivative can be calculated, too).

And, such is called a Differential equation.

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  • $\begingroup$ Thanks, I do not know many math terms, so it is polite when I am told math terms. $\endgroup$
    – yiyi
    Oct 29, 2012 at 1:40
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    $\begingroup$ Specifically, it's a second-order, nonlinear, ordinary differential equation. In general, it is not always possible to solve these (solve means find $f(x)$ such that $f(x)=e^{f''(x)}$). I thought solving this would clarify your question, but I could only get so far. If $y=f(x)$ then I got $y'=2ylny-2y-C_1$ but that's where I got stuck. Maybe wiser minds than mine can finish it (or do it right from the beginning). $\endgroup$
    – Todd Wilcox
    Oct 29, 2012 at 2:05
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$\frac{d}{dx} [f(x)] = \frac{d}{dx}\exp (f''(x))$ and by the chain rule its derivative is $\frac{d}{dx}f''(x) \times \frac{d}{d(f''(x))} \exp(f''(x)) = f'''(x) \times \exp(f''(x))$

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  • $\begingroup$ but how can you have something depend upon its own 3rd derivative? $\endgroup$
    – yiyi
    Oct 29, 2012 at 1:26
  • $\begingroup$ Since you use $f$ in the answer, you could also simply say $f'(x)$. $\endgroup$
    – Berci
    Oct 29, 2012 at 1:26
  • $\begingroup$ That happens when the function depends on its second derivative $\endgroup$
    – Henrique Tyrrell
    Nov 8, 2012 at 0:34

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