Oscillating derivatives?

Is there a function or a class of functions whose derivatives oscillate? Something like $${d\over dx}\left({df(x)\over dx}\right) =f(x)$$ And hence, the higher order derivatives will only be the function and the first derivative.

($f(x)$ and $f'(x)$ have to be different. Example: $f(x)=e^x$ is not a valid answer.) I don't know much about different types of functions yet, so there might be trivial ones I have missed. Also, I think $\sqrt{x}$ might lead to an answer.

• What about hyperbolic cosine and sine? $\cosh(x)=\frac{e^x+e^{-x}}{2}$ and $\sinh(x)=\frac{e^x-e^{-x}}{2}$. Or any linear combination of these two functions. This is actually the set of solutions. You should read about differential equations too. – Augustin Feb 25 '16 at 12:58
• Yes, they work. Do you think any polynomial functions work? – SS_C4 Feb 25 '16 at 13:03
• Apart from the zero polynomial, no. It's easy to see as the higher derivatives of a polynomial are 0 if the order is high enough. – Augustin Feb 25 '16 at 13:04
• What about negative powers? – SS_C4 Feb 25 '16 at 13:06
• It won't work either. As as said, the only solutions of the differential equation $f''=f$ are the functions of the form $A\cosh(x)+B\sinh(x)$. – Augustin Feb 25 '16 at 13:09

$$f(x) = \sinh(x)$$

$$\frac{\text{d}}{\text{d}x}\sinh(x) = \cosh(x)$$

$$\frac{\text{d}}{\text{d}x}\left(\frac{\text{d cosh(x)}}{\text{d}x}\right) = \sinh(x)$$

The same for $f(x) = \cosh(x)$

Then you may also find other functions, just solve the associate differential equation

$$f''(x) - f(x) = 0$$

$$y''=y$$ when $$y'y''=yy'.$$ Integrating, $$y'^2=y^2+C,$$ or

$$\frac{y'}{\sqrt{y^2+C}}=\pm1.$$

Integrating again,

$$\ln\left(y+\sqrt{y^2+C}\right)=\pm x+C'$$

This is the most general solution.

Fortunately, the relation can be inverted with the use of hyperbolic functions to yield one of

$$y=a\cosh(x+b),\\y=a\sinh(x+b),$$ or any linear combination.