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Is it possible to do some kind of simplifications on an expression like

$$ f : x, y \to \mathbb{R} \\ \frac{\frac{\partial^2 f}{\partial y^2}}{\frac{\partial f}{\partial y}} = \frac{\partial f}{\partial y} $$

So that $\frac{\partial f}{\partial y}$ in the denominator reduces $\frac{\partial^2 f}{\partial y^2}$ to $\frac{\partial f}{\partial y}$ like $\frac{a^2}{a}=a$, without knowing the function $f$.

I would say no, that's not allowed, but I'm just wondering.

Thank you.

WolfgangP
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  • I think that the closest you have to it is the chain rule: $$ \frac{\partial f}{\partial u} = \frac{\partial x}{\partial u} \frac{\partial f}{\partial x} $$ – rafa11111 Nov 26 '18 at 22:59
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    No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified. –  Nov 26 '18 at 23:00

4 Answers4

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$$\frac{\frac{\partial^2 f}{\partial y^2}}{\frac{\partial f}{\partial y}}=\frac{\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)}{\frac{\partial f}{\partial y}}\ne\frac{\partial f}{\partial y}$$

just as

$$ \frac{y^{\prime\prime}}{y^\prime}\ne y^\prime $$

John Wayland Bales
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That is indeed not allowed, even for single-variable derivatives. Take for example $f(x) = x^2$. Then $$ \frac{f''}{f'} = \frac{2}{2x} = \frac{1}{x} \not= 2x = f' $$

MSDG
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    Actually the case in the question is single-variable, as $x$ plays no role. –  Nov 26 '18 at 23:01
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    @YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked. – MSDG Nov 26 '18 at 23:02
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No. Remember the second derivative is just the derivative of the first derivative, so using $g$ for the first derivative, your equation is equivalent to $\frac{\partial g}{\partial y}=g^2$, which of course can't always be true.

BallBoy
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No, usually these kinds of analogies don't work. Take for example $f(x,y)=y$.

Scientifica
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