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This may be an obvious question but I'm just not thinking straight, thanks

The answer must be no

Kola B.
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Jeff
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    It is true if and only if $f(x) = \log(C-x) + D$ where $C,D$ are constants (of if $f$ is constant). – Winther Nov 28 '14 at 00:32
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    When you suspect something to be true, proving so may be difficult or even unreachable. Fortunately, when you suspect something to be false, all you have to do is find a single counterexample. In this case, it would be difficult to find a function for which this was true! – andrepd Nov 28 '14 at 00:34

5 Answers5

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Of course not ! Take for exemple $$f(x,y)=x,$$ then $$\left(\frac{\partial f}{\partial x}\right)^2=1,$$ but $$\frac{\partial^2 f}{\partial x^2}=0.$$

idm
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As others have noted, nope. But let's explore for which functions it is true. If the function is not linear, it has a second derivative that is not always zero, so the first step here is valid. (And if the function is linear, we can work out that it would have to be a constant function.)

$$ \begin{align} \frac{\partial^2f}{\partial x^2} &=\left(\frac{\partial f}{\partial x}\right)^2\\ \left(\frac{\partial f}{\partial x}\right)^{-2}\frac{\partial^2f}{\partial x^2} &=1&\text{anitdifferntiate on both sides}\\ -\left(\frac{\partial f}{\partial x}\right)^{-1} &=x+C_1\\ \frac{\partial f}{\partial x} &=\frac{-1}{x+C_1}&\text{anitdifferntiate on both sides}\\ f(x)&=-\ln\left|x+C_1\right|+C_2(x) \end{align} $$ where $C_2$ is some function that is a constant value on $\left(-C_1,\infty\right)$ and a possibly different constant value on $\left(-\infty,-C_1\right)$. You could also write it as $$f(x)=-\ln\left|x+C\right|+D\frac{x+C}{|x+C|}+E$$

2'5 9'2
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Absolutely not. Pick any function you like and work out the RHS and LHS, and you will almsot always get two different solutions.

preferred_anon
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This is not true. To see this, let's consider an example (there are simpler examples...) $$ f(x,y) = x^2y^2 $$ Then $$ \frac{\partial f}{\partial x} = 2xy^2 \\ \frac{\partial^2 f}{\partial x^2} = 2y^2. $$ And $$\left(\frac{\partial f}{\partial x}\right)^2 = (2xy^2)^2 = 4x^2y^4. $$

Thomas
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No. Let $f(x,y)=x$. Verification is left to you.

Kola B.
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