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If $\frac{df(x,y)}{dx} = a$, does $\frac{1}{a} = \frac{dx}{df(x,y)}$?

Consider $f(x,y) = x^2y \Rightarrow \frac{df(x,y)}{dx} = 2xy \equiv a$, than $\frac{1}{a} = \frac{1}{2xy}$.

Now calculate $\frac{dx}{df(x,y)} = \frac{dx}{d(x^2y)}$ via substitution $x^2y = u \Rightarrow x = \pm \sqrt{\frac{u}{y}} \Rightarrow \frac{dx}{d(x^2y)} = \frac{\sqrt{\frac{u}{y}}}{du} = \frac{1}{2} (\frac{u}{y})^{-\frac{1}{2}} \frac{1}{y} = \frac{1}{2} \frac{1}{\sqrt{x^2}y} = \frac{1}{2xy}$

So it seems my claim is true, but it still feels strange to me. Could I be just lucky with my example?

Some info about my background. I'm a physics major and started pondering about thermodynamics, where the definition of absolute temperature is given as $\frac{1}{T} = \frac{dS(E,\lambda)}{dE}$, with temperature $T$, entropy $S$, energy $E$ and some other (fixed) system variable $\lambda$. Reason for my confusion is that I can't make much (physical) sense out of $\frac{dE}{dS(E,\lambda)}$, which with my claim would equal to $T$.

Emilio Novati
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rtime
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1 Answers1

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Well actually you're both correct and not correct. Had it been just f(x) and not f(x,y) your logic would have been correct and it follows from the inverse function theorem.

For multi variable functions we need to look at the Jacobian and see if it holds true. Basically your example was a special case.

Here's the link to the wiki page which explains the math if you're interested :)

https://en.m.wikipedia.org/wiki/Inverse_function_theorem

spandan madan
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  • so, if the determinant of the Jacobian is non-zero, function F is invertible? I think I have to read into the topic a bit more, to get a definite grasp on it. I feel like I understand the application of the inverse function theorem, but not the meaning behind it – rtime Feb 09 '16 at 23:29