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Is the reciprocal of a partial derivative always equal to 1 divided by that partial derivative? I.e. Is it true that;

$$\frac{1}{\frac{\partial x_1}{\partial x_2}} = \frac{\partial x_2}{\partial x_1} $$

JDoe2
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1 Answers1

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Not always. E.g. if $x \equiv x(r,\theta)=r\cos \theta$ and $y \equiv y(r,\theta)= r\sin \theta$ (polar coordinates), then $r \equiv r(x,y)= \sqrt{x^2+y^2}$, so $$\frac{\partial x}{\partial r} = \cos \theta = \frac{x}{r}$$ and also $$\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2+y^2}} = \frac{x}{r} = \frac{\partial x}{\partial r}.$$

Minus One-Twelfth
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  • I see... in general is there a rule that you know of for switching between $\frac{\partial x_1}{\partial x_2}$ and $\frac{\partial x_2}{\partial x_1}$ – JDoe2 May 11 '20 at 14:38
  • Do you know, when does the reciprocal rule hold/not hold? – JDoe2 May 11 '20 at 14:39