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The example I am using is polar coordinates, however my confusion can be extended.

so we have $f(x,y)$ where $x = x(r,\theta)$ and $y = y(r,\theta) $

The first derivative I understand:

$$\left(\frac{\partial f}{\partial x}\right)_y=\left(\frac{\partial f}{\partial r}\right)_\theta\left(\frac{\partial r}{\partial x}\right)_y+\left(\frac{\partial f}{\partial \theta}\right)_r\left(\frac{\partial \theta}{\partial x}\right)_y$$

now if we differentiate with respect to x again - I have used the product rule on both functions in each term:

$$\left(\frac{\partial^2 f}{\partial x^2}\right)_y= \left(\frac{\partial^2 f}{\partial r^2}\right)_\theta\left(\frac{\partial r}{\partial x}\right)_y^2 + \left(\frac{\partial f}{\partial r}\right)_\theta\left(\frac{\partial^2 r}{\partial x^2}\right)_y+ \left(\frac{\partial^2 f}{\partial \theta^2}\right)_r\left(\frac{\partial \theta}{\partial x}\right)_y^2 + \left(\frac{\partial f}{\partial \theta}\right)_r\left(\frac{\partial^2 \theta}{\partial x^2}\right)_y +$$

$$ +2\left(\frac{\partial^2 f}{\partial r \partial \theta}\right)_{\theta r}\,\left(\frac{\partial r}{\partial x}\right)_y \,\left(\frac{\partial \theta}{\partial x}\right)_y $$

However, I believe this to be wrong as we can take

$$\left(\frac{\partial }{\partial x}\right)_y=\left(\frac{\partial }{\partial r}\right)_\theta\left(\frac{\partial r}{\partial x}\right)_y+\left(\frac{\partial }{\partial \theta}\right)_r\left(\frac{\partial \theta}{\partial x}\right)_y$$

Applying this to $(\frac{\partial f}{\partial x})_y$ we get:

$$(\frac{\partial^2 f}{\partial x^2})_y= (\frac{\partial^2 f}{\partial r^2})_\theta(\frac{\partial r}{\partial x})_y^2 + (\frac{\partial^2 f}{\partial \theta^2})_r(\frac{\partial \theta}{\partial x})_y^2 + 2(\frac{\partial^2 f}{\partial r \partial \theta})_{\theta r}(\frac{\partial r}{\partial x})_y (\frac{\partial \theta}{\partial x})_y $$

Where did the

$$ (\frac{\partial f}{\partial r})_\theta(\frac{\partial^2 r}{\partial x^2})_y\;\;,\;\;\;\; (\frac{\partial f}{\partial \theta})_r(\frac{\partial^2 \theta}{\partial x^2})_y$$

terms go? Its as if the product rule doesn't hold, the $\; (\frac{\partial r}{\partial x})_y\;\;,\;\;\; (\frac{\partial \theta}{\partial x})_y\;$ didn't get differentiated.

Where is my mistake? thank you

  • 0
    Use \left( , \right) to put parentheses of the right size around your expressions – 2017-01-01
  • 0
    With the first derivative I already have problems: is that the second order partial *mixed* derivative? – 2017-01-01

1 Answers 1

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First, I think we have:

$$f'_x=f'_r\cdot r'_x+f'_\theta\cdot\theta'_x=f'_r\cos\theta-f'_\theta\frac{\sin\theta}r$$

so your first (second order mixed) derivative is:

$$f'_{xy}=\left(f'_r\right)'_y\,r'_x+f'_r\,(r'_x)'_y+(f'_\theta)'_y\,\theta'_x+f'_\theta\,(\theta_x)'_y$$

For example:

$$\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies r=\sqrt{x^2+y^2}\implies r'_x=\frac x{\sqrt{x^2+y^2}}=\sin\theta\;,\;\;$$

$$(r'_x)'_y=-\frac{xy}{(x^2+y^2)^{3/2}}=-\frac{\cos\theta\sin\theta}r=-\frac1{2r}\,\sin2\theta$$

and etc.