1

I asked a question about multivariate implicit differentiation a while back and was told that the result

$$\frac{\text{d}\alpha}{\text{d}\beta} = \frac{1}{\frac{\text{d}\beta}{\text{d}\alpha}}$$

where $\frac{\text{d}\beta}{\text{d}\alpha} \neq 0$ holds for partial derivatives as well, i.e.

$$\frac{\partial\alpha}{\partial\beta} = \frac{1}{\frac{\partial\beta}{\partial\alpha}}$$

This was a huge relief to me, as it is much simpler than the method involving Jacobians. However, today I was trying to derive the expression for the gradient in polar coordinates and tried exploiting the result for $\frac{\partial r}{\partial x}$:

$$\frac{\partial r}{\partial x} = \frac{1}{\frac{\partial x}{\partial r}} = \frac{1}{\cos\theta}$$

However, this is not what I obtain when I use the method of Jacobians, or even the inverse mapping between polar and Cartesian coordinates (i.e. $r = \sqrt{x^2 + y^2}$):

$$\frac{\partial r}{\partial x} = \frac{r \cos \theta}{r \cos^2\theta + r\sin^2\theta} = \cos\theta = \frac{x}{\sqrt{x^2+y^2}}$$

Can anyone help me understand what it is I am missing? I would like to use the $\frac{\partial r}{\partial x} = \frac{1}{\frac{\partial x}{\partial r}} = \frac{1}{\cos\theta}$ result if possible, as it is a more direct computation, but I don't see how the results are equivalent.

Rax Adaam
  • 1,176
  • 2
    See this: https://math.stackexchange.com/questions/1090061/manipulating-partial-derivatives-of-inverse-function – Hans Lundmark Nov 21 '17 at 16:17
  • @HansLundmark so, in this case, $\frac{\partial r}{\partial x} = \bigl( \frac{\partial r}{\partial x} \bigr){y} \neq \frac{1}{ \bigl(\frac{\partial x}{\partial r}\bigr){\theta} }$? I'm confused by the answers' discussion of "which is the case when we take the derivative of the inverse map" - isn't that what I'm trying to do here? – Rax Adaam Nov 21 '17 at 21:15
  • Yes, they are different. See my comment below the answer; it seems to me that there is a “not” missing in that sentence. – Hans Lundmark Nov 21 '17 at 23:32

0 Answers0