Partial Derivative of Tan(y/x)

Asked Aug 04 '20 at 23:35

Active Aug 04 '20 at 23:35

Viewed 751 times

Given that =tan/ by Cylindrical Coordinates, I have that ... But how do you go from that to ?

In particular, why does the sec just disappear? Please assist. Thank you.

asked Aug 04 '20 at 23:35

DarkRunner

1

Please clarify: $\phi=\tan(y/x)$ or $\phi=\arctan(y/x)$? The former is not how we define cylindrical coordinates. – user10354138 Aug 04 '20 at 23:38
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The second image is the derivative of $\arctan( \tfrac yx)$ instead of $\tan( \tfrac yx)$. – azif00 Aug 04 '20 at 23:38
@Azif00 Now I'm confused! I guess I mean arctan(y/x) then – DarkRunner Aug 05 '20 at 00:03

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