# Differentiation dr/dx in cylindrical coordinates from cartesian

For transformation from cartesian to cylindrical coordinates,

$$\ x=r~cos \theta ,~~ y=r~ sin \theta , ~~ r=(x^2+y^2)^{1/2}$$ Then, $$\frac {dr} {dx} =\frac {1} {2} (x^2+y^2)^{-1/2}~2x~=~\frac {x} {(x^2+y^2)^{1/2} }~=~\frac {r~cos \theta} {r}~=~ cos \theta$$ Is it possible to obtain the same result by differentiating the below relationship? $$r~=~\frac {x} {cos \theta}$$

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hint use quotient rule: https://en.wikipedia.org/wiki/Quotient_rule

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Thank you very much for your kind and fast answer. Yes. I do get the same answer!! You helped me a lot!!! –  Tony Mar 10 '12 at 4:26
@tony - You are welcome –  Victor Mar 10 '12 at 14:27