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[SOLVED] chain rule partial derivatives

dwsmith

Well-known member
Feb 1, 2012
1,673
$x = r\cos\theta$ and $y=r\sin\theta$
$$
\frac{\partial u}{\partial\theta} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial\theta} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial\theta} = -r\sin\theta\frac{\partial u}{\partial x} + r\cos\theta\frac{\partial u}{\partial y}
$$
So taking the second derivative.
$$
\frac{\partial u}{\partial\theta} = r\left[\frac{\partial u}{\partial x}\frac{\partial }{\partial\theta}\left(-\sin\theta\frac{\partial u}{\partial x} + \cos\theta\frac{\partial u}{\partial y}\right)+\frac{\partial u}{\partial y}\frac{\partial }{\partial\theta}\left(-\sin\theta\frac{\partial u}{\partial x} + \cos\theta\frac{\partial u}{\partial y}\right)\right]
$$
What is the next step? I keep getting it wrong.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$$
\frac{\partial u}{\partial\theta} = r\left[\frac{\partial u}{\partial x}\frac{\partial }{\partial\theta}\left(-\sin\theta\frac{\partial u}{\partial x} + \cos\theta\frac{\partial u}{\partial y}\right)+\frac{\partial u}{\partial y}\frac{\partial }{\partial\theta}\left(-\sin\theta\frac{\partial u}{\partial x} + \cos\theta\frac{\partial u}{\partial y}\right)\right]
$$
I don't understand how you obtained the above equation. It should read,

\[\frac{\partial^2 u}{\partial\theta^2} =r\frac{\partial}{\partial\theta}\left(-\sin \theta\frac{\partial u}{\partial x} + \cos\theta\frac{\partial u}{\partial y}\right)\]