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I'm blanking on the simplest thing ever:

$L = x$ $\frac{dL}{dx} = 1$

But if I do:

$x = m + n$ $\frac{dL}{dx} = \frac{\partial L}{\partial m}\frac{\partial m}{\partial x} + \frac{\partial L}{\partial n}\frac{\partial n}{\partial x}$

$\frac{dL}{dx} = (1)(1) + (1)(1)$ $\frac{dL}{dx} = 2$

Yeah, this is embarrassing.

1 Answers 1

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The problem is, you assume that $\frac{\partial m}{\partial x} = \frac{\partial n}{\partial x} = 1$

However in reality $\frac{\partial m}{\partial x} = 1 - \frac{\partial n}{\partial x}$

And vice versa. So now the result checks out.

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    D'oh. Slaps forehead. Thank you.2012-10-31