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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.

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1 Answer 1

up vote 5 down vote accepted

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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1  
D'oh. Slaps forehead. Thank you. –  Nick Oct 31 '12 at 3:17

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