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Define $D_q=\frac{dD}{d_q}$

Is the following true or have I taken too many liberties with the Leibniz quasi-algebra?

$\frac{dD_q}{dq}\frac{q}{D_q}=\frac{d}{dq}\frac{dD}{dq}\frac{dq}{dD}q=\frac{d}{dq}q=\frac{dq}{dq}=0$

Thanks

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    Shouldn't it be $\frac{dq}{dq}=1$?2012-02-04

1 Answers 1

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Nope. You need to parenthesize more carefully. $\begin{align} \frac{dD_q}{dq}\cdot\frac{q}{D_q} &= \left(\frac{d}{dq}\frac{dD}{dq}\right)\cdot\left(\frac{dq}{dD}\right)\cdot q \\ & \ne \frac{d}{dq} \left(\frac{dD}{dq}\cdot\frac{dq}{dD}\right)\cdot q. \end{align}$ You're trying to say that $\dfrac{dx}{dy}\cdot z = \dfrac{d(xz)}{dy}$, but that's not true. You can't simply rearrange what the differentiation operator applies to, any more than you can say $f(x)y = f(xy)$.