One of the things I never really understood properly, is $$D \log(x) = \frac{1}{x}.$$ This essentially comes from two properties: derivatives of inverses are related through multiplicative inverses, and the exponential function is an eigenfunction of the derivative operator. The expression of the derivative of the inverse $Df^{-1}$ can be obtained from the chain rule as $$D (f \circ f^{-1}) = [(D f) \circ f^{-1}] D f^{-1}.$$ Since this also expresses the derivative of the identity function, we have $$D f^{-1} = \frac{1}{(D f)\circ f^{-1}}.$$ This relation is obvious for a linear function, e.g. $y=ax \Rightarrow x=a^{-1}y.$ Also, following the composition of the 3 operations $f^{-1}$, $Df$ and $x \to 1/x$ on a plot makes it fairly obvious what is going on is just obtaining the tangent line and expressing it in the proper coordinate system. The expression for $D\log$ follows straight from the rule of the derivative of the inverse of $\exp$. $$D \log(x) = \frac{1}{\exp [ \log (x)] } = \frac{1}{x}.$$