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15. Note, however, that this assumes that $\ln x$ is differentiable. (That is required if you want to use the chain rule) So unless you have proved that $\ln x$ is differentiable, this proof cannot work. As far as I can see, there is no better way to prove that $\ln x$ is differentiable that to calculate the derivative explicitly.
Here I say 'We first note that for the case where the elements of X are independent, a constructive proof involving cofactor expansion and adjoint matrices can be made to show that $\frac{\partial ln|X|}{\partial X} = X^{-T}$ (Harville). This is not always equal to $2X^{-1}-diag(X^{-1})$.
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A simple expression can be derived by manipulating the Taylor series $\ln X = \sum_{n=1}^\infty -\frac{(-1)^n}{n}(X-1)^n$ with the result $$\frac{d}{ds}\ln X(s ...
The derivative of a function of a discrete variable doesn't really make sense in the typical calculus setting. However, there is a continuous variant of the factorial function called the Gamma function, for which you can take derivatives and evaluate the derivative at integer values. In particular, since n! = Γ(n + 1), there is a nice formula ...
Derivative Of $\ln(x)$ Ask Question Asked 9 years, 2 months ago. Modified 9 years, 2 months ago. Viewed ...
One of the answers to the problems I'm doing had straight lines: $$ \\ln|y^2-25|$$ versus another problem's just now: $$ \\ln(1+e^r) $$ I know this is probably to do with the absolute value...
1. Consider the function $\ln |x|$, since it is locally integrable we can form the distribution. Now, I want to show that in the sense of distributions we have $\ln |x|' = \operatorname {Pv}\frac {1} {x}$. My obvious try was to substitute directly the definition of the derivative for distributions:
$\begingroup$ Then the derivative of $\ln c$ is $0$ because it is a constant. $\endgroup$ – Ross Millikan.
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