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February 1st, 2019, 09:57 PM  #1 
Member Joined: Aug 2018 From: România Posts: 45 Thanks: 2  Derivatives of a function
Hello, Calculate the derivatives of the function: 1) $\displaystyle f(x)=x!$ 2) $\displaystyle f(x)=(2\sqrt{2})\cdot (2\sqrt[3]{2})\cdot \dots \cdot (2\sqrt[x] {2})$. All the best, Integrator 
February 1st, 2019, 10:59 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,633 Thanks: 2080 
Neither function is continuous, so neither function is differentiable.

February 2nd, 2019, 12:54 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,759 Thanks: 696 
$x!=\Gamma(x+1)=\int_0^\infty u^xe^{u}du$. You can get the derivative from this. The second expression is defined only for integers, so you can't get a derivative.

February 2nd, 2019, 03:31 PM  #4  
Senior Member Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
On the other hand, the second function, which also can be analytically extended from the positive integers, has a natural choice for this extension which makes its derivative unique. This choice comes from the continuous extension of $f(x) = a^x$ from the integers to the rationals. This automatically determines a unique extension to the positive reals and this extension happens to be analytic. In a nutshell, the first expression has no "correct" answer without more assumptions and the second one does.  
February 2nd, 2019, 09:45 PM  #5  
Member Joined: Aug 2018 From: România Posts: 45 Thanks: 2  Quote:
1) From "WolframAlpha" reading: https://www.wolframalpha.com/input/?i=x!%27. Is it correct what WolframAlpha says? 2) I do not know how to use "WolframAlpha" for the first derivative of the function $\displaystyle f(x)=(2\sqrt{2})\cdot (2\sqrt[3]{2})\cdot \dots \cdot (2\sqrt[x]{2})$. Thank you very much! All the best, Integrator  
February 3rd, 2019, 04:46 AM  #6  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,158 Thanks: 878 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  

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