October 29th, 2016, 12:08 PM  #1 
Senior Member Joined: Feb 2014 Posts: 110 Thanks: 1  Derivative of logarithm
Hello, Could you give me, how can simplified fraction? 
October 29th, 2016, 12:16 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,109 Thanks: 983  let $k=\log_2{e}$ $\dfrac{\frac{k}{n}}{\frac{1}{2\sqrt{n}}}= \dfrac{k}{n} \cdot \dfrac{2\sqrt{n}}{1}=\dfrac{2k}{\sqrt{n}}$ 
October 29th, 2016, 12:51 PM  #3 
Senior Member Joined: Feb 2014 Posts: 110 Thanks: 1 
Thank you, but this derivative is correct?

October 29th, 2016, 12:58 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,109 Thanks: 983 
Yes.

October 29th, 2016, 01:02 PM  #5 
Senior Member Joined: Feb 2014 Posts: 110 Thanks: 1  
October 29th, 2016, 01:10 PM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,109 Thanks: 983 
$\dfrac{1}{\ln{2}} = \dfrac{\ln{e}}{\ln{2}} = \log_2{e}$

October 29th, 2016, 01:12 PM  #7 
Senior Member Joined: Feb 2014 Posts: 110 Thanks: 1 
Very helpful. Thanks

October 30th, 2016, 03:31 PM  #8 
Senior Member Joined: Feb 2014 Posts: 110 Thanks: 1 
How can solve it?

October 30th, 2016, 03:59 PM  #9 
Math Team Joined: Jul 2011 From: Texas Posts: 2,109 Thanks: 983  In future, start a new problem with a new thread. Also, derivative problems belong in the Calculus forum. chain rule ... $\bigg[(\ln{u})^3\bigg]' = 3(\ln{u})^2 \cdot \dfrac{u'}{u}$ 

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