November 14th, 2016, 05:21 PM  #1 
Member Joined: Nov 2016 From: Kansas Posts: 65 Thanks: 0  Differentiation
Anyone??

November 15th, 2016, 05:48 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,822 Thanks: 750 
The problem is: given two differentiable functions, f and g, with f(x) > 0, show that $\displaystyle e^{g(x)\ln(f(x))}$ is differentiable. Looks to me that the best way to do that is to differentiate it! The derivative is, of course, $\displaystyle e^{g(x)\ln(f(x))}$ times the derivative of g(x)ln(f(x)). And that is, by the product rule and chain rule, $\displaystyle g'(x)\ln(f(x))+ \frac{g(x)f'(x)}{f(x)}$. Since f and g are differentiable, f' and g' exist and since f(x) > 0, f is never 0 so that clearly exists. Last edited by skipjack; November 15th, 2016 at 07:02 AM. 

Tags 
differentiation, real analysis 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Difference between vector differentiation and parametric differentiation  Mr Davis 97  Calculus  1  March 9th, 2015 05:50 AM 
I really need help with these ... more Differentiation  Donna  Calculus  1  November 13th, 2014 10:19 AM 
Differentiation  Mathproblem123  Calculus  6  March 13th, 2014 05:17 AM 
Differentiation  jiasyuen  Calculus  2  January 3rd, 2014 10:57 AM 
Differentiation  Zynoakib  Calculus  1  October 26th, 2013 07:24 AM 