November 14th, 2016, 04:21 PM  #1 
Member Joined: Nov 2016 From: Kansas Posts: 73 Thanks: 1  Differentiation
Anyone??

November 15th, 2016, 04:48 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,199 Thanks: 873 
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 06:02 AM. 

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differentiation, real analysis 
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