December 27th, 2016, 07:50 PM  #1 
Newbie Joined: Dec 2016 From: Lehigh valley, pa Posts: 5 Thanks: 0  Need help with chain rule
Could someone please tell me what it is that I'm not getting.. I attached an image, I hope it loaded properly.

December 27th, 2016, 08:38 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,311 Thanks: 1137 
$y=(52x)^{3}+\dfrac{1}{8}\left(\dfrac{2}{x}+1\right)^4$ $\dfrac{dy}{dx}=3(52x)^{4}\cdot (2) + \dfrac{1}{2}\left(\dfrac{2}{x}+1\right)^3 \cdot \left(\dfrac{2}{x^2}\right)$ $\dfrac{dy}{dx}=\dfrac{6}{(52x)^4}  \dfrac{1}{x^2} \left(\dfrac{2}{x}+1\right)^3$ 
December 27th, 2016, 08:45 PM  #3 
Newbie Joined: Dec 2016 From: Lehigh valley, pa Posts: 5 Thanks: 0 
Thank you so much.

December 28th, 2016, 12:14 AM  #4  
Global Moderator Joined: Dec 2006 Posts: 16,604 Thanks: 1201  Quote:
Let's consider y = (5  2x)^(3). Let u = 5  2x, so that du/dx = 0  2 = 2. As y = u^(3), dy/du = 3u^(4) = 3/u^4, and so dy/dx = (dy/du)(du/dx) = (3/u^4)(2) = 6/u^4 = 6/(5  2x)^4.  
December 30th, 2016, 05:28 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,279 Thanks: 570 
Generally speaking, using the "power" rule, $\displaystyle \frac{du^{n}}{dx}= nu^{n1}\frac{du}{dx}$ with a negative power is far easier than using the "quotient" rule with $\displaystyle \frac{1}{u}$


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