My Math Forum Need help with chain rule

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December 27th, 2016, 07:50 PM   #1
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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.
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 December 27th, 2016, 08:38 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,678 Thanks: 1339 $y=(5-2x)^{-3}+\dfrac{1}{8}\left(\dfrac{2}{x}+1\right)^4$ $\dfrac{dy}{dx}=-3(5-2x)^{-4}\cdot (-2) + \dfrac{1}{2}\left(\dfrac{2}{x}+1\right)^3 \cdot \left(-\dfrac{2}{x^2}\right)$ $\dfrac{dy}{dx}=\dfrac{6}{(5-2x)^4} - \dfrac{1}{x^2} \left(\dfrac{2}{x}+1\right)^3$ Thanks from greg1313
 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
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Quote:
 Originally Posted by Marcos tavarez Could someone please tell me what it is that I'm not getting...
You made a series of simple errors. You should be able to spot them if you compare your steps in detail with a correct solution. In particular, take care with exponents.

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: 2,878 Thanks: 766 Generally speaking, using the "power" rule, $\displaystyle \frac{du^{-n}}{dx}= -nu^{-n-1}\frac{du}{dx}$ with a negative power is far easier than using the "quotient" rule with $\displaystyle \frac{1}{u}$

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