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 Calculus Calculus Math Forum

December 27th, 2016, 06:50 PM   #1
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From: Lehigh valley, pa

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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.
Attached Images IMG_20161227_224305.jpg (89.1 KB, 15 views) December 27th, 2016, 07:38 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,034 Thanks: 1621 $y=(5-2x)^{-3}+\dfrac{1}{8}\left(\dfrac{2}{x}+1\right)^4$ $\dfrac{dy}{dx}=-3(5-2x)^{-4}\cdot (-2) + \dfrac{1}{2}\left(\dfrac{2}{x}+1\right)^3 \cdot \left(-\dfrac{2}{x^2}\right)$ $\dfrac{dy}{dx}=\dfrac{6}{(5-2x)^4} - \dfrac{1}{x^2} \left(\dfrac{2}{x}+1\right)^3$ Thanks from greg1313 December 27th, 2016, 07:45 PM #3 Newbie   Joined: Dec 2016 From: Lehigh valley, pa Posts: 5 Thanks: 0 Thank you so much. December 27th, 2016, 11:14 PM   #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, 04:28 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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 $\displaystyle \frac{1}{u}$ Tags chain, rule Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jake6390 Calculus 4 May 17th, 2014 04:21 PM ungeheuer Calculus 1 July 30th, 2013 05:10 PM unwisetome3 Calculus 4 October 19th, 2012 01:21 PM arron1990 Calculus 5 May 19th, 2012 06:04 AM Peter1107 Calculus 1 September 8th, 2011 10:25 AM

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