Calculus Calculus Math Forum

 December 31st, 2014, 06:44 PM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Differentiation I'm given $\displaystyle y=\frac{cos2x}{1-sinx}$ to differentiate. My answer is $\displaystyle \frac{cosxcos2x-2(sin2x)(1-sinx)}{(1-sinx)^2}$ But the given answer is $\displaystyle -\frac{4sin(2x)-3cos(x)+cos(3x)}{2(sin(x)-1)^2}$. January 1st, 2015, 12:04 AM #2 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,974 Thanks: 1156 Math Focus: Elementary mathematics and beyond Post your work and we'll help you identify and correct any errors. Thanks from szz January 1st, 2015, 01:10 AM #3 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 $\displaystyle y=\frac{cos2x}{1-sinx}$ $\displaystyle \frac{dy}{dx}=\frac{(1-sinx)(-2sin2x)-(cos2x)(-cosx)}{(1-sinx)^2}$ $\displaystyle =\frac{-2sin2x+2sin^22x+cosx(cos2x)}{(1-sinx)^2}$ $\displaystyle =\frac{(-2)(sin2x)\left [ 1-sinx \right ]+cosx(cos2x)}{(1-sinx)^2}$ $\displaystyle =\frac{cosxcos2x-2(sin2x)(1-sinx)}{(1-sinx)^2}$ The given answer is $\displaystyle -\frac{4sin(2x)-3cos(x)+cos(3x)}{2(sin(x)-1)^2}$ January 1st, 2015, 10:43 AM #4 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,974 Thanks: 1156 Math Focus: Elementary mathematics and beyond The given answer is correct. Your mistake is in assuming $\displaystyle (-\sin(x))(-2\sin(2x))=2\sin^2(2x)$ Correct that, then use the identities $\displaystyle \cos(\theta)-\cos(\varphi)= -2\sin\left(\frac{\theta+\varphi}{2}\right)\sin \left(\frac{\theta-\varphi}{2}\right)$ and $\displaystyle \cos(\theta)+\cos(\varphi)=2\cos\left(\frac{\theta +\varphi}{2}\right)\cos\left(\frac{\theta-\varphi}{2}\right)$ to simplify further. Thanks from szz January 3rd, 2015, 09:31 AM #5 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 What this symbol $\displaystyle \varphi$ means? January 3rd, 2015, 10:31 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra It's a variable, phi, like theta is a variable. Tags differentiation 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 Pudding Calculus 3 February 15th, 2013 05:25 AM maximia Calculus 4 January 31st, 2013 08:48 PM najaa Calculus 2 November 17th, 2012 11:11 AM gspimoli Calculus 1 November 12th, 2012 07:40 AM florian101 Calculus 1 December 22nd, 2009 07:42 AM

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