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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 December 5th, 2016, 08:05 AM #1 Member   Joined: Sep 2016 From: zambia Posts: 31 Thanks: 0 differentiation help me with the following question differentiate: y=cos inverse [(2cosx+3)/(2+3cosx)] Last edited by jeho; December 5th, 2016 at 08:12 AM. Reason: the question was not correct December 5th, 2016, 12:44 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,034 Thanks: 1621 if $y = \cos^{-1}{u}$ where $u$ is a function of $x$, then $\dfrac{dy}{dx} = -\dfrac{\frac{du}{dx}}{\sqrt{1 - u^2}}$ $u = \dfrac{2\cos{x}+3}{2+3\cos{x}}$ $\dfrac{du}{dx} = \dfrac{(2+3\cos{x})(-2\sin{x}) - (2\cos{x}+3)(-3\sin{x})}{(2+3\cos{x})^2}$ or ... $\dfrac{d}{dx} \bigg[\cos{y} = \dfrac{2\cos{x}+3}{2+3\cos{x}}\bigg]$ $-\sin{y} \dfrac{dy}{dx} = \dfrac{(2+3\cos{x})(-2\sin{x}) - (2\cos{x}+3)(-3\sin{x})}{(2+3\cos{x})^2}$ either way, finish it ... Thanks from topsquark and jeho Last edited by skeeter; December 5th, 2016 at 01:22 PM. December 5th, 2016, 03:34 PM #3 Member   Joined: Sep 2016 From: zambia Posts: 31 Thanks: 0 thank you l now see where l was making a mistake 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 Mr Davis 97 Calculus 1 March 9th, 2015 04:50 AM Soulanoid Calculus 1 September 29th, 2014 08:56 PM jiasyuen Calculus 3 September 25th, 2014 03:34 AM arron1990 Calculus 6 February 12th, 2012 01:15 PM MyNameIsVu Calculus 1 June 3rd, 2009 05:18 AM

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