My Math Forum differentiation

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 December 5th, 2016, 09:05 AM #1 Newbie   Joined: Sep 2016 From: zambia Posts: 26 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 09:12 AM. Reason: the question was not correct
 December 5th, 2016, 01:44 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,204 Thanks: 1049 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 02:22 PM.
 December 5th, 2016, 04:34 PM #3 Newbie   Joined: Sep 2016 From: zambia Posts: 26 Thanks: 0 thank you l now see where l was making a mistake

 Thread Tools Display Modes Linear 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 02:15 PM MyNameIsVu Calculus 1 June 3rd, 2009 05:18 AM

 Contact - Home - Forums - Top