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December 5th, 2016, 09:05 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 09:12 AM. Reason: the question was not correct 
December 5th, 2016, 01:44 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,692 Thanks: 1351 
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 ... Last edited by skeeter; December 5th, 2016 at 02:22 PM. 
December 5th, 2016, 04:34 PM  #3 
Member Joined: Sep 2016 From: zambia Posts: 31 Thanks: 0 
thank you l now see where l was making a mistake 

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