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December 5th, 2016, 09:05 AM   #1
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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
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December 5th, 2016, 01:44 PM   #2
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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.
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December 5th, 2016, 04:34 PM   #3
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thank you

l now see where l was making a mistake
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