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differentiationhelp me with the following question differentiate: y=cos inverse [(2cosx+3)/(2+3cosx)] |

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 ... |

thank you l now see where l was making a mistake |

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