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27 September, 00:11

Cos (inv) { (a+bcosx) / (b+acosx) }

Differentiate ...

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  1. 27 September, 00:22
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    Differentiation of Cos (inv) { (a+bcosx) / (b+acosx) } = - (((a + b Cos[x]) / (Sqrt[1 - x^2] (b + ArcCos[x]) ^2) - (b Sin[x]) / (b + ArcCos[x])) / Sqrt[1 - (a + b Cos[x]) ^2 / (b + ArcCos[x]) ^2])
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