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Question Number 18906 by Arnab Maiti last updated on 01/Aug/17

$$\mathrm{calculate}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{where}$$ $$\mathrm{y}={\cos }^{-1}\frac{\mathrm{a}+\mathrm{b}\mathrm{cosx}}{\mathrm{b}+\mathrm{a}\mathrm{cosx}}(\mathrm{b}>\mathrm{a})$$

Answered by ajfour last updated on 01/Aug/17

$$\mathrm{y}={\cos }^{-1}\left(\frac{\mathrm{a}+\mathrm{bcos}\mathrm{x}}{\mathrm{b}+\mathrm{acos}\mathrm{x}}\right)$$ $$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{\sin \mathrm{y}}.\frac{(-\mathrm{bsin}\mathrm{x})(\mathrm{b}+\mathrm{acos}\mathrm{x})-(-\mathrm{asin}\mathrm{x})(\mathrm{a}+\mathrm{bcos}\mathrm{x})}{{(\mathrm{b}+\mathrm{acos}\mathrm{x})}^2}$$ $$=-\frac{\mathrm{b}+\mathrm{acos}\mathrm{x}}{\sqrt{{(\mathrm{b}+\mathrm{acos}\mathrm{x})}^2-{(\mathrm{a}+\mathrm{bcos}\mathrm{x})}^2}}\times \frac{({\mathrm{a}}^2-{\mathrm{b}}^2)\sin \mathrm{x}}{{(\mathrm{b}+\mathrm{acos}\mathrm{x})}^2}$$ $$=\frac{({\mathrm{b}}^2-{\mathrm{a}}^2)\sin \mathrm{x}}{(\mathrm{b}+\mathrm{acos}\mathrm{x})\sqrt{({\mathrm{b}}^2-{\mathrm{a}}^2)(1-{\cos }^2\mathrm{x})}}$$ $$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{{\mathrm{b}}^2-{\mathrm{a}}^2}}{(\mathrm{b}+\mathrm{acos}\mathrm{x})}\times \frac{\sin \mathrm{x}}{\mid \sin \mathrm{x}\mid }$$ $$\mathrm{and}\mathrm{if}\sin \mathrm{x}>0$$ $$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{{\mathrm{b}}^2-{\mathrm{a}}^2}}{\mathrm{b}+\mathrm{acos}\mathrm{x}}.$$

Commented byArnab Maiti last updated on 01/Aug/17

$$\mathrm{But}\mathrm{the}\mathrm{answer}\mathrm{is}\mathrm{only}\frac{\sqrt{{\mathrm{b}}^2-{\mathrm{a}}^2}}{\mathrm{b}+\mathrm{a}\mathrm{cosx}}$$

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