Answer
$y'=\frac{\sqrt{(a^2-b^2)}}{(a+bcosx)}$
Work Step by Step
$y=arccos(\frac{b+acosx}{a+bcosx})\\
y'=-\frac{1}{\sqrt{1-(\frac{b+acosx}{a+bcosx})^2}}\times \frac{(a+bcosx)(-asinx)-(b+acosx)(-bsinx)}{(a+bcosx)^2}\\
y'=-\frac{1}{\sqrt{1-(\frac{b+acosx}{a+bcosx})^2}}\times \frac{(b+acosx)(bsinx)-(a+bcosx)(asinx)}{(a+bcosx)^2}\\
y'=-\frac{(b+acosx)(bsinx)-(a+bcosx)(asinx)}{(\sqrt{1-(\frac{b+acosx}{a+bcosx})^2})(a+bcosx)^2}\\
y'=-\frac{b^2sinx+abcosxsinx-a^2sinx-abcosxsinx}{(\sqrt{\frac{(a+bcosx)^2-(b+acosx)^2}{(a+bcosx)^2}})(a+bcosx)^2}\\
y'=-\frac{sinx(b^2-a^2)}{(\sqrt{{(a+bcosx)^2-(b+acosx)^2}})(a+bcosx)}\\
y'=-\frac{sinx(b^2-a^2)}{(\sqrt{{(a^2-b^2)(1-cos^2x)}})(a+bcosx)}\\
y'=-\frac{sinx(b^2-a^2)}{\sqrt{{(a^2-b^2)}}\sqrt{(1-cos^2x)}(a+bcosx)}\\
y'=-\frac{sinx(b^2-a^2)}{\sqrt{{(a^2-b^2)}}(sinx)(a+bcosx)}\\
y'=\frac{a^2-b^2}{\sqrt{{(a^2-b^2)}}(a+bcosx)}\\
y'=\frac{\sqrt{(a^2-b^2)}}{(a+bcosx)}$
