The derivative of {cos ^{ - 1}}left( {frac{{1 | vedclass

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(C) Let ${y_1} = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} ight)$.Using the substitution $x = 	an 	heta$,we get ${y_1} = {\cos ^{ - 1

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$\frac{2}{3}$

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(C) Let ${y_1} = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} ight)$.Using the substitution $x = 	an 	heta$,we get ${y_1} = {\cos ^{ - 1}}(\cos 2	heta) = 2	heta = 2{	an ^{ - 1}}x$.Thus,$\frac{{d{y_1}}}{{dx}} = \frac{2}{{1 + {x^2}}}$.Let ${y_2} = {\cot ^{ - 1}}\left( {\frac{{1 - 3{x^2}}}{{3x - {x^3}}}} ight)$.Using the substitution $x = 	an 	heta$,we get ${y_2} = {\cot ^{ - 1}}\left( {\frac{{1 - 3	an^2 	heta}}{{3	an 	heta - 	an^3 	heta}}} ight) = {\cot ^{ - 1}}(\cot 3	heta) = 3	heta = 3{	an ^{ - 1}}x$.Thus,$\frac{{d{y_2}}}{{dx}} = \frac{3}{{1 + {x^2}}}$.Therefore,the derivative of ${y_1}$ with respect to ${y_2}$ is $\frac{{d{y_1}}}{{d{y_2}}} = \frac{{d{y_1}/dx}}{{d{y_2}/dx}} = \frac{2/(1 + {x^2})}{3/(1 + {x^2})} = \frac{2}{3}$.

The derivative of ${\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ with respect to ${\cot ^{ - 1}}\left( {\frac{{1 - 3{x^2}}}{{3x - {x^3}}}} \right)$ is

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