If f(x) = operatorname{Sec}^{-1}left(frac{1}{ | vedclass

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(D) Let $x = \cos 	heta$. Then $f(x) = \operatorname{Sec}^{-1}\left(\frac{1}{2\cos^2 	heta - 1}ight) = \operatorname{Sec}^{-1}\left(\frac{1}{\cos

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$1/2$

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(D) Let $x = \cos 	heta$. Then $f(x) = \operatorname{Sec}^{-1}\left(\frac{1}{2\cos^2 	heta - 1}ight) = \operatorname{Sec}^{-1}\left(\frac{1}{\cos 2	heta}ight) = \operatorname{Sec}^{-1}(\sec 2	heta) = 2	heta = 2\cos^{-1}x$. Thus,$\frac{df}{dx} = 2 	imes \left(-\frac{1}{\sqrt{1-x^2}}ight) = -\frac{2}{\sqrt{1-x^2}}$. Now,let $x = 	an \phi$. Then $g(x) = \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+	an^2 \phi}-1}{	an \phi}ight) = \operatorname{Tan}^{-1}\left(\frac{\sec \phi - 1}{	an \phi}ight) = \operatorname{Tan}^{-1}\left(\frac{1-\cos \phi}{\sin \phi}ight) = \operatorname{Tan}^{-1}\left(	an \frac{\phi}{2}ight) = \frac{\phi}{2} = \frac{1}{2}	an^{-1}x$. Thus,$\frac{dg}{dx} = \frac{1}{2} 	imes \frac{1}{1+x^2} = \frac{1}{2(1+x^2)}$. The derivative of $f(x)$ with respect to $g(x)$ is $\frac{df/dx}{dg/dx} = \frac{-2/\sqrt{1-x^2}}{1/(2(1+x^2))} = -\frac{4(1+x^2)}{\sqrt{1-x^2}}$.

If $f(x) = \operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}\right)$ and $g(x) = \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$,then the derivative of $f(x)$ with respect to $g(x)$ is

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