If y = sec^{-1} left( frac{2x}{1 + x^2} right | vedclass

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(C) Let $y = \sec^{-1} \left( \frac{2x}{1 + x^2} \right) + \sin^{-1} \left( \frac{x - 1}{x + 1} \right)$.For the term $\sec^{-1} \left( \frac{2x}{1 +

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Does not exist

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(C) Let $y = \sec^{-1} \left( \frac{2x}{1 + x^2} \right) + \sin^{-1} \left( \frac{x - 1}{x + 1} \right)$.For the term $\sec^{-1} \left( \frac{2x}{1 + x^2} \right)$,the domain of $\sec^{-1}(u)$ is $|u| \ge 1$.Thus,we must have $\left| \frac{2x}{1 + x^2} \right| \ge 1$.This implies $2|x| \ge 1 + x^2$,which simplifies to $x^2 - 2|x| + 1 \le 0$,or $(|x| - 1)^2 \le 0$.Since a square cannot be negative,this is only possible if $|x| - 1 = 0$,which means $x = 1$ or $x = -1$.Since the function $y$ is defined only at discrete points $x = 1$ and $x = -1$,it is not defined on any interval.Therefore,the derivative $\frac{dy}{dx}$ does not exist.

If $y = \sec^{-1} \left( \frac{2x}{1 + x^2} \right) + \sin^{-1} \left( \frac{x - 1}{x + 1} \right)$,then $\frac{dy}{dx}$ is equal to

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