Question

If $y = \sec^{-1}(\frac{x+x^{-1}}{x-x^{-1}})$, then $\frac{dy}{dx} =$

• A. $\frac{-2}{1+x^2}$
• B. $\frac{-1}{1+x^2}$
• C. $\frac{2}{1-x^2}$
• D. $\frac{1}{1+x^2}$

Answer 1

Answer: (A)

$\frac{-2}{1+x^2}$

Answer 2

The given function is $y = \sec^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)$.

First, simplify the argument of the $\sec^{-1}$ function:

$\frac{x+x^{-1}}{x-x^{-1}} = \frac{x+\frac{1}{x}}{x-\frac{1}{x}} = \frac{\frac{x^2+1}{x}}{\frac{x^2-1}{x}} = \frac{x^2+1}{x^2-1}$.

So, $y = \sec^{-1}\left(\frac{x^2+1}{x^2-1}\right)$.

We can use the identity $\sec^{-1}(u) = \cos^{-1}\left(\frac{1}{u}\right)$ for $|u| \ge 1$.

The argument is $u = \frac{x^2+1}{x^2-1}$. For the function to be defined, we must have $|u| \ge 1$, i.e., $\left|\frac{x^2+1}{x^2-1}\right| \ge 1$.

Since $x^2+1 > 0$, this is equivalent to $\frac{x^2+1}{|x^2-1|} \ge 1$, which means $x^2+1 \ge |x^2-1|$.

If $x^2-1 > 0$, i.e., $|x| > 1$, then $x^2+1 \ge x^2-1$, which is true ($1 \ge -1$).

If $x^2-1 < 0$, i.e., $|x| < 1$ and $x \neq 0$, then $x^2+1 \ge -(x^2-1) = 1-x^2$, which means $2x^2 \ge 0$, which is true for $x \neq 0$.

The function is defined for $x \in \mathbb{R} \setminus \{-1, 0, 1\}$.

Case 1: $|x| > 1$.

In this case, $x^2-1 > 0$, and $\frac{x^2+1}{x^2-1} > 1$.

$y = \sec^{-1}\left(\frac{x^2+1}{x^2-1}\right) = \cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)$.

Let $x = \cot \theta$. Since $|x| > 1$, we have $0 < \theta < \frac{\pi}{4}$ (if $x > 1$) or $\frac{3\pi}{4} < \theta < \pi$ (if $x < -1$).

$\frac{x^2-1}{x^2+1} = \frac{\cot^2 \theta - 1}{\cot^2 \theta + 1} = \frac{\frac{\cos^2 \theta}{\sin^2 \theta} - 1}{\frac{\cos^2 \theta}{\sin^2 \theta} + 1} = \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta + \sin^2 \theta} = \cos(2\theta)$.

So, $y = \cos^{-1}(\cos(2\theta))$.

If $x > 1$, then $0 < \theta < \frac{\pi}{4}$, so $0 < 2\theta < \frac{\pi}{2}$. In this range, $\cos^{-1}(\cos(2\theta)) = 2\theta$.

$y = 2\theta = 2\cot^{-1}(x)$.

$\frac{dy}{dx} = 2 \cdot \frac{-1}{1+x^2} = \frac{-2}{1+x^2}$.

If $x < -1$, then $\frac{3\pi}{4} < \theta < \pi$, so $\frac{3\pi}{2} < 2\theta < 2\pi$.

In this range, $\cos(2\theta) = \cos(2\pi - 2\theta)$. Also, $0 < 2\pi - 2\theta < \frac{\pi}{2}$.

So $\cos^{-1}(\cos(2\theta)) = \cos^{-1}(\cos(2\pi - 2\theta)) = 2\pi - 2\theta$.

$y = 2\pi - 2\theta = 2\pi - 2\cot^{-1}(x)$.

$\frac{dy}{dx} = 0 - 2 \cdot \frac{-1}{1+x^2} = \frac{2}{1+x^2}$.

The derivative depends on whether $x > 1$ or $x < -1$. However, the options are single expressions. Let's check the other case.

Case 2: $|x| < 1$ and $x \neq 0$.

In this case, $x^2-1 < 0$, and $\frac{x^2+1}{x^2-1} < -1$.

$y = \sec^{-1}\left(\frac{x^2+1}{x^2-1}\right)$.

Let $x = \tan \theta$. Since $|x| < 1, x \neq 0$, we have $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, $\theta \neq 0$.

$\frac{x^2+1}{x^2-1} = \frac{\tan^2 \theta + 1}{\tan^2 \theta - 1} = \frac{1+\tan^2 \theta}{-(1-\tan^2 \theta)} = \frac{\sec^2 \theta}{-\cos(2\theta)/\cos^2 \theta} = \frac{1/\cos^2 \theta}{-\cos(2\theta)/\cos^2 \theta} = -\frac{1}{\cos(2\theta)} = -\sec(2\theta)$.

So, $y = \sec^{-1}(-\sec(2\theta))$.

If $-\frac{\pi}{4} < \theta < \frac{\pi}{4}, \theta \neq 0$, then $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}, 2\theta \neq 0$.

In this range, $\cos(2\theta) > 0$, so $\sec(2\theta) > 0$.

Let $v = -\sec(2\theta)$. Then $v < 0$.

We need to evaluate $\sec^{-1}(v)$ where $v < -1$.

Let $v = -\sec(2\theta)$. Since $\sec^{-1}(-u) = \pi - \sec^{-1}(u)$ for $u \ge 1$.

$y = \sec^{-1}(-\sec(2\theta))$. Here $u = \sec(2\theta) = \frac{1}{\cos(2\theta)}$.

Since $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}, 2\theta \neq 0$, we have $\cos(2\theta) > 0$.

If $0 < |x| < 1$, then $0 < x^2 < 1$, so $-1 < x^2-1 < 0$. $x^2+1 > 1$.

$\frac{x^2+1}{x^2-1} < -1$.

So $-\sec(2\theta) < -1$, which means $\sec(2\theta) > 1$. This is true since $0 < |2\theta| < \frac{\pi}{2}$.

$y = \pi - \sec^{-1}(\sec(2\theta))$.

Since $0 < |2\theta| < \frac{\pi}{2}$, we have $2\theta \in (-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2})$.

If $0 < 2\theta < \frac{\pi}{2}$ (i.e., $0 < \theta < \frac{\pi}{4}$, which means $0 < x < 1$), then $\sec^{-1}(\sec(2\theta)) = 2\theta$.

$y = \pi - 2\theta = \pi - 2\tan^{-1}(x)$.

$\frac{dy}{dx} = 0 - 2 \cdot \frac{1}{1+x^2} = \frac{-2}{1+x^2}$.

If $-\frac{\pi}{2} < 2\theta < 0$ (i.e., $-\frac{\pi}{4} < \theta < 0$, which means $-1 < x < 0$), then $\sec^{-1}(\sec(2\theta)) = -2\theta$.

$y = \pi - (-2\theta) = \pi + 2\theta = \pi + 2\tan^{-1}(x)$.

$\frac{dy}{dx} = 0 + 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}$.

The derivative is $\frac{-2}{1+x^2}$ for $|x| > 1$ and for $0 < x < 1$.

The derivative is $\frac{2}{1+x^2}$ for $-1 < x < 0$.

Let's look at the options again. Option (A) is $\frac{-2}{1+x^2}$. This matches the derivative for $x \in (0, 1) \cup (|x| > 1)$.

Option (D) is $\frac{1}{1+x^2}$.

Option (B) is $\frac{-1}{1+x^2}$.

Option (C) is $\frac{2}{1-x^2}$.

It seems there might be a domain restriction implied or a standard convention used in the problem. Often in such problems, the principal values are considered, and the domain is chosen such that the simplification holds directly.

Let's re-examine the substitution $x = \cot \theta$.

$y = \sec^{-1}(\sec(2\theta))$.

The principal range of $\sec^{-1}(u)$ is $[0, \pi] - \{\frac{\pi}{2}\}$.

We need $2\theta$ to be in this range.

If $x > 0$, then $0 < \theta < \frac{\pi}{2}$. So $0 < 2\theta < \pi$.

If $2\theta \neq \frac{\pi}{2}$, i.e., $\theta \neq \frac{\pi}{4}$, i.e., $x \neq \cot(\frac{\pi}{4}) = 1$.

For $x > 0, x \neq 1$, we have $2\theta \in (0, \pi) - \{\frac{\pi}{2}\}$.

In this range, $\sec^{-1}(\sec(2\theta)) = 2\theta$.

So for $x > 0, x \neq 1$, $y = 2\theta = 2\cot^{-1}(x)$.

$\frac{dy}{dx} = 2 \cdot \frac{-1}{1+x^2} = \frac{-2}{1+x^2}$.

If $x < 0$, then $\frac{\pi}{2} < \theta < \pi$. So $\pi < 2\theta < 2\pi$.

We need to express $\sec(2\theta)$ as $\sec(\phi)$ where $\phi \in [0, \pi] - \{\frac{\pi}{2}\}$.

Let $\phi = 2\pi - 2\theta$. Then $0 < \phi < \pi$.

$\sec(\phi) = \sec(2\pi - 2\theta) = \sec(2\theta)$.

So $y = \sec^{-1}(\sec(2\theta)) = \sec^{-1}(\sec(2\pi - 2\theta))$.

If $2\pi - 2\theta \in [0, \pi] - \{\frac{\pi}{2}\}$, then $y = 2\pi - 2\theta$.

$2\pi - 2\theta = \frac{\pi}{2}$ implies $2\theta = \frac{3\pi}{2}$, $\theta = \frac{3\pi}{4}$, $x = \cot(\frac{3\pi}{4}) = -1$. The function is undefined at $x=-1$.

For $x < 0, x \neq -1$, we have $\frac{\pi}{2} < \theta < \pi, \theta \neq \frac{3\pi}{4}$.

So $\pi < 2\theta < 2\pi, 2\theta \neq \frac{3\pi}{2}$.

$0 < 2\pi - 2\theta < \pi, 2\pi - 2\theta \neq \frac{\pi}{2}$.

So $2\pi - 2\theta \in (0, \pi) - \{\frac{\pi}{2}\}$.

Thus, for $x < 0, x \neq -1$, $y = 2\pi - 2\theta = 2\pi - 2\cot^{-1}(x)$.

$\frac{dy}{dx} = 0 - 2 \cdot \frac{-1}{1+x^2} = \frac{2}{1+x^2}$.

The derivative is $\frac{-2}{1+x^2}$ for $x > 0, x \neq 1$.

The derivative is $\frac{2}{1+x^2}$ for $x < 0, x \neq -1$.

Let's consider the substitution $x = \tan \theta$.

$y = \sec^{-1}(-\sec(2\theta))$.

Principal range of $\sec^{-1}$ is $[0, \pi] - \{\frac{\pi}{2}\}$.

If $0 < x < 1$, then $0 < \theta < \frac{\pi}{4}$, so $0 < 2\theta < \frac{\pi}{2}$. $\sec(2\theta) > 1$. $-\sec(2\theta) < -1$.

Let $u = \sec(2\theta)$. $y = \sec^{-1}(-u) = \pi - \sec^{-1}(u) = \pi - \sec^{-1}(\sec(2\theta)) = \pi - 2\theta = \pi - 2\tan^{-1}(x)$.

$\frac{dy}{dx} = 0 - 2 \cdot \frac{1}{1+x^2} = \frac{-2}{1+x^2}$.

If $x > 1$, then $\frac{\pi}{4} < \theta < \frac{\pi}{2}$, so $\frac{\pi}{2} < 2\theta < \pi$. $\sec(2\theta) < -1$.

$y = \sec^{-1}(-\sec(2\theta))$. Let $v = -\sec(2\theta)$. $v > 1$.

Let $\phi = 2\theta - \pi$. Then $-\frac{\pi}{2} < \phi < 0$. $\sec(\phi) = \sec(2\theta - \pi) = \sec(\pi - 2\theta) = -\sec(2\theta)$.

So $y = \sec^{-1}(\sec(\phi))$. Since $-\frac{\pi}{2} < \phi < 0$, $\sec(\phi) > 1$.

$y = \sec^{-1}(\sec(\phi)) = -\phi = -(2\theta - \pi) = \pi - 2\theta = \pi - 2\tan^{-1}(x)$.

$\frac{dy}{dx} = \frac{-2}{1+x^2}$.

If $-1 < x < 0$, then $-\frac{\pi}{4} < \theta < 0$, so $-\frac{\pi}{2} < 2\theta < 0$. $\sec(2\theta) > 1$. $-\sec(2\theta) < -1$.

$y = \sec^{-1}(-\sec(2\theta))$. Let $u = \sec(2\theta)$. $y = \pi - \sec^{-1}(u) = \pi - \sec^{-1}(\sec(2\theta)) = \pi - (-2\theta) = \pi + 2\theta = \pi + 2\tan^{-1}(x)$.

$\frac{dy}{dx} = 0 + 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}$.

If $x < -1$, then $-\frac{\pi}{2} < \theta < -\frac{\pi}{4}$, so $-\pi < 2\theta < -\frac{\pi}{2}$. $\sec(2\theta) < -1$.

$y = \sec^{-1}(-\sec(2\theta))$. Let $v = -\sec(2\theta)$. $v > 1$.

Let $\phi = 2\theta + \pi$. Then $\frac{\pi}{2} < \phi < \pi$. $\sec(\phi) = \sec(2\theta + \pi) = -\sec(2\theta)$.

$y = \sec^{-1}(\sec(\phi))$. Since $\frac{\pi}{2} < \phi < \pi$, $\sec(\phi) < -1$.

$y = \phi = 2\theta + \pi = 2\tan^{-1}(x) + \pi$.

$\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}$.

Summary of derivatives:

If $x \in (0, 1) \cup (1, \infty)$, $\frac{dy}{dx} = \frac{-2}{1+x^2}$.

If $x \in (-\infty, -1) \cup (-1, 0)$, $\frac{dy}{dx} = \frac{2}{1+x^2}$.

Since option (A) $\frac{-2}{1+x^2}$ is given, it is likely that the domain considered in the question is $x > 0$. In many standard problems involving inverse trigonometric functions, a specific domain is implicitly assumed to get a single answer. If the question is from a single-choice context, and option (A) is the intended answer, the domain $x>0$ is implied.

Let's verify if there is any property that leads to a single derivative.

Consider $y = \cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)$.

Let $x^2 = t$. $y = \cos^{-1}\left(\frac{t-1}{t+1}\right)$.

Let $\frac{t-1}{t+1} = -\cos \alpha$.

$\frac{1-t}{1+t} = \cos \alpha$.

If $t = \tan^2 \beta$, $\frac{1-\tan^2 \beta}{1+\tan^2 \beta} = \cos(2\beta)$.

So $\frac{x^2-1}{x^2+1} = -\cos(2\beta)$.

$y = \cos^{-1}(-\cos(2\beta)) = \pi - \cos^{-1}(\cos(2\beta))$.

If $x > 0$, let $x = \tan \theta$, $x^2 = \tan^2 \theta$. Let $\beta = \theta$.

$y = \pi - \cos^{-1}(\cos(2\theta))$.

If $x > 0$, $0 < \theta < \frac{\pi}{2}$. $0 < 2\theta < \pi$.

If $0 < 2\theta < \pi$, $\cos^{-1}(\cos(2\theta)) = 2\theta$.

$y = \pi - 2\theta = \pi - 2\tan^{-1}(x)$.

$\frac{dy}{dx} = -2 \frac{1}{1+x^2}$. This is valid for $x > 0$.

If $x < 0$, let $x = -\tan \phi$, where $\phi > 0$.

$x^2 = \tan^2 \phi$.

$y = \cos^{-1}(\cos(2\phi))$.

If $x < 0$, $x = -\tan \phi$. We need to relate $\phi$ to $x$. $\phi = \tan^{-1}(-x)$.

$y = \cos^{-1}(\cos(2\tan^{-1}(-x)))$.

If $x < 0$, let $x = -a$ where $a > 0$. $y = \cos^{-1}\left(\frac{a^2-1}{a^2+1}\right)$.

Let $a = \cot \theta$, $0 < \theta < \frac{\pi}{2}$.

$y = \cos^{-1}(\cos(2\theta))$. $0 < 2\theta < \pi$.

$y = 2\theta = 2\cot^{-1}(a) = 2\cot^{-1}(-x)$.

$\frac{dy}{dx} = 2 \cdot \frac{-1}{1+(-x)^2} \cdot (-1) = \frac{2}{1+x^2}$.

The derivative is $\frac{-2}{1+x^2}$ for $x > 0$ and $\frac{2}{1+x^2}$ for $x < 0$.

The options suggest a single answer. Option (A) is $\frac{-2}{1+x^2}$. This is the result when $x > 0$.

Let's check if the question implies a specific domain. The form $\frac{x+x^{-1}}{x-x^{-1}}$ suggests $x \neq 0$. The term $x-x^{-1}$ in the denominator suggests $x \neq \pm 1$.

The domain of $\sec^{-1}(u)$ is $|u| \ge 1$. We found that this is true for all $x \in \mathbb{R} \setminus \{-1, 0, 1\}$.

Consider the derivative of $\sec^{-1}(u)$ is $\frac{1}{|u|\sqrt{u^2-1}}$.

If we are required to choose one option, and option (A) appears as a valid derivative for a significant part of the domain ($x>0$), it is likely the intended answer.

Let's check the derivative formula for $\sec^{-1}(u)$ again.

$\frac{d}{du}(\sec^{-1} u) = \frac{1}{|u|\sqrt{u^2-1}}$.

$u = \frac{x^2+1}{x^2-1}$. $\frac{du}{dx} = \frac{-4x}{(x^2-1)^2}$.

$\sqrt{u^2-1} = \frac{2|x|}{|x^2-1|}$.

$\frac{dy}{dx} = \frac{1}{\left|\frac{x^2+1}{x^2-1}\right| \frac{2|x|}{|x^2-1|}} \cdot \frac{-4x}{(x^2-1)^2} = \frac{|x^2-1|^2}{(x^2+1) 2|x|} \frac{-4x}{(x^2-1)^2} = \frac{(x^2-1)^2}{(x^2+1) 2|x|} \frac{-4x}{(x^2-1)^2} = \frac{-4x}{(x^2+1) 2|x|} = \frac{-2x}{(x^2+1)|x|}$.

If $x > 0$, $|x| = x$, $\frac{dy}{dx} = \frac{-2x}{(x^2+1)x} = \frac{-2}{x^2+1}$.

If $x < 0$, $|x| = -x$, $\frac{dy}{dx} = \frac{-2x}{(x^2+1)(-x)} = \frac{2}{x^2+1}$.

The derivative is $\frac{-2}{1+x^2}$ for $x > 0$ and $\frac{2}{1+x^2}$ for $x < 0$.

Assuming the question implicitly considers the domain $x>0$ (as is common when a single option is correct), the derivative is $\frac{-2}{1+x^2}$.