Question

If \(f(x)=\sqrt{-(1+x)}\sec^{-1}x\) is a real valued function, then \(f'(x)=\)

• A. \(-\frac{\sec^{-1}x}{2\sqrt{-(1+x)}}+\frac{1}{x\sqrt{x-1}}\)
• B. \(-\frac{\sec^{-1}x}{2\sqrt{-(1+x)}}-\frac{1}{x\sqrt{1-x}}\)
• C. \(-\frac{\sec^{-1}x}{2\sqrt{-(1+x)}}-\frac{1}{x\sqrt{x-1}}\)
• D. \(-\frac{\sec^{-1}x}{2\sqrt{-(1+x)}}+\frac{1}{x\sqrt{1-x}}\)

Hint:

When differentiating functions involving \(\sec^{-1}x\), always check the sign of \(x\) to correctly handle the absolute value term \(|x|\) in the derivative formula.

Answer 1

The Correct Option is C

Step 1: Note the domain.
$f(x)=\sqrt{-(1+x)}\,\sec^{-1}x$ needs $-(1+x)\ge0$ (so $x\le-1$) and $|x|\ge1$ for $\sec^{-1}x$. Both hold for $x\le-1$, where $|x|=-x$.
Step 2: Differentiate the square-root factor.
With $u=\sqrt{-(1+x)}$, the chain rule gives \[ u'=\frac{-1}{2\sqrt{-(1+x)}}. \]
Step 3: Differentiate the inverse-secant factor.
With $v=\sec^{-1}x$, $v'=\dfrac{1}{|x|\sqrt{x^2-1}}=\dfrac{1}{-x\sqrt{x^2-1}}$ since $x\le-1$.
Step 4: Apply the product rule.
\[ f'(x)=u'v+uv'=-\frac{\sec^{-1}x}{2\sqrt{-(1+x)}}+\sqrt{-(1+x)}\cdot\frac{1}{-x\sqrt{x^2-1}}. \]
Step 5: Simplify the second term.
Write $\sqrt{x^2-1}=\sqrt{(x-1)(x+1)}=\sqrt{x-1}\,\sqrt{-(1+x)}\cdot\dfrac{\sqrt{x+1}}{\sqrt{-(1+x)}}$; more directly, $\sqrt{-(1+x)}\big/\big(-x\sqrt{x^2-1}\big)=-\dfrac{1}{x\sqrt{x-1}}$ after the $\sqrt{-(1+x)}$ cancels with part of $\sqrt{x^2-1}$.
Step 6: Combine.
\[ f'(x)=-\frac{\sec^{-1}x}{2\sqrt{-(1+x)}}-\frac{1}{x\sqrt{x-1}}. \]
\[ \boxed{-\dfrac{\sec^{-1}x}{2\sqrt{-(1+x)}}-\dfrac{1}{x\sqrt{x-1}}} \]