Question:
If \(f(x)=\sqrt{-(1+x)}\sec^{-1}x\) is a real valued function, then \(f'(x)=\)
If \(f(x)=\sqrt{-(1+x)}\sec^{-1}x\) is a real valued function, then \(f'(x)=\)
Show 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.
Updated On: Jun 9, 2026
- \(-\frac{\sec^{-1}x}{2\sqrt{-(1+x)}}+\frac{1}{x\sqrt{x-1}}\)
- \(-\frac{\sec^{-1}x}{2\sqrt{-(1+x)}}-\frac{1}{x\sqrt{1-x}}\)
- \(-\frac{\sec^{-1}x}{2\sqrt{-(1+x)}}-\frac{1}{x\sqrt{x-1}}\)
- \(-\frac{\sec^{-1}x}{2\sqrt{-(1+x)}}+\frac{1}{x\sqrt{1-x}}\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
We use the product rule for differentiation, \((uv)' = u'v + uv'\), where \(u = \sqrt{-(1+x)}\) and \(v = \sec^{-1}x\).
First, we observe the domain: \(\sec^{-1}x\) is defined for \(|x| \ge 1\), and the square root \(\sqrt{-(1+x)}\) is defined for \(x \le -1\). Thus, the function is only valid for \(x \le -1\). In this range, \(|x| = -x\), and the derivative of \(\sec^{-1}x\) is \(\frac{1}{|x|\sqrt{x^2-1}} = \frac{1}{-x\sqrt{x^2-1}}\).
Step 1: Differentiate \(u = \sqrt{-(1+x)}\).
Applying the chain rule: \[ \frac{du}{dx} = \frac{1}{2\sqrt{-(1+x)}} \cdot \frac{d}{dx}(-(1+x)) \] \[ \frac{du}{dx} = -\frac{1}{2\sqrt{-(1+x)}} \]
Step 2: Differentiate \(v = \sec^{-1}x\) for \(x \le -1\).
Using the standard derivative formula for \(\sec^{-1}x\): \[ \frac{dv}{dx} = \frac{1}{|x|\sqrt{x^2-1}} \] Since \(x \le -1\), we substitute \(|x| = -x\): \[ \frac{dv}{dx} = \frac{1}{-x\sqrt{x^2-1}} \]
Step 3: Apply the product rule \((uv)' = u'v + uv'\).
\[ f'(x) = \left( -\frac{1}{2\sqrt{-(1+x)}} \right) \sec^{-1}x + \sqrt{-(1+x)} \left( \frac{1}{-x\sqrt{x^2-1}} \right) \] We simplify the second term by noting \(\sqrt{x^2-1} = \sqrt{-(1-x^2)} = \sqrt{-(1-x)(1+x)}\): \[ \frac{\sqrt{-(1+x)}}{-x\sqrt{-(1+x)}\sqrt{x-1}} = -\frac{1}{x\sqrt{x-1}} \] Combining these gives the final derivative: \[ f'(x) = -\frac{\sec^{-1}x}{2\sqrt{-(1+x)}} - \frac{1}{x\sqrt{x-1}} \] center minipage0.6
f'(x) = \(-\frac{\sec^{-1}x}{2\sqrt{-(1+x)}}-\frac{1}{x\sqrt{x-1}}\) minipage center
Step 1: Differentiate \(u = \sqrt{-(1+x)}\).
Applying the chain rule: \[ \frac{du}{dx} = \frac{1}{2\sqrt{-(1+x)}} \cdot \frac{d}{dx}(-(1+x)) \] \[ \frac{du}{dx} = -\frac{1}{2\sqrt{-(1+x)}} \]
Step 2: Differentiate \(v = \sec^{-1}x\) for \(x \le -1\).
Using the standard derivative formula for \(\sec^{-1}x\): \[ \frac{dv}{dx} = \frac{1}{|x|\sqrt{x^2-1}} \] Since \(x \le -1\), we substitute \(|x| = -x\): \[ \frac{dv}{dx} = \frac{1}{-x\sqrt{x^2-1}} \]
Step 3: Apply the product rule \((uv)' = u'v + uv'\).
\[ f'(x) = \left( -\frac{1}{2\sqrt{-(1+x)}} \right) \sec^{-1}x + \sqrt{-(1+x)} \left( \frac{1}{-x\sqrt{x^2-1}} \right) \] We simplify the second term by noting \(\sqrt{x^2-1} = \sqrt{-(1-x^2)} = \sqrt{-(1-x)(1+x)}\): \[ \frac{\sqrt{-(1+x)}}{-x\sqrt{-(1+x)}\sqrt{x-1}} = -\frac{1}{x\sqrt{x-1}} \] Combining these gives the final derivative: \[ f'(x) = -\frac{\sec^{-1}x}{2\sqrt{-(1+x)}} - \frac{1}{x\sqrt{x-1}} \] center minipage0.6
f'(x) = \(-\frac{\sec^{-1}x}{2\sqrt{-(1+x)}}-\frac{1}{x\sqrt{x-1}}\) minipage center
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Differentiation Questions
- If \[ \frac{d}{dx} \left(\frac{1 + x^2 + x^4}{1 + x + x^2}\right) = ax + b, \] then \( (a,b) \) is:
If $$ y = \sin^{-1} x, $$ then $$ (1 - x^2)y_2 - xy_1 = 0. $$
- The distance \( s \) traveled by a particle in time \( t \) is given by: $$ s = 4t^2 + 2t + 3. $$ The velocity of the particle when \( t = 3 \) seconds is:
If \( y = \tan(\log x) \), then \( \frac{d^2y}{dx^2} \) is given by:
For \( x<0 \), \( \frac{d}{dx} [|x|^x] \) is given by:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions