Question

Which one of the following is false?

• A. \( \frac{d}{dx} \left[ {Sec}^{-1} (\cosh x) \right] = {sech } x \)
• B. \( \frac{d}{dx} \left[ {Cos}^{-1} ({sech } x) \right] = {sech } x \)
• C. \( \frac{d}{dx} \left[ {Tan}^{-1} (\sinh x) \right] = {sech } x \)
• D. \( \frac{d}{dx} \left[ {Tan}^{-1} (\tan \frac{x}{2}) \right] = {sech } x \)

Hint:

To differentiate inverse trigonometric and hyperbolic functions, use their standard derivative formulas and simplify carefully.

Answer 1

The Correct Option is D

Step 1: Differentiating each expression 
1. For \( \frac{d}{dx} \left[ {Sec}^{-1} (\cosh x) \right] \): \[ \frac{d}{dx} \left[ {Sec}^{-1} (\cosh x) \right] = \frac{1}{\cosh x \sqrt{\cosh^2 x - 1}} \cdot \sinh x. \] Since \( \cosh^2 x - 1 = \sinh^2 x \), we get: \[ = \frac{\sinh x}{\cosh x \sinh x} = \frac{1}{\cosh x} = {sech } x. \] Thus, (1) is true. 2. For \( \frac{d}{dx} \left[ {Cos}^{-1} ({sech } x) \right] \): \[ \frac{d}{dx} \left[ {Cos}^{-1} ({sech } x) \right] = \frac{-1}{\sqrt{1 - {sech}^2 x}} \cdot (-{sech } x \tanh x). \] Since \( 1 - {sech}^2 x = -\tanh^2 x \), we get: \[ = \frac{{sech } x \tanh x}{\tanh x} = {sech } x. \] Thus, (2) is true. 3. For \( \frac{d}{dx} \left[ {Tan}^{-1} (\sinh x) \right] \): \[ \frac{1}{1 + \sinh^2 x} \cdot \cosh x. \] Since \( 1 + \sinh^2 x = \cosh^2 x \), we get: \[ = \frac{\cosh x}{\cosh^2 x} = \frac{1}{\cosh x} = {sech } x. \] Thus, (3) is true. 4. For \( \frac{d}{dx} \left[ {Tan}^{-1} (\tan \frac{x}{2}) \right] \): Since \( {Tan}^{-1} (\tan \frac{x}{2}) \) simplifies to \( \frac{x}{2} \) for values where it is defined, its derivative is: \[ \frac{d}{dx} \left( \frac{x}{2} \right) = \frac{1}{2}. \] Since \( \frac{1}{2} \neq {sech } x \), this statement is false. 
Step 2: Conclusion The false statement is: \[ \frac{d}{dx} \left[ {Tan}^{-1} (\tan \frac{x}{2}) \right] = {sech } x. \]