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: \[ \frac{d}{dx} \left[ {Tan}^{-1} (\tan \frac{x}{2}) \right] = {sech } x. \]

Answer 2

To determine which of the given derivatives is false, we need to evaluate each expression:

• \(\frac{d}{dx} \left[ \text{sec}^{-1} (\cosh x) \right]\)

  The derivative of \(\text{sec}^{-1}(x)\) is \(\frac{1}{|x|\sqrt{x^2-1}}\), but since \(\cosh x\ge 1\) and secant identity is valid, the derivative using the chain rule is \(\frac{\sinh x}{|\cosh x| \sqrt{\cosh^2 x -1}}=\text{sech} x\).

• \(\frac{d}{dx} \left[ \text{cos}^{-1} (\text{sech } x) \right]\)

  The derivative of \(\text{cos}^{-1}(x)\) is \(-\frac{1}{\sqrt{1-x^2}}\) and using the chain rule gives \(-\frac{\text{sech}\,x\,\tanh x}{\sqrt{1-\text{sech}^2 x}}\), simplifying independently does not result in \(\text{sech } x\).

• \(\frac{d}{dx} \left[ \text{tan}^{-1} (\sinh x) \right]\)

  The derivative of \(\text{tan}^{-1}(x)\) is \(\frac{1}{1+x^2}\), therefore using the chain rule leads to \(\frac{\cosh x}{1+\sinh^2 x}\), simplifying becomes \(\text{sech } x\).

• \(\frac{d}{dx} \left[ \text{tan}^{-1} (\tan \frac{x}{2}) \right] \)

  The derivative of this expression simplifies to \( \frac{1}{2}\), as this is the derivative of \(\tan^{-1}(\tan(y))\) with respect to \(y\), not related to \(\text{sech } x\).

The false statement is:

\(\frac{d}{dx} \left[ \text{tan}^{-1} (\tan \frac{x}{2}) \right] = \text{sech } x\)