Question

If \( y = \cos^{-1}(\mathrm{e}^x) \), then \( \frac{dy}{dx} \) is:

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

Hint:

To differentiate inverse trigonometric functions, always apply the chain rule and simplify carefully. Pay attention to the domain of the functions to ensure valid results.

Answer 1

The Correct Option is D

Step 1: {Differentiate \( y = \cos^{-1}(\mathrm{e}^x) \)}
The derivative of \( \cos^{-1}(u) \) with respect to \( x \) is: \[ \frac{d}{dx} \cos^{-1}(u) = \frac{-1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}. \] Here, \( u = \mathrm{e}^x \), so: \[ \frac{du}{dx} = \mathrm{e}^x. \] Step 2: {Substitute \( u = \mathrm{e}^x \)}
Using the formula: \[ \frac{dy}{dx} = \frac{-1}{\sqrt{1 - (\mathrm{e}^x)^2}} \cdot \mathrm{e}^x. \] Step 3: {Simplify the denominator}
Simplify \( 1 - (\mathrm{e}^x)^2 \): \[ 1 - (\mathrm{e}^x)^2 = 1 - \mathrm{e}^{2x} = \mathrm{e}^{-2x} - 1. \] Thus, \[ \frac{dy}{dx} = \frac{-1}{\sqrt{\mathrm{e}^{-2x} - 1}}. \]