Question

At \( x = \frac{\pi^2}{4} \), \( \frac{d}{dx} \left( \tan^{-1}(\cos\sqrt{x}) + \sec^{-1}(e^x) \right) = \)

• A. \( \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} - \frac{1}{\pi} \)
• B. \( \frac{\pi}{4} + \frac{1}{\sqrt{e^{\pi^2} + e^{\frac{\pi^2}{2}}}} \)
• C. \( \frac{1}{\sqrt{e^{\pi^2} + e^{\frac{\pi^2}{2}}}} + \frac{2}{\pi} \cot \left( \frac{\pi}{2} \right) \)
• D. \( \frac{1}{\sqrt{e^{\pi}}} + \frac{1}{\pi} \)

Hint:

Remember the derivatives of inverse trigonometric functions and use the chain rule appropriately. Also, remember that \( \sec^{-1}(x) \) is defined for \( |x| \ge 1 \), and its derivative is given by \( \frac{1}{|x|\sqrt{x^2 - 1}} \).

Answer 1

The Correct Option is A

Step 1: Differentiate the given expression with respect to \( x \)
Let \( y = \tan^{-1}(\cos\sqrt{x}) + \sec^{-1}(e^x) \). We want to find \( \frac{dy}{dx} \). Using the chain rule, we have: \[ \frac{dy}{dx} = \frac{d}{dx} \tan^{-1}(\cos\sqrt{x}) + \frac{d}{dx} \sec^{-1}(e^x) \] \[ = \frac{1}{1 + (\cos\sqrt{x})^2} \cdot \frac{d}{dx} (\cos\sqrt{x}) + \frac{1}{|e^x|\sqrt{e^{2x} - 1}} \cdot \frac{d}{dx} (e^x) \] \[ = \frac{1}{1 + \cos^2\sqrt{x}} \cdot (-\sin\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} + \frac{e^x}{e^x\sqrt{e^{2x} - 1}} \] \[ = -\frac{\sin\sqrt{x}}{2\sqrt{x}(1 + \cos^2\sqrt{x})} + \frac{1}{\sqrt{e^{2x} - 1}} \] Step 2: Evaluate the derivative at \( x = \frac{\pi^2}{4} \)
Substitute \( x = \frac{\pi^2}{4} \) into the derivative: \[ \frac{dy}{dx} \Bigg|_{x = \frac{\pi^2}{4}} = -\frac{\sin\sqrt{\frac{\pi^2}{4}}}{2\sqrt{\frac{\pi^2}{4}}(1 + \cos^2\sqrt{\frac{\pi^2}{4}})} + \frac{1}{\sqrt{e^{2(\frac{\pi^2}{4})} - 1}} \] \[ = -\frac{\sin\frac{\pi}{2}}{2(\frac{\pi}{2})(1 + \cos^2\frac{\pi}{2})} + \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} \] \[ = -\frac{1}{\pi(1 + 0)} + \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} \] \[ = -\frac{1}{\pi} + \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} \] \[ = \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} - \frac{1}{\pi} \] Therefore, the value of the given expression at \( x = \frac{\pi^2}{4} \) is \( \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} - \frac{1}{\pi} \), which matches option (A).

Answer 2

Step 1: Find the derivative of the given expression
We are given the function \( f(x) = \tan^{-1}(\cos(\sqrt{x})) + \sec^{-1}(e^x) \). To find the derivative at \( x = \frac{\pi^2}{4} \), we need to differentiate each term of the function using standard derivative rules.

Step 2: Derivative of \( \tan^{-1}(\cos(\sqrt{x})) \)
Using the chain rule, the derivative of \( \tan^{-1}(u) \) is: \[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \] Let \( u = \cos(\sqrt{x}) \), then: \[ \frac{du}{dx} = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \] Thus, the derivative of the first term is: \[ \frac{d}{dx} \tan^{-1}(\cos(\sqrt{x})) = \frac{1}{1 + \cos^2(\sqrt{x})} \cdot \left( -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \right) \] This simplifies to: \[ \frac{-\sin(\sqrt{x})}{2\sqrt{x}(1 + \cos^2(\sqrt{x}))} \]

Step 3: Derivative of \( \sec^{-1}(e^x) \)
Using the chain rule again, the derivative of \( \sec^{-1}(v) \) is: \[ \frac{d}{dx} \sec^{-1}(v) = \frac{1}{|v| \sqrt{v^2 - 1}} \cdot \frac{dv}{dx} \] Let \( v = e^x \), then: \[ \frac{dv}{dx} = e^x \] Thus, the derivative of the second term is: \[ \frac{d}{dx} \sec^{-1}(e^x) = \frac{1}{e^x \sqrt{e^{2x} - 1}} \cdot e^x \] This simplifies to: \[ \frac{1}{\sqrt{e^{2x} - 1}} \]

Step 4: Combine the derivatives
The derivative of the entire function is: \[ \frac{d}{dx} \left( \tan^{-1}(\cos(\sqrt{x})) + \sec^{-1}(e^x) \right) = \frac{-\sin(\sqrt{x})}{2\sqrt{x}(1 + \cos^2(\sqrt{x}))} + \frac{1}{\sqrt{e^{2x} - 1}} \]

Step 5: Evaluate at \( x = \frac{\pi^2}{4} \)
At \( x = \frac{\pi^2}{4} \), we first calculate the necessary values: - \( \cos(\sqrt{x}) = \cos\left(\frac{\pi}{2}\right) = 0 \) - \( e^x = e^{\frac{\pi^2}{4}} \) - \( \sin\left(\frac{\pi}{2}\right) = 1 \) Thus, the derivative at \( x = \frac{\pi^2}{4} \) becomes: \[ \frac{-0}{2\sqrt{\frac{\pi^2}{4}}(1 + 0)} + \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} = \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} - \frac{1}{\pi} \] Therefore, the answer is:

\( \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} - \frac{1}{\pi} \)