Question

At $ x = \frac{\pi^2}{4} $, $ \frac{d}{dx} \left( \operatorname{Tan}^{-1}(\cos \sqrt{x}) + \operatorname{Sec}^{-1}(e^x) \right) = $
At $ x = \frac{\pi^2}{4} $, $ \frac{d}{dx} \left( \operatorname{Tan}^{-1}(\cos \sqrt{x}) + \operatorname{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^{\pi^2/2}}} $
• C. $ \frac{1}{\sqrt{e^{\pi^2} + e^{\pi^2/2}}} + \frac{2}{\pi} \cot \left( \frac{\sqrt{\pi^2}}{2} \right) $
• D. $ \frac{1}{\sqrt{e^{\pi^2}}} + \frac{1}{\pi} $

Hint:

Be careful with the argument of the square root in the derivative of $ \operatorname{Sec}^{-1}(u) $.

Answer 1

The Correct Option is A

Step 1: Differentiate $ \operatorname{Tan}^{-1}(\cos \sqrt{x}) $.} $ \frac{d}{dx}(\operatorname{Tan}^{-1}(\cos \sqrt{x})) = \frac{-\sin \sqrt{x}}{1 + \cos^2 \sqrt{x}} \cdot \frac{1}{2\sqrt{x}} $ At $ x = \frac{\pi^2}{4} $: $ \frac{-\sin(\pi/2)}{1 + \cos^2(\pi/2)} \cdot \frac{1}{2(\pi/2)} = \frac{-1}{1 + 0} \cdot \frac{1}{\pi} = -\frac{1}{\pi} $ 
Step 2: Differentiate $ \operatorname{Sec}^{-1}(e^x) $.} $ \frac{d}{dx}(\operatorname{Sec}^{-1}(e^x)) = \frac{1}{|e^x|\sqrt{e^{2x} - 1}} \cdot e^x = \frac{1}{e^x\sqrt{e^{2x} - 1}} \cdot e^x = \frac{1}{\sqrt{e^{2x} - 1}} $ At $ x = \frac{\pi^2}{4} $: $ \frac{1}{\sqrt{e^{2(\pi^2/4)} - 1}} = \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} $ 
Step 3: Add the derivatives.
$ -\frac{1}{\pi} + \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} = \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} - \frac{1}{\pi} $ 
Step 4: Conclusion.
The value of the derivative at $ x = \frac{\pi^2}{4} $ is $ \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} - \frac{1}{\pi} $.