Question

If $f(x) = \sqrt{1 + \cos^2(x^2)}$, then $f'(\frac{\sqrt{\pi}}{2})$ is ______.

• A. $\frac{\sqrt{\pi}}{6}$
• B. $-\frac{\sqrt{\pi}}{6}$
• C. $\frac{\pi}{\sqrt{6}}$
• D. $\sqrt{\frac{\pi}{6}}$

Hint:

Be extremely meticulous with the Chain Rule! Working from the very outside layer inwards ($(\text{Box})^{1/2} \rightarrow \text{Box}^2 \rightarrow \cos(\text{Box}) \rightarrow x^2$) prevents missing crucial multipliers.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Apply the Chain Rule: $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$.

Step 2: Formula Application:
$f'(x) = \frac{1}{2\sqrt{1+\cos^2(x^2)}} \cdot [2\cos(x^2) \cdot (-\sin(x^2)) \cdot 2x]$. $f'(x) = \frac{-2x \sin(2x^2)}{2\sqrt{1+\cos^2(x^2)}} = \frac{-x \sin(2x^2)}{\sqrt{1+\cos^2(x^2)}}$.

Step 3: Explanation:
Plug in $x = \frac{\sqrt{\pi}}{2} \implies x^2 = \frac{\pi}{4}$. $\sin(2 \cdot \frac{\pi}{4}) = \sin(\frac{\pi}{2}) = 1$. $\cos^2(\frac{\pi}{4}) = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$. $f'(\frac{\sqrt{\pi}}{2}) = \frac{-(\frac{\sqrt{\pi}}{2}) \cdot 1}{\sqrt{1 + 1/2}} = \frac{-\sqrt{\pi}/2}{\sqrt{3/2}} = \frac{-\sqrt{\pi}}{2} \cdot \frac{\sqrt{2}}{\sqrt{3}} = -\frac{\sqrt{\pi}}{\sqrt{6}}$.

Step 4: Final Answer:
The derivative value is $-\frac{\sqrt{\pi}}{\sqrt{6}}$.