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 Question:
We must apply the chain rule sequentially (multiple times) to find the derivative of a nested composite trigonometric function, and then evaluate it at a specific value.

Step 2: Detailed Explanation:
The function is $f(x) = \sqrt{1 + \cos^2(x^2)}$.
Let's rewrite it with explicit fractional exponents to track the chain rule layers easily:
$f(x) = [1 + (\cos(x^2))^2]^{1/2}$
Now, differentiate layer by layer from the outside in:
Layer 1 (The outer square root):
$f'(x) = \frac{1}{2} [1 + \cos^2(x^2)]^{-1/2} \times \frac{d}{dx}[1 + \cos^2(x^2)]$
Layer 2 (The squared cosine):
The derivative of $1$ is $0$. The derivative of $[\cos(x^2)]^2$ brings down the $2$:
$\frac{d}{dx}[\cos^2(x^2)] = 2\cos(x^2) \times \frac{d}{dx}[\cos(x^2)]$
Layer 3 (The cosine function):
$\frac{d}{dx}[\cos(x^2)] = -\sin(x^2) \times \frac{d}{dx}[x^2]$
Layer 4 (The innermost polynomial):
$\frac{d}{dx}[x^2] = 2x$
Multiply all the links of the chain rule together:
$f'(x) = \frac{1}{2\sqrt{1 + \cos^2(x^2)}} \cdot 2\cos(x^2) \cdot (-\sin(x^2)) \cdot 2x$
Cancel the leading $\frac{1}{2}$ and $2$, and rearrange:
$f'(x) = \frac{-2x \cos(x^2) \sin(x^2)}{\sqrt{1 + \cos^2(x^2)}}$
Use the double angle identity $2\sin A\cos A = \sin(2A)$:
$f'(x) = \frac{-x \sin(2x^2)}{\sqrt{1 + \cos^2(x^2)}}$
Now evaluate this derivative at exactly $x = \frac{\sqrt{\pi}}{2}$:
First, calculate the common argument $x^2$:
$x^2 = \left(\frac{\sqrt{\pi}}{2}\right)^2 = \frac{\pi}{4}$
Substitute $x$ and $x^2$ into the derivative:
$f'\left(\frac{\sqrt{\pi}}{2}\right) = \frac{- \left(\frac{\sqrt{\pi}}{2}\right) \cdot \sin\left(2 \cdot \frac{\pi}{4}\right)}{\sqrt{1 + \cos^2\left(\frac{\pi}{4}\right)}}$
Evaluate the trigonometric terms:
$\sin(\pi/2) = 1$
$\cos(\pi/4) = \frac{1}{\sqrt{2}} \implies \cos^2(\pi/4) = \frac{1}{2}$
Plug these back in:
$f'\left(\frac{\sqrt{\pi}}{2}\right) = \frac{-\frac{\sqrt{\pi}}{2} \cdot (1)}{\sqrt{1 + \frac{1}{2}}} = \frac{-\frac{\sqrt{\pi}}{2}}{\sqrt{\frac{3}{2}}}$
Simplify the compound fraction:
$f'\left(\frac{\sqrt{\pi}}{2}\right) = -\frac{\sqrt{\pi}}{2} \times \frac{\sqrt{2}}{\sqrt{3}} = -\frac{\sqrt{\pi} \cdot \sqrt{2}}{2\sqrt{3}}$
Since $2 = \sqrt{2} \cdot \sqrt{2}$, we can cancel a $\sqrt{2}$:
$f'\left(\frac{\sqrt{\pi}}{2}\right) = -\frac{\sqrt{\pi}}{\sqrt{2}\sqrt{3}} = -\frac{\sqrt{\pi}}{\sqrt{6}} = -\sqrt{\frac{\pi}{6}}$

Step 3: Final Answer:
The derivative evaluates to $-\frac{\sqrt{\pi}}{6}$, matching option (b).