If $\mathrm{f}(x)=\sqrt{1+\cos ^2\left(x^2\right)}$, then $\mathrm{f}^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)$ is

The derivative of $f(x) = \sqrt{1 + \cos^2(x^2)}$ follows from the chain rule:

$f'(x) = \frac{1}{2\sqrt{1 + \cos^2(x^2)}} \cdot \frac{d}{dx}[1 + \cos^2(x^2)]$

Differentiating the inner function yields $\frac{d}{dx}[\cos^2(x^2)] = 2\cos(x^2)(-\sin(x^2))(2x) = -4x\sin(x^2)\cos(x^2)$

Applying the identity $2\sin\theta\cos\theta = \sin(2\theta)$ gives $\frac{d}{dx}[\cos^2(x^2)] = -2x\sin(2x^2)$

Thus $f'(x) = \frac{-x\sin(2x^2)}{\sqrt{1 + \cos^2(x^2)}}$

Evaluating at $x = \frac{\sqrt{\pi}}{2}$:

$x^2 = \frac{\pi}{4}$ and $2x^2 = \frac{\pi}{2}$
Numerator: $-\frac{\sqrt{\pi}}{2} \cdot \sin\left(\frac{\pi}{2}\right) = -\frac{\sqrt{\pi}}{2}$
Denominator: $\sqrt{1 + \cos^2\left(\frac{\pi}{4}\right)} = \sqrt{1 + \left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{3}{2}}$

$f'\left(\frac{\sqrt{\pi}}{2}\right) = \frac{-\frac{\sqrt{\pi}}{2}}{\sqrt{\frac{3}{2}}} = -\frac{\sqrt{\pi}}{2} \cdot \frac{\sqrt{2}}{\sqrt{3}} = -\frac{\sqrt{2\pi}}{2\sqrt{3}}$

Rationalizing the denominator gives $-\frac{\sqrt{6\pi}}{6} = -\sqrt{\frac{\pi}{6}}$

Final answer: $\boxed{\text{B}}$