Given function

$f(x)=x\cos^{-1}(-\sin|x|), \qquad x\in\left[-\frac\pi2,\frac\pi2\right]$

We first simplify $\cos^{-1}(-\sin|x|)$ on the given interval.

Simplify the inverse cosine part

Since $|x|\in\left[0,\frac\pi2\right]$, we have $\sin|x|\in[0,1]$ so $-\sin|x|\in[-1,0]$.

Now use the identity $\cos\left(\frac\pi2+\theta\right)=-\sin\theta$ for $\theta\in\left[0,\frac\pi2\right]$.

Therefore, $\cos^{-1}(-\sin|x|)=\frac\pi2+|x|$ because $\frac\pi2+|x|\in\left[\frac\pi2,\pi\right]$, which is the principal range of $\cos^{-1}$.

Hence, $f(x)=x\left(\frac\pi2+|x|\right)$

Write $f(x)$ piecewise

• For $x\ge 0$, $|x|=x$, so $f(x)=x\left(\frac\pi2+x\right)=\frac\pi2x+x^2$

• For $x<0$, $|x|=-x$, so $f(x)=x\left(\frac\pi2-x\right)=\frac\pi2x-x^2$

Thus,

Differentiate on each interval

• For $x<0$, $f'(x)=\frac\pi2-2x$

• For $x>0$, $f'(x)=\frac\pi2+2x$

Check differentiability at $x=0$

Compute one-sided derivatives:

$f'_-(0)=\lim_{x\to 0^-}\left(\frac\pi2-2x\right)=\frac\pi2$ $f'_+(0)=\lim_{x\to 0^+}\left(\frac\pi2+2x\right)=\frac\pi2$

Since both are equal, $f'(0)=\frac\pi2$

So $f$ is differentiable at $x=0$.

Check monotonicity of $f'$

On $\left(-\frac\pi2,0\right)$: $f'(x)=\frac\pi2-2x$ This is a linear function with slope $-2<0$, so $f'$ is decreasing.

On $\left(0,\frac\pi2\right)$: $f'(x)=\frac\pi2+2x$ This is a linear function with slope $2>0$, so $f'$ is increasing.

Evaluate options

• A: $f'$ is decreasing in $\left(-\frac\pi2,0\right)$ and increasing in $\left(0,\frac\pi2\right)$

  True.

• B: $f'(0)=-\frac\pi2$

  False, since $f'(0)=\frac\pi2$

• C: $f$ is not differentiable at $x=0$

  False.

• D: $f'$ is increasing in $\left(-\frac\pi2,0\right)$ and decreasing in $\left(0,\frac\pi2\right)$

  False.

Final answer

The correct option is: $\boxed{A}$