Let $$f(x) = x\cos^{-1}(-\sin|x|)$$, $$x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$, then which of the following is true?

We have the function

$$f(x)=x\cos^{-1}(-\sin|x|),\qquad x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].$$

Because of the absolute value, the expression inside the inverse cosine changes its sign on the two sides of the origin. We therefore write the function in two separate pieces, one for positive $$x$$ and one for negative $$x$$.

For $$x\ge 0$$ we have $$|x|=x,$$ so

$$f(x)=x\cos^{-1}(-\sin x),\qquad 0\le x\le\dfrac{\pi}{2}.$$

For $$x\le 0$$ we have $$|x|=-x,$$ hence

$$\sin|x|=\sin(-x)=-\sin x,$$ so

$$-\sin|x|=+\sin x,$$

and therefore

$$f(x)=x\cos^{-1}(\sin x),\qquad -\dfrac{\pi}{2}\le x\le 0.$$

We now differentiate both branches. First we recall the standard derivative

$$\dfrac{d}{dx}\left[\cos^{-1}(u)\right]=-\dfrac{u'}{\sqrt{1-u^{2}}}.$$

Right-hand branch, $$x\gt 0$$.

For $$x\gt 0$$ $$$ f(x)=x\cos^{-1}(-\sin x). $$$ Let $$u(x)=-\sin x,$$ so $$u'(x)=-\cos x.$$ Applying the formula:

$$\dfrac{d}{dx}\cos^{-1}(-\sin x)=-\dfrac{-\cos x}{\sqrt{1-( -\sin x)^2}} =\dfrac{\cos x}{\sqrt{1-\sin^{2}x}} =\dfrac{\cos x}{|\cos x|}.$$ For $$x\in(0,\frac{\pi}{2})$$ we have $$\cos x\gt 0,$$ hence $$|\cos x|=\cos x$$ and

$$\dfrac{d}{dx}\cos^{-1}(-\sin x)=1.$$

Therefore

$$f'(x)=\cos^{-1}(-\sin x)+x\cdot 1 =\cos^{-1}(-\sin x)+x,\qquad 0\lt x\lt \dfrac{\pi}{2}.$$

Differentiating once more to study the monotonicity of $$f'$$ :

$$f''(x)=\dfrac{d}{dx}\bigl[\cos^{-1}(-\sin x)\bigr]+\dfrac{d}{dx}(x)=1+1=2\gt 0.$$ Since the second derivative is positive throughout the interval, $$f'(x)$$ is increasing on $$\left(0,\dfrac{\pi}{2}\right).$$

Left-hand branch, $$x\lt 0$$.

For $$x\lt 0$$ $$$ f(x)=x\cos^{-1}(\sin x). $$$ Take $$v(x)=\sin x,$$ so $$v'(x)=\cos x.$$ Using the same formula,

$$\dfrac{d}{dx}\cos^{-1}(\sin x)=-\dfrac{\cos x}{\sqrt{1-\sin^{2}x}} =-\dfrac{\cos x}{|\cos x|}.$$ For $$x\in\left(-\dfrac{\pi}{2},0\right)$$ we again have $$\cos x\gt 0,$$ hence $$|\cos x|=\cos x$$ and

$$\dfrac{d}{dx}\cos^{-1}(\sin x)=-1.$$

Thus

$$f'(x)=\cos^{-1}(\sin x)+x(-1)=\cos^{-1}(\sin x)-x,\qquad -\dfrac{\pi}{2}\lt x\lt 0.$$

Differentiate once more:

$$f''(x)=\dfrac{d}{dx}\bigl[\cos^{-1}(\sin x)\bigr]-1=-1-1=-2\lt 0.$$

Because the second derivative is negative everywhere in this interval, $$f'(x)$$ is decreasing on $$\left(-\dfrac{\pi}{2},0\right).$$

Differentiability at $$x=0$$. We must compute the two one-sided derivatives.

Right-hand limit:

For small positive $$x$$, use $$\sin x\approx x,$$ so $$-\sin x\approx -x.$$ Around $$t=0$$ we know $$\cos^{-1}(t)\approx\dfrac{\pi}{2}-t.$$ Therefore $$$ \cos^{-1}(-\sin x)\approx\dfrac{\pi}{2}+x. $$$ Hence $$$ f'(x)=\cos^{-1}(-\sin x)+x\approx\left(\dfrac{\pi}{2}+x\right)+x=\dfrac{\pi}{2}+2x \rightarrow[x\to0^{+}]{}\dfrac{\pi}{2}. $$$

Left-hand limit:

For small negative $$x,$$ $$\sin x\approx x,$$ so $$$ \cos^{-1}(\sin x)\approx\cos^{-1}(x)\approx\dfrac{\pi}{2}-x. $$$ Thus $$$ f'(x)=\cos^{-1}(\sin x)-x\approx\left(\dfrac{\pi}{2}-x\right)-x=\dfrac{\pi}{2}-2x \rightarrow[x\to0^{-}]{}\dfrac{\pi}{2}. $$$

The two limits coincide, so $$f'(0)$$ exists and equals $$\dfrac{\pi}{2}.$$ Consequently the function is differentiable at $$x=0,$$ and $$f'(0)=-\dfrac{\pi}{2}$$ is incorrect.

Summary of our findings.

• $$f'(x)$$ is decreasing on $$\left(-\dfrac{\pi}{2},0\right).$$

• $$f'(x)$$ is increasing on $$\left(0,\dfrac{\pi}{2}\right).$$

• $$f'(0)=\dfrac{\pi}{2}.$$

Among the given statements, only Option D matches these conclusions.

Hence, the correct answer is Option D.