Animated Solution for Mathematics - Differentiation: 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?

The function provided is $f(x)=x{\cos }^{-1}(-\sin |x|)$. This expression contains a modulus, an inverse trigonometric function, and a negative argument, which often serves as a trap in JEE Advanced problems.

To simplify, we utilize the identity ${\cos }^{-1}(-y)=\pi -{\cos }^{-1}(y)$. Applying this to our function, we get:

By removing the negative sign, the expression becomes significantly more manageable.

Next, we address the term ${\cos }^{-1}(\sin |x|)$. We recall the fundamental identity ${\cos }^{-1}(\theta )+{\sin }^{-1}(\theta )=\frac{\pi }{2}$.

This allows us to rewrite the term as:

Assuming the standard domain restriction where $|x|\in [0,\frac{\pi }{2}]$, the expression ${\sin }^{-1}(\sin |x|)$ simplifies directly to $|x|$. Substituting this back into our function yields:

This collapses into the elegant form:

With the function simplified to $f(x)=x(\frac{\pi }{2}+|x|)$, we resolve the modulus by considering the two possible cases for $x$:

For $x\ge 0$:

For $x<0$:

We now differentiate the piecewise components. For $x>0$, the derivative is $f^{\prime }(x)=\frac{\pi }{2}+2x$. For $x<0$, the derivative is $f^{\prime }(x)=\frac{\pi }{2}-2x$.

Checking the differentiability at $x=0$, we find that the left-hand derivative and the right-hand derivative both equal $\frac{\pi }{2}$. Therefore, the function is differentiable at $x=0$.

The final behavior of the function is determined by these derivatives: for $x<0$, the slope is decreasing, and for $x>0$, the slope is increasing. Always remember the golden rule of JEE calculus: simplify first, differentiate later.