Animated Solution for Mathematics - Limits, Continuity and Differentiability: The domain of the derivative of the function $f(x)=\{\begin{array}{ll}{\tan }^{-1}x& \text{if }|x|\le 1\\ \frac{1}{2}(|x|-1)& \text{if }|x|>1\end{array}$ is

Welcome, fellow explorers of the mathematical landscape! Today, we are going to dissect a function that might look intimidating at first glance, but reveals its secrets once we peel back the layers.

We are looking at the function:

Our mission is to find the domain of its derivative, $f^{\prime }(x)$.

Before we rush into calculating slopes, let us pause and remember the golden rule: Differentiability implies continuity.

Imagine you are walking along the graph of a function. If you suddenly hit a wall or a cliff—a jump discontinuity—you cannot possibly define a slope at that exact moment of the jump. The path is broken.

Therefore, if a function is discontinuous at a point, it is strictly not differentiable there. Our strategy is simple: identify the boundary points, check for continuity, and if we find a break, we exclude those points from the domain of the derivative.

Our function changes its behavior at the boundaries $|x|=1$, which means $x=1$ and $x=-1$. Let us put $x=1$ under the microscope first.

First, the exact value: Since $1\le 1$, we use the first definition, $f(x)={\tan }^{-1}(x)$. Thus:

Next, the Right Hand Limit (RHL): As we approach $1$ from the right, we are in the territory where $x>1$. Here, the function is $f(x)=\frac{1}{2}(|x|-1)$.

Since $x>1$, $|x|=x$, so $f(x)=\frac{1}{2}(x-1)$. As $x\to 1^{+}$, the limit is:

Comparing the two, we see that $f(1)=\frac{\pi }{4}$, but the limit is $0$. They are not equal! There is a jump discontinuity at $x=1$. Consequently, the function is not differentiable at $x=1$.

Checking $x=-1$

Now, let us turn our attention to $x=-1$. Again, we find the exact value: Since $|-1|=1$, the condition $|x|\le 1$ is satisfied. Thus:

Now, the Left Hand Limit (LHL): As we approach $-1$ from the left, we are in the territory where $x<-1$. Here, $|x|>1$, so we use $f(x)=\frac{1}{2}(|x|-1)$.

Since $x$ is negative, $|x|=-x$. So, $f(x)=\frac{1}{2}(-x-1)$. As $x\to -1^{-}$, the limit is:

Once again, $f(-1)=-\frac{\pi }{4}$, while the limit is $0$. Another jump discontinuity! The function is broken at $x=-1$ as well.

Inside the open intervals $(-1,1)$, $(-\infty ,-1)$, and $(1,\infty )$, the function is composed of smooth, continuous, and differentiable pieces. The only roadblocks to differentiability are the points where the function is discontinuous.

We have proven that these points are $x=1$ and $x=-1$. Therefore, the derivative $f^{\prime }(x)$ exists for all real numbers except these two.

The domain of the derivative is $\mathbb{R}-\{-1,1\}$.

Remember, in the JEE, the most dangerous traps are often the ones that look like simple differentiation problems but are actually tests of your conceptual understanding of continuity. Stay sharp, keep visualizing, and keep loving the math!