Animated Solution for Mathematics - Limits, Continuity and Differentiability: The set of points where $f(x)=\frac{x}{1+|x|}$ is differentiable is

Welcome, future IITian! Today, we are embarking on a journey to demystify a function that often trips up even the most prepared students. We are looking at $f(x)=\frac{x}{1+|x|}$.

At first glance, it looks like a standard rational function, but that modulus sign in the denominator is a signal—a warning light that tells us to be careful. In the world of JEE Advanced, the modulus function is not just a symbol; it is a boundary that marks the transition between two different mathematical realities.

The modulus function $|x|$ is defined as $x$ when $x\ge 0$ and $-x$ when $x<0$. This means our function $f(x)$ is actually two functions living under one roof:

Imagine you are walking along the graph. As you approach the origin from the right, you are following the curve $y=\frac{x}{1+x}$. As you approach from the left, you are following $y=\frac{x}{1-x}$.

The question that keeps us up at night is: what happens exactly at $x=0$? Is there a sharp, jagged corner that breaks the smoothness of our curve, or does it glide through the origin gracefully?

To answer this, we must use our most powerful tool: the derivative. We need to check if the slope of the tangent line is the same from both sides.

Let us calculate the Right-Hand Derivative (RHD) first. For $x>0$, we differentiate $f(x)=\frac{x}{1+x}$ using the quotient rule:

Now, we evaluate this as $x$ approaches $0$ from the right. Plugging in $x=0$, we get:

The slope on the right is $1$.

Now, let us look at the Left-Hand Derivative (LHD). For $x<0$, we differentiate $f(x)=\frac{x}{1-x}$:

Evaluating this as $x$ approaches $0$ from the left, we plug in $x=0$ to get:

The slope on the left is also $1$.

Look at that! The RHD is $1$ and the LHD is $1$. They are identical.

This is the moment of beauty in calculus. Despite the modulus function's reputation for creating sharp corners, this specific combination of terms forces the slopes to align perfectly at the origin.

The function is not just continuous; it is differentiable at $x=0$. The tangent line at the origin exists and has a slope of $1$.

Since we already know the function is a smooth rational function for all $x eq 0$, and we have just proven it is smooth at $x=0$, we can conclude with absolute certainty that the function is differentiable for all real numbers $x\in (-\infty ,\infty )$.

You have successfully navigated the trap! This problem teaches us a vital lesson: never trust your intuition blindly. Always rely on the rigorous definitions of calculus. When you see a modulus, don't panic—just split it, differentiate it, and let the math reveal the truth.