The set of points where \(\rm f(x)=\frac{x}{1+|x|}\) is differentiable, is:

Concept:

• Differentiability of a Function: A function f(x) is differentiable at x = a in its domain if its derivative is continuous at a.

  This means that f'(a) must exist, or equivalently: \(\rm \lim_{x\to a^+}f'(x)=\lim_{x\to a^-}f'(x)=\lim_{x\to a}f'(x)=f'(a)\).

• The Modulus Function '| |' is defined as: \(\rm |x|=\left\{\begin{matrix}\rm \ \ \ x, &\rm x \geq 0\\ \rm -x, &\rm x<0\\\end{matrix}\right.\).

Calculation:

By using the definition of modulus function, the given function can be written as: \(\rm f(x)=\left\{\begin{matrix}\rm \frac{x}{1+x}, &\rm x > 0\\ \ \ \ \ 0,&\rm x=0\\ \rm \frac{x}{1-x}, &\rm x<0\\\end{matrix}\right.\).

Since the expressions for f(x) change for x > 0 and x < 0, let us compare the limits of the derivatives as x → 0.

For x > 0, \(\rm f(x)=\frac{x}{1+x}\).

⇒ \(\rm f'(x)=x\left[\frac{d}{dx}\left(\frac{1}{1+x}\right)\right]+\left(\frac{d}{dx}x\right)\frac{1}{1+x}\)

⇒ \(\rm f'(x)=x\frac{(-1)}{(1+x)^2}+\frac{1}{1+x}\)

⇒ \(\rm f'(x)=\frac{1}{(1+x)^2}\)

⇒ \(\rm \lim_{x\to0^+} f'(x)=1\).

Similarly, for x < 0, \(\rm f(x)=\frac{x}{1-x}\).

⇒ \(\rm \lim_{x\to0^-} f'(x)=\lim_{x\to0^-} \frac{1}{(1-x)^2}=1\).

Since \(\rm \lim_{x\to 0^+}f'(x)=\lim_{x\to 0^-}f'(x)=1\), the function f(x) is differentiable at x = 0, and f'(0) = 1.

Also, \(\rm \lim_{x\to \infty^+}f'(x)=\lim_{x\to \infty^-}f'(x)=0\).

∴ The function is differentiable in (-∞, ∞), i.e. it is differentiable everywhere.