JEE Main · 2006 · 30 Apr 2006 · Offline · Continuity and Differentiability

Remove the absolute value by cases. For $x\ge 0$, $f(x)=\frac{x}{1+x}$ with $f^{\prime }(x)=\frac{1}{(1+x{)}^2}$; for $x<0$, $f(x)=\frac{x}{1-x}$ with $f^{\prime }(x)=\frac{1}{(1-x{)}^2}$. Away from the origin each piece is a smooth rational function, so differentiability is only in doubt at $x=0$. There the right-hand derivative tends to $\frac{1}{(1+0{)}^2}=1$ and the left-hand derivative tends to $\frac{1}{(1-0{)}^2}=1$; the two agree, so $f$ is differentiable at $0$ as well. The function is thus differentiable on all of $(-\infty ,\infty )$.