STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

$g(x) = 1 + |x|,\,\,x \in ( - \infty ,\infty )$

Here $h$ is differentiable in $( - \infty ,\infty )$ but $|x|$ is not differentiable at $x = 0$.

Therefore $g$ is differentiable in $( - \infty ,0) \cup (0,\infty )$ and $g(x) \ne 0,\rlap{--} Vx \in $ $( - \infty ,\infty )$,

therefore $f(x) = \frac{{h(x)}}{{g(x)}} = \frac{x}{{1 + |x|}}$

It is differentiable in $( - \infty ,0) \cup (0,\infty )$ for $x = 0$

$\mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{{h - 0}} = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{h}{{1 + |h|}} - 0}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{1}{{1 + |h|}} = 1$

Therefore $f$ is differentiable at $x = 0$,

so $f$ is differentiable in $( - \infty ,\infty )$.