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The set points where \[f(x)=\dfrac{x}{1+|x|}\] is differentiable is
(a) \[(-\infty ,0)\cup (0,\infty )\]
(b) \[(-\infty ,1)\cup (-1,\infty )\]
(c) \[(-\infty ,\infty )\]
(d)\[(0,\infty )\]

Answer
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Hint: Separate the modulus part. Then find out the left hand derivative and right hand derivative. And compare them.

Complete step-by-step answer:
The given function is \[f(x)=\dfrac{x}{1+|x|}.\]
Here we can observe a modulus function in the denominator, so we can rewrite this function as,
\[f(x)=\left\{ \begin{matrix}
   \dfrac{x}{1+x},x\ge 0 \\
   \dfrac{x}{1-x},x<0 \\
\end{matrix} \right.\]
Now we will check the continuity and differentiability at \[x=0\].
First of all let us check the continuity at \[x=0\].
We know for a function f(x) to be continuous at \[x=0\] its left hand limit (LHL) should be equal to right hand limit (RHL).
So, let us consider the LHL first.
\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{1-x}\]
Applying the limits, we have
\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{0}{1-0}=0\]
So, $LHL=0........(i)$
Now, we will find the RHL.
\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{1+x}=\dfrac{0}{1+0}=0\]
Applying the limits, we have
\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{0}{1+0}=0\]
So, $RHL=0........(ii)$
From equation (i) and (ii), we have
 LHL = RHL
So, the given function f(x) is continuous at \[x=0\].
Now, we shall check the differentiability.
We know for a function to be differentiable at \[x=0\] its left hand derivative (LHD) should be equal to its right hand derivative (RHD).
First, we can find LHD.
\[{{\underset{x\to {{0}^{-}}}{\mathop{\lim f}}\,}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{d}{dx}\left( \dfrac{x}{1-x} \right)\]
For differentiating this function we will use the quotient rule, i.e., \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}(u)-u\dfrac{d}{dx}(v)}{{{v}^{2}}}\]
So the LHD becomes,
\[\begin{align}
  & {{\underset{x\to {{0}^{-}}}{\mathop{\lim f}}\,}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1-x)\dfrac{d}{dx}(x)-x\dfrac{d}{dx}(1-x)}{{{(1-x)}^{2}}} \\
 & \Rightarrow {{\underset{x\to {{0}^{-}}}{\mathop{\lim f}}\,}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1-x)(1)-x(-1)}{{{(1-x)}^{2}}} \\
 & \Rightarrow {{\underset{x\to {{0}^{-}}}{\mathop{\lim f}}\,}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1-x+x}{{{(1-x)}^{2}}} \\
 & \Rightarrow {{\underset{x\to {{0}^{-}}}{\mathop{\lim f}}\,}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{{{(1-x)}^{2}}} \\
\end{align}\]
Now, applying the limit, we have
\[{{\underset{x\to {{0}^{-}}}{\mathop{\lim f}}\,}^{'}}(x)=\dfrac{1}{{{(1-0)}^{2}}}=1\]
\[\therefore LHD=1.......(iii)\]
Now, we will find the RHD.
\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{d}{dx}\left( \dfrac{x}{1+x} \right)\]
Again, applying the quotient rule, we have
\[\begin{align}
  & \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1+x)\dfrac{d}{dx}(x)-x\dfrac{d}{dx}(1+x)}{{{(1+x)}^{2}}} \\
 & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1+x)(1)-x(1)}{{{(1+x)}^{2}}} \\
 & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1+x-x}{{{(1+x)}^{2}}} \\
 & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{{{(1+x)}^{2}}} \\
\end{align}\]
Now, by applying the limits we get,
\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\dfrac{1}{{{(1+0)}^{2}}}=1\]
\[\therefore RHD=1.......(iv)\]
So, from equation (iii) and (iv), we see that $LHD=RHD$ .
Hence, f(x) is differentiable at $x=0$.
Hence, we conclude that the given function is differentiable from \[+\infty \] to \[-\infty \].
Hence, the correct answer is option (c).
Answer is Option (c).

 Note: For finding the left hand and right hand derivative we can use the formula,
$\begin{align}
  & LHD=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(a-x)-f(a)}{-h} \\
 & RHD=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(a+x)-f(a)}{h} \\
\end{align}$
Using these formulas also we will get the same result.