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More # 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)$(0,\infty )$

Last updated date: 18th Mar 2023
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Hint: Separate the modulus part. Then find out the left hand derivative and right hand derivative. And compare them.

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).
\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}