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**Hint:**For the given function we are given to check the differentiability at x=0. We have to consider the equation as equation (1) and then write the function in all possible cases of x. Differentiate the function and then check whether left hand derivative and right hand derivative are equal or not.

**Complete step by step answer:**

We are given to find the differentiability of the function \[\text{f(x)=x }\!\!|\!\!\text{ x }\!\!|\!\!\text{ }\] for all\[x\in R\] at \[x=0\].

Now let us consider the given equation as equation (1).

\[\text{f(x)=x }\!\!|\!\!\text{ x }\!\!|\!\!\text{ }..................\left( 1 \right)\]

As we can see modulus in the given equation so the function will vary at greater and less than 0.

Therefore equation (1) at greater and less than 0 will be

\[\begin{align}

& \Rightarrow \text{f(x)=x }\!\!|\!\!\text{ x }\!\!|\!\!\text{ } \\

& \text{ = }\!\!\{\!\!\text{ }{{\text{x}}^{2}};\text{ x}\ge \text{0} \\

& \text{ = }\!\!\{\!\!\text{ -}{{\text{x}}^{\text{2}}}\text{; x < 0} \\

\end{align}\]

Now we have to differentiate the function at greater than 0 and less than 0.

Let us differentiate the given function at greater than 0 and less than 0.

\[\begin{align}

& \Rightarrow {{\text{f}}^{'}}\text{(x) = 2x};\text{ x}\ge \text{0} \\

& \text{ -2x; x < 0 } \\

\end{align}\]

Now let us find the differentiability at \[x=0\].

Now let us find the Left hand derivative of \[\text{f}\left( x \right)\] at \[\left( x={{0}^{-}} \right)\].

\[\begin{align}

& \Rightarrow -2\left( 0 \right) \\

& \Rightarrow 0 \\

\end{align}\]

Let us consider the above equation as equation (1).

\[\text{LHD = 0}....................\left( 1 \right)\]

Now for the right hand derivative of \[\text{f}\left( x \right)\] at \[\left( x={{0}^{+}} \right)\].

\[\begin{align}

& \Rightarrow 2\left( 0 \right) \\

& \Rightarrow 0 \\

\end{align}\]

Let us consider the above equation as equation (2).

\[\text{RHD = 0}....................\left( 2 \right)\]

Therefore, by comparing the equation (1) and equation (2) we can say that

\[\text{LHD = RHD}\]

Hence we can say that the given equation is differentiable at \[x=0\].

**Note:**We have to note a point that modulus function will change at greater than, less than and equal to 0. So we have to write all the cases to differentiate easily. Any simple or toughest problem can be solved using this method only.

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