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**Hint:**In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument, or we can say it is the rate of change of a function. Derivation of two functions multiplied with each other can be executed by simplifying them into a single function using Product rule.

**Complete step by step answer:**

Here we have the function as \[y=x{{e}^{x}}\]

As we can see the function is a multiplication of two function , that is \[x\] and \[{{e}^{x}}\]

The Product rule of derivative is given as,

\[\Rightarrow \dfrac{d(uv)}{dx}=u\dfrac{d(v)}{dx}+v\dfrac{d(u)}{dx}\]

Where \[u\] and \[v\] are two functions/variables and \[\dfrac{d}{dx}\] is derivative with respect to \[x\].

In the above question the two functions can be \[x\] and \[{{e}^{x}}\] can be taken as \[u\] and \[v\].

We will apply the product rule to find the derivative of the above equation.

That is, \[y=(x)({{e}^{x}})\Leftrightarrow y=(u)(v)\]

Applying product rule we get,

\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d(x{{e}^{x}})}{dx}\]

\[\Rightarrow \dfrac{d(x{{e}^{x}})}{dx}=x\dfrac{d({{e}^{x}})}{dx}+{{e}^{x}}\dfrac{d(x)}{dx}\]

Since we know that \[\dfrac{d(x)}{dx}=1\] and \[\dfrac{d({{e}^{x}})}{dx}={{e}^{x}}\] , derivative of \[x\] with respect to \[x\] is 1.

And we also know that the exponential function has the special property that its derivative is the function itself.

Substituting the values of derivatives in the above equation we get,

\[\Rightarrow \dfrac{d(x{{e}^{x}})}{dx}=x({{e}^{x}})+{{e}^{x}}(1)\]

\[\Rightarrow \dfrac{d(x{{e}^{x}})}{dx}=x{{e}^{x}}+{{e}^{x}}\]

Taking \[{{e}^{x}}\] as common

\[\Rightarrow \dfrac{d(x{{e}^{x}})}{dx}={{e}^{x}}(x+1)\]

**Hence the derivative of the function \[y=x{{e}^{x}}\] is \[{{e}^{x}}(x+1)\].**

**Note:**In product rule the order of function does not matter, that is the values of \[u\] and \[v\] can be interchanged and we will still get the same answer. For example, in the above question, we took \[u\] and \[v\] as \[x\] and \[{{e}^{x}}\] respectively, now interchanging the values , \[u={{e}^{x}},v=x\] we get

\[\Rightarrow \dfrac{d({{e}^{x}}x)}{dx}\]

\[\Rightarrow \dfrac{d({{e}^{x}}x)}{dx}={{e}^{x}}\dfrac{d(x)}{dx}+x\dfrac{d({{e}^{x}})}{dx}\]

\[\Rightarrow \dfrac{d({{e}^{x}}x)}{dx}={{e}^{x}}(1)+x({{e}^{x}})\]

Taking \[{{e}^{x}}\] as common

\[\Rightarrow \dfrac{d({{e}^{x}}x)}{dx}={{e}^{x}}(1+x)\] , which is the same answer as obtained above.

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