How do you find the derivative of $y = x{e^x}$?

Hint: To differentiate means to find the derivative of a function or rate of change of a function with respect to some variable. If a function $y$ is to be differentiated with respect to $x$, then it will be written as $\dfrac{{dy}}{{dx}}$. Since the given expression to be differentiated is the product of two expressions or functions, therefore we will use the product rule for differentiation. The product rule states that if $y = f(x) \times g(x)$, then$\dfrac{{dy}}{{dx}} = f(x)\dfrac{{dg(x)}}{{dx}} + g(x)\dfrac{{df(x)}}{{dx}}$.

Complete step by step solution:
It is given that $y = x{e^x}$.
Here $f(x) = x$ and $g(x) = {e^x}$. On substituting these values in the product rule of differentiation, we will get
$ \Rightarrow \dfrac{{dy}}{{dx}} = f(x)\dfrac{{dg(x)}}{{dx}} + g(x)\dfrac{{df(x)}}{{dx}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = x\dfrac{{d{e^x}}}{{dx}} + {e^x}\dfrac{{dx}}{{dx}}$
For the first part of the expression at the right-hand side of the equation, it must be known that the differentiation or derivative of ${e^x}$ with respect to $x$ is ${e^x}$, i.e. $\dfrac{{d{e^x}}}{{dx}} = {e^x}$ On substituting these values, we will get
$ \Rightarrow \dfrac{{dy}}{{dx}} = (x \times {e^x}) + {e^x}\dfrac{{dx}}{{dx}}$
Now for the second part of the expression at the right hand side of the equation, one of the rule of differentiation must be remembered that states $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$. In our case $n = 1$, so $\dfrac{{d{x^1}}}{{dx}} = 1 \times {x^{1 - 1}} = {x^0} = 1$, i.e. $\dfrac{{dx}}{{dx}} = 1$ On substituting these values, we will get
$ \Rightarrow \dfrac{{dy}}{{dx}} = (x \times {e^x}) + ({e^x} \times 1)$
On further simplifying, we will get
$ \Rightarrow \dfrac{{dy}}{{dx}} = x{e^x} + {e^x}$

Hence, when we find the derivative of $y = x{e^x}$, we get $x{e^x} + {e^x}$ as the answer.

Note:
If a function $y$ is to be differentiated with respect to $x$, then it is written as $\dfrac{{dy}}{{dx}}$. But it can also be expressed as $Dy$ as $D$ is sometimes used in place of $\dfrac{d}{{dx}}$.