Consider the expression $x{{e}^{x}}-1=y$ , then show that
$\dfrac{dy}{dx}={{e}^{x}}+y+1$.

Hint: Differentiate the given equation in both the sides with respect to $'x'$ .Use the rule of multiplication for differentiation, wherever required. It is given as
$\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ , where u and v are two functions in multiplication.

Complete step-by-step answer:
Given expression in the problem is
$x{{e}^{x}}-1=y.......(i)$
And hence, we need to prove the following expression with the help of the above equation given in the question.
$\dfrac{dy}{dx}={{e}^{x}}+y+1........(ii)$
So, let us differentiate the equation (i) to both sides w.r.t $'x'$
So, we get
$\begin{align}
  & \dfrac{d}{dx}\left( x{{e}^{x}}-1 \right)=\dfrac{dy}{dx} \\
 & \dfrac{d}{dx}\left( x{{e}^{x}} \right)-\dfrac{d}{dx}\left( 1 \right)=\dfrac{dy}{dx}..........\left( iii \right) \\
\end{align}$
Now, as we know, the derivative of any constant term is 0. So, we get
$\dfrac{d}{dx}\left( \text{constant} \right)=0...........\left( iv \right)$
Now, for differentiating $x{{e}^{x}}$ w.r.t $x$ , we need to use the multiplication rule of differentiation. It is given as
$\dfrac{d}{dx}\left( u.v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}..........\left( v \right)$
Now, we can find differentiation of $x{{e}^{x}}$ , using the above rule, by taking $u=x$ and $v={{e}^{x}}$ .So, we get
$\dfrac{d}{dx}\left( x{{e}^{x}} \right)=x\dfrac{d}{dx}{{e}^{x}}+{{e}^{x}}\dfrac{d}{dx}x$
We know
$\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}\text{ and }\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$
So, we get
$\dfrac{d}{dx}\left( x{{e}^{x}} \right)=x{{e}^{x}}+{{e}^{x}}..........\left( vi \right)$
Now, we can use equation (iv) and (vi) to simplify equation (iii). So, we get
\[\begin{align}
  & x{{e}^{x}}+{{e}^{x}}-0=\dfrac{dy}{dx} \\
 & \dfrac{dy}{dx}=x{{e}^{x}}+{{e}^{x}}.........\left( vii \right) \\
\end{align}\]
Now, we know the value of $x{{e}^{x}}$ from the equation (i) in terms of $'y'$ .So, we get
$\begin{align}
  & x{{e}^{x}}-1=y \\
 & x{{e}^{x}}=y+1..........\left( viii \right) \\
\end{align}$
Now, we can replace $x{{e}^{x}}$ by $y+1$ from the equation (viii) to the equation (vii). So, we get
$\begin{align}
  & \dfrac{dy}{dx}=y+1+{{e}^{x}} \\
 & \dfrac{dy}{dx}={{e}^{x}}+y+1 \\
\end{align}$
Hence, the given expression is proved

Note: One may differentiate the given expression with respect to $'y'$ as well but he/she will get the value of $\dfrac{dx}{dy}$ from this approach. Use relation
$\dfrac{dy}{dx}=\dfrac{1}{\left( \dfrac{dy}{dx} \right)}$ to get $\dfrac{dy}{dx}$ .Don’t apply $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ for calculating derivative of ${{e}^{x}}$ . Both are different type of function i.e. ${{x}^{n}}$ is of type ${{\left( \text{variable} \right)}^{\text{constant}}}$ and ${{e}^{x}}$ is of type ${{\left( \text{constant} \right)}^{\text{variable}}}$ .So, we cannot use same formula for getting derivative of them. Both identities are given as $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ and $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}$.