$\begin{align}  & y=x{{e}^{x}} \\ & y+\delta y=(x+\delta x){{e}^{(x+\delta x)}} \\ & \delta y=(x+\delta x){{e}^{(x+\delta x)}}-x{{e}^{x}} \\ & \delta y=x{{e}^{(x+\delta x)}}+(\delta x){{e}^{(x+\delta x)}}-x{{e}^{x}} \\ & \delta y=x{{e}^{x}}({{e}^{\delta x}}-1)+(\delta x){{e}^{(x+\delta x)}} \\ & \frac{\delta y}{\delta x}=\frac{x{{e}^{x}}({{e}^{\delta x}}-1)+(\delta x){{e}^{(x+\delta x)}}}{\delta x} \\ & \frac{\delta y}{\delta y}=\frac{x{{e}^{x}}({{e}^{\delta x}}-1)}{\delta x}+\frac{(\delta x){{e}^{(x+\delta x)}}}{\delta x} \\ & \frac{\delta y}{\delta x}=x{{e}^{x}}\frac{({{e}^{\delta x}}-1)}{\delta x}+{{e}^{(x+\delta x)}} \\ & \text{Note: }\underset{\delta x\to 0}{\mathop{\lim }}\,\left( \frac{{{e}^{x}}-1}{x} \right)=1\text{ } \\ & \text{which implies that }\frac{({{e}^{\delta x}}-1)}{\delta x}=1 \\ & \frac{dy}{dx}=\underset{\delta x\to 0}{\mathop{\lim }}\,\left[ x{{e}^{x}}\frac{({{e}^{\delta x}}-1)}{\delta x}+{{e}^{(x+\delta x)}} \right]=x{{e}^{x}}+{{e}^{x}} \\\end{align}$