Answer
$$\frac{d}{d x}(x f(x)) =x f^{\prime}(x)+f(x)$$
Work Step by Step
Since $$ f'(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$
Then
\begin{aligned}
\frac{d}{d x}(x f(x)) &=\lim _{h \rightarrow 0} \frac{(x+h) f(x+h)-f(x)}{h}\\
&=\lim _{h \rightarrow 0} \frac{x f(x+h)+hf(x+h)-f(x)}{h}\\
&=\lim _{h \rightarrow 0}\left(x \frac{f(x+h)-f(x)}{h}+f(x+h)\right) \\
&=x \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}+\lim _{h \rightarrow 0} f(x+h) \\
&=x f^{\prime}(x)+f(x)
\end{aligned}
