differentiate xe^x using first principle

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{xe}}^{\mathrm{x}} \\ \mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)=\left(\mathrm{x}+\mathrm{h}\right){\mathrm{e}}^{\mathrm{x}+\mathrm{h}} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\lim } \frac{\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}} \\ =\underset{\mathrm{h}\to 0}{\lim } \frac{\left(\mathrm{x}+\mathrm{h}\right){\mathrm{e}}^{\mathrm{x}+\mathrm{h}}-{\mathrm{xe}}^{\mathrm{x}}}{\mathrm{h}} \\ =\underset{\mathrm{h}\to 0}{\lim } \frac{\left(\mathrm{x}+\mathrm{h}\right){\mathrm{e}}^{\mathrm{x}}.{\mathrm{e}}^{\mathrm{h}}-{\mathrm{xe}}^{\mathrm{x}}}{\mathrm{h}} \\ ={\mathrm{e}}^{\mathrm{x}} \underset{\mathrm{h}\to 0}{\lim } \frac{\left(\mathrm{x}+\mathrm{h}\right) {\mathrm{e}}^{\mathrm{h}}-\mathrm{x}}{\mathrm{h}} \\ ={\mathrm{e}}^{\mathrm{x}} \underset{\mathrm{h}\to 0}{\lim } \frac{{\mathrm{xe}}^{\mathrm{h}}-\mathrm{x}+{\mathrm{he}}^{\mathrm{h}}}{\mathrm{h}} \\ ={\mathrm{e}}^{\mathrm{x}} \left\{\underset{\mathrm{h}\to 0}{\lim } \left(\frac{\mathrm{x}\left({\mathrm{e}}^{\mathrm{h}}-1\right)}{\mathrm{h}}+\underset{\mathrm{h}\to 0}{\lim } \frac{{\mathrm{he}}^{\mathrm{h}}}{\mathrm{h}}\right)\right\} \\ ={\mathrm{e}}^{\mathrm{x}} \left\{\mathrm{x} \underset{\mathrm{h}\to 0}{\lim } \left(\frac{\left({\mathrm{e}}^{\mathrm{h}}-1\right)}{\mathrm{h}}+\underset{\mathrm{h}\to 0}{\lim } {\mathrm{e}}^{\mathrm{h}}\right)\right\} \\ \mathrm{We} \mathrm{know} \mathrm{that} \underset{\mathrm{h}\to 0}{\lim } \frac{\left({\mathrm{e}}^{\mathrm{h}}-1\right)}{\mathrm{h}}=1 \\ ={\mathrm{e}}^{\mathrm{x}} \left\{\mathrm{x}\times 1+{\mathrm{e}}^0\right\} \\ ={\mathrm{e}}^{\mathrm{x}}\left(\mathrm{x}+1\right) \left[\mathrm{Answer}\right] \end{gathered}$$

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards