Differentiate the following from first principle:xe^x - Vidyadip

Question

Differentiate the following from first principle:$\text{x}\text{e}^\text{x}$

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Answer

We have,$\text{f(x)}=\text{xe}^\text{x}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x+h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x+h})\text{e}^{(\text{x+h})}-\text{xe}^\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{xe}^\text{x}.\text{e}^\text{h}+\text{he}^\text{x}.\text{e}^\text{h}-\text{xe}^\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{xe}^\text{x}\Big(\frac{\text{e}^\text{h}-1}{\text{h}}\Big)+\frac{\text{he}^{\text{x}+\text{h}}}{\text{h}}$
$=\text{xe}^\text{x}+\text{e}^\text{x}$
$=\text{e}^\text{x}\big(\text{x}+1\big)$

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