Differentiate the following from the first principlex^2e ^x - Vidyadip

Question

Differentiate the following from the first principle$\text{x}^2\text{e}{^\text{x}}$

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Answer

We have, $\text{f}(\text{x})=\text{x}^2\text{e}^\text{x}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}^2\text{e}^\text{x}\text{e}^\text{h}+\text{h}^2\text{e}^\text{x}\text{e}^\text{h}+2\text{xhe}^\text{x}\text{e}^\text{h}-\text{x}^2\text{e}^\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\text{x}^2\text{e}^\text{x}\frac{(\text{e}^\text{h}-1)}{\text{h}}+\text{e}^\text{x}\text{e}^\text{h}\frac{(\text{h}^2+\text{2xh})}{\text{h}}\ \Big[\because\frac{\text{e}^\text{h}-1}{\text{h}}-1\Big]$
$\therefore\text{x}^2\text{e}^\text{x}+\text{e}^\text{x}(0+\text{2x})$
$=\text{x}^2\text{e}^\text{x}+\text{2xe}^\text{x}$
$=\text{e}^\text{x}(\text{x}^2+\text{2x})$

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