Differentiate the following functions from first principles: x2ex… - Vidyadip

Question

Differentiate the following functions from first principles:
x²e^x.

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Answer

Let f(x) = x²e^x
⇒ f(x + h) = (x + h)² e^(x+h)
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}+\text{h})^2\text{e}^{(\text{x}+\text{h})}-\text{x}^2\text{e}^{\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{x}^2\text{e}^{(\text{x}+\text{h})}-\text{x}^2\text{e}^\text{x}}{\text{h}}+\frac{2\text{xhe}^{(\text{x}+\text{h})}}{\text{h}}+\frac{\text{h}^2\text{e}^{(\text{x}+\text{h})}}{\text{h}}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\bigg(\frac{\text{x}^2\text{e}^\text{x}\big(\text{e}^{(\text{x}+\text{h}-\text{x})}-1\big)}{\text{x}}+2\text{xe}^{(\text{x}+\text{h})}+\text{he}^{(\text{x}+\text{h})}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{x}^2\text{e}^{\text{x}}\frac{\big(\text{e}^\text{h}-1\big)}{\text{h}}+2\text{xe}^{(\text{x}+\text{h})}+\text{h}^{\text{e}}(\text{x}+\text{h})\bigg]$
$=\text{x}^2\text{e}^\text{x}+2\text{xe}^\text{x}+0\times\text{e}^\text{x}\ \Big[\text{Since,}\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
So,
$\frac{\text{d}}{\text{dx}}(\text{x}^2\text{e}^\text{x})=\text{e}^\text{x}(\text{x}^2+2\text{x})$

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