Differentiate $x e^{x}$ using first principle.

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Solution

Let $f (x) = x e^{x},$ then $f (x + h) = (x + h) e^{x + h}$

$∴ f^{'} (x) = l i m h \to 0 \frac{f (x + h) - f (x)}{h}$

= $l i m h \to 0 \frac{(x + h) e^{x + h} - x e^{x}}{h} = l i m h \to 0 \frac{x e^{x + h} - x e^{x} + h e^{x + h}}{h}$

= $l i m h \to 0 \frac{x e^{x} (e^{h} - 1)}{h} + l i m h \to 0 \frac{h e^{x + h}}{h}$

= $x e^{x} + e^{x} [∵ l i m h \to 0 \frac{e^{h} - 1}{h} = 1]$

= $e^{x} (x + 1)$

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