Differentiate from first principle:

$(i v) x e^{x}$

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Solution

Given:

$f (x) = x e^{x}$

The derivative of a function $f (x)$ is defined as:

$f^{'} (x) = lim h \to 0 \frac{f (x + h) - f (x)}{h}$

Putting $f (x)$ in the above expression, we get:

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{(x + h) e^{(x + h)} - x e^{x}}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{(x + h) e^{x} e^{h} - x e^{x}}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{x e^{x} e^{h} + h e^{x} e^{h} - x e^{x}}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{x e^{x} e^{h} - x e^{x}}{h} + lim h \to 0 \frac{h e^{x} e^{h}}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{x e^{x} (e^{h} - 1)}{h} + lim h \to 0 e^{x} e^{h}$

$\Rightarrow f^{'} (x) = x e^{x} (1) + e^{x} (e^{0}) (∵ lim x \to 0 \frac{e^{x} - 1}{x} = 1)$

$\Rightarrow f^{'} (x) = x e^{x} + e^{x} .1$

$\Rightarrow f^{'} (x) = (x + 1) e^{x}$

Hence, the derivative of $x e^{x}$ is $(x + 1) e^{x}$

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