Differentiate the following from first principle: $e^{x^2 + 1}$

Solution:

$\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}$
$\frac{d}{dx}\left( e^{x^2 + 1} \right) = \lim_{h \to 0} \frac{e^{(x + h )^2 + 1} - e^{x^2 + 1}}{h}$
$ = \lim_{h \to 0} \frac{e^{x^2 + h^2 + 2xh + 1} - e^{x^2 + 1}}{h}$
$ = \lim_{h \to 0} \frac{e^{x^2 + 1} e^{h^2 + 2xh} - e^{x^2 + 1}}{h}$
$= \lim_{h \to 0} \frac{e^{x^2 + 1} \left( e^{h\left( h + 2x \right)} - 1 \right)}{h} \times \frac{\left( h + 2x \right)}{\left( h + 2x \right)}$
$ = e^{x^2 + 1} \lim_{h \to 0} \frac{e^{h\left( h + 2x \right)} - 1}{h\left( h + 2x \right)} \lim_{h \to 0} \left( h + 2x \right)$
$ = e^{x^2 + 1} \left( 1 \right) \left( 2x \right)$
$= 2x e^{x^2 + 1}$