곱의 미분

$\displaystyle \lim\limits _{x \rightarrow 1} {\frac{f \left( x \right) - 2}{x - 1}} = 12$에서 극한값이 존재하고 $x \rightarrow 1$일 때, (분모)$\rightarrow 0$이므로 (분자)$\rightarrow 0$ $\displaystyle \lim\limits _{x \rightarrow 1} {\left\{ f \left( x \right) - 2 \right\}} = 0$에서 $f \left( 1 \right) = 2$이다. $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{f \left( x \right) - 2}{x - 1}} = \lim\limits _{x \rightarrow 1} {\frac{f \left( x \right) - f \left( 1 \right)}{x - 1}} = f' \left( 1 \right) = 12$ $g' ( x ) = 2 xf \left( x \right) + \left( x ^{2} + 1 \right) f' ( x )$이므로 $g' ( 1 )$$= 2 f \left( 1 \right) + 2 f' ( 1 )$ $= 2 \times 2 + 2 \times 12$$= 4 + 24 = 28$