微分(1)の1．(2)の解答

$3$\displaystyle{ f $x$ ={x}^{2} }$
$3$\displaystyle{ \begin{array}{lll} {f}^{\prime } $x$ &=& { \lim }\limits_{ {\Delta}x\rightarrow 0 }\frac{\hspace{2} { $ {\Delta}x+x $ }^{2}- {x}^{2} \hspace{2}}{\hspace{2} {\Delta}x \hspace{2}} & \vspace{6}\\ &=& { \lim }\limits_{ {\Delta}x\rightarrow 0 }\frac{\hspace{2} {\Delta}x $ 2x+{\Delta}x $ \hspace{2}}{\hspace{2} {\Delta}x \hspace{2}} & \vspace{6}\\ &=& { \lim }\limits_{ {\Delta}x\rightarrow 0 }2x+{\Delta}x & \vspace{6}\\ &=& 3x & \vspace{6}\\ \end{array} }$
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