微分の計算問題

${\displaystyle \frac{dy}{du}=\frac{1}{u}}$

${\displaystyle =\frac{{\left(e^{3x}-3\right)}^{\prime }·\left(e^{3x}+1\right)-\left(e^{3x}-3\right)·{\left(e^{3x}+1\right)}^{\prime }}{{\left(e^{3x}+1\right)}^2}}$

${\displaystyle =\frac{3e^{3x}\left(e^{3x}+1\right)-3e^{3x}\left(e^{3x}-3\right)}{{\left(e^{3x}+1\right)}^2}}$

${\displaystyle =\frac{3e^{3x}\left\{\left(e^{3x}+1\right)-\left(e^{3x}-3\right)\right\}}{{\left(e^{3x}+1\right)}^2}}$

${\displaystyle =\frac{12e^{3x}}{{\left(e^{3x}+1\right)}^2}}$

${\displaystyle y^{\prime }={\left\{\log \left(e^{3x}-3\right)\right\}}^{\prime }-{\left\{\log \left(e^{3x}+1\right)\right\}}^{\prime }}$

${\displaystyle =\frac{1}{e^{3x}-3}{\left(e^{3x}-3\right)}^{\prime }-\frac{1}{e^{3x}+1}{\left(e^{3x}+1\right)}^{\prime }}$

${\displaystyle =\frac{3e^{3x}}{e^{3x}-3}-\frac{3e^{3x}}{e^{3x}+1}}$

${\displaystyle =\frac{3e^{3x}\left\{\left(e^{3x}+1\right)-\left(e^{3x}-3\right)\right\}}{\left(e^{3x}-3\right)\left(e^{3x}+1\right)}}$

${\displaystyle =\frac{3e^{3x}\cdot 4}{\left(e^{3x}-3\right)\left(e^{3x}+1\right)}}$

${\displaystyle =\frac{12e^{3x}}{\left(e^{3x}-3\right)\left(e^{3x}+1\right)}}$