例：已知y=(x+1)(x+2)/(x+3)，求y'解：两边取自然对数：lny=ln(x+1)+ln(x+2)-ln(x+3);两边对x取导数得：y'/y=1/(x+1)+1/(x+2)-1/(x+3)故y'=y[1/(x+1)+1/(x+2)-1/(x+3)]=[(x+1)(x+2)/(x+3)][1/(x+1)+1/(x+2)-1/(x+3)]这样计算可以使计算大为简化。
