作业帮求导：y=[e^x-e^(-x)]/[e^x+e^(-x)]
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/17 08:25:13
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求导：y=[e^x-e^(-x)]/[e^x+e^(-x)]
求导：y=[e^x-e^(-x)]/[e^x+e^(-x)]
求导：y=[e^x-e^(-x)]/[e^x+e^(-x)]
先注意,(e^x)'=e^x,(e^(-x))'=-e^(-x).
再利用相除的求导法则,即可求得
y'={[e^x-e^(-x)]/[e^x+e^(-x)]}'
={[e^x-e^(-x)]'[e^x+e^(-x)]-[e^x-e^(-x)][e^x+e^(-x)]'}/[e^x+e^(-x)]^2
={[e^x+e^(-x)]^2-[e^x-e^(-x)]^2}/[e^x+e^(-x)]^2
=4/[e^x+e^(-x)]^2
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