x=ln(1+t^2)y=arctant求一阶导和二阶导

2025-01-24 05:44:33
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回答1:

x=ln√(1+t²)=½ln(1+t²) y=arctant dx/dt=t/(1+t²) dy/dt=1/(1+t²) ∴dy/dx=1/t d²y/dx²=-(1/t²)/[t/(1+t²)]=-(1+t²)/t³