高等数学题目?
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(4). 方程 x^y=y^x能确定函数y=y(x),求dy/dx;
解:两边取对数得:ylnx=xlny;
两边对x取导数得:y'lnx+(y/x)=lny+(x/y)y'
移项得:[lnx-(x/y)]y'=lny-(y/x)
∴y'=[lny-(y/x)]/[lnx-(x/y)]=(xylnx-y²)/(xylnx-x²)
(5). y=1+xe^y;求d²y/dx²;
解:dy/dx=e^y+x(e^y)(dy/dx);故dy/dx=(e^y)/(1-xe^y);
d²y/dx²=[(1-xe^y)(e^y)y'-(e^y)(-e^y-xe^y•y')]/(1-xe^y)²
=[(e^y)y'+e^(2y)]/(1-xe^y)²
=[e^(2y)/(1-xe^y)+e^(2y)]/(1-xe^y)²
=[2e^(2y)-xe^(3y)]/(1-xe^y)³;
(6). ∫[1/x²(1+x²)]dx=∫[(1/x²)-1/(1+x²)]dx=∫(1/x²)dx-∫[1/(1+x²)]dx
=-(1/x)-arctanx+C ;
解:两边取对数得:ylnx=xlny;
两边对x取导数得:y'lnx+(y/x)=lny+(x/y)y'
移项得:[lnx-(x/y)]y'=lny-(y/x)
∴y'=[lny-(y/x)]/[lnx-(x/y)]=(xylnx-y²)/(xylnx-x²)
(5). y=1+xe^y;求d²y/dx²;
解:dy/dx=e^y+x(e^y)(dy/dx);故dy/dx=(e^y)/(1-xe^y);
d²y/dx²=[(1-xe^y)(e^y)y'-(e^y)(-e^y-xe^y•y')]/(1-xe^y)²
=[(e^y)y'+e^(2y)]/(1-xe^y)²
=[e^(2y)/(1-xe^y)+e^(2y)]/(1-xe^y)²
=[2e^(2y)-xe^(3y)]/(1-xe^y)³;
(6). ∫[1/x²(1+x²)]dx=∫[(1/x²)-1/(1+x²)]dx=∫(1/x²)dx-∫[1/(1+x²)]dx
=-(1/x)-arctanx+C ;
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