Question

求解两道导函数问题

求dy/dx
1.cos（xy^2)=y
2.√ y-sin x=2
3. (xy^3)/ (1+sec y) = 1+y^4

写一下解题步骤吧...谢谢了！

Answer 1

1.y'=-sin(xy²)(xy²)'=-sin(xy²)(y²+2xyy')
[1+2xysin(xy²)]y'=-y²sin(xy²)
y‘=-y²sin(xy²)/[1+2xysin(xy²)]
2.√ y-sin x=2 两边对x求导
1/2(√ y-sin x)(y-sinx)'=0
1/2(√ y-sin x)(y'-cosx)=0
y'=cosx
3.(xy^3)/ (1+sec y) = 1+y^4 两边对x求导
[(xy³)'(1+sec y)-(1+sec y)'(xy³)]/(1+sec y)²=4y³y'
y'=y(1+sec y)/[4y(1+sec y)²-3x(1+sec y)+xytany]