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问题: ycosx=e^2y y=cosylnx y=5^x+2^y

求y'

解答:

ycosx=e^2y y=cosylnx y=5^x+2^y
1.ycosx=e^2y
两边导数得:
yˊcosx-ysinx=2yˊe^2y
yˊcosx-2yˊe^2y=ysinx
yˊ(cosx-2e^2y)=ysinx
yˊ=ysinx/(cosx-2e^2y)
2.y=cosylnx
两边导数得:
yˊ=-yˊsinylnx+cosy/x
yˊ+yˊsinylnx=cosy/x
yˊ(1+sinylnx)=cosy/x
yˊ=cosy/x(1+sinylnx)
3.y=5^x+2^y
两边导数得:
yˊ=5^xln5+yˊ2^ylnz
yˊ-yˊ2^ylnz=5^xln5
yˊ(1-2^ylnz)=5^xln5
yˊ=5^xln5/(1-2^ylnz)