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隐函数求导训练题

First

y=f(x)

y2=f(x)f(x)y2f(x)(y2)=[f2(x)]f(x)

(y2)=2yy


Second

对数求导法:先取对数,再求导数

y=x1x,y(x)dydx=?

y=x1xlny=1xlnx

y视作x的函数,对两边求导:

(lny)=(lnxx)

隐函数求导之术与复合函数求导之术通用.则根据链式法则: ddx=ddydydx

d(lny)dxdydx=d(lnxx)dx

1ydydx=lnxxxlnxx2

(x)=(x1)=1

dydx=1lnxx2y

dydx=1lnxx2x1x

dydx=x2x1x(1lnx)

dydx=x1x2(1lnx)

y(x)=x1x2(1lnx)


Second

x4+y5+y31=0,y(x)dydx

y5+y3=1x4

y看成x的函数,同时对两侧求导: (y5+y3)=(1x4)

根据链式法则得:

d(y5+y3)dxdydx=d(1x4)dx

(5y4+3y2)dydx=4x3

dydx=4x35y4+3y2

y(x)=4x35y4+3y2


Third

exy=sin(x+y),dydx=?

xy=k,x+y=u

(ek)=(sinu)

because they both are composite function, therefore:

(ek)(k)=(sinu)(u)

(ek)(k)ek[(x)(y)]=ek[xy+yx]=ek(y+xdydx)

(sinu)(u)=cos(x+y)(1+dydx)

exy(y+xdydx)=cos(x+y)(1+dydx)

exyy+exyxdydx=cos(x+y)+dydxcos(x+y)

exyxdydxdydxcos(x+y)=cos(x+y)exyy

dydx[exyxcos(x+y)]=cos(x+y)exyy

dydx=cos(x+y)exyyexyxcos(x+y)

y(x)=cos(x+y)exyyexyxcos(x+y)


Fourth

:x2y24xy=02:(x2)(y2)(4xy)=0

(x2)=2x(y2)=(y2)y=2yy(4xy)=(4x)y+4x(y)=4y+4xy

2x2yy4y4xy=02yy4xy=2x+4yy(2y4x)=2x+4y

y=2x+4y2y4x

y=2(x2y)2(y+2x)

y=x2y2x+y


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