设函数y=f(x)由方程sin(x^2+y)=xy 确定,求dy\dx
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/06 12:54:42
![设函数y=f(x)由方程sin(x^2+y)=xy 确定,求dy\dx](/uploads/image/z/1988609-41-9.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0y%3Df%28x%29%E7%94%B1%E6%96%B9%E7%A8%8Bsin%28x%5E2%2By%29%3Dxy+%E7%A1%AE%E5%AE%9A%2C%E6%B1%82dy%5Cdx)
xJ0 _ v[-F/2'Ef{ +eEj.4^$&_ZV'$(b;慹V2d @nA@v?jVuwwrㄍl=bsX#4nA
Gd;r-[/KD$A
I
su&E[ C6$dYTA*l<_\ah_V7%[=2MckNYk1$r(4TDSS=O-jjޔ
(t~
M^.:
设函数y=f(x)由方程sin(x^2+y)=xy 确定,求dy\dx
设函数y=f(x)由方程sin(x^2+y)=xy 确定,求dy\dx
设函数y=f(x)由方程sin(x^2+y)=xy 确定,求dy\dx
这个题目要利用隐函数的求导法则.
则sin(x^2+y)=xy (两边同时求导,还要结合复合函数的求导法则)
cos(x^2+ y)*(2x+y′)=y+xy′
2xcos(x^2+y)-y=xy′-y′ cos(x^2+ y)
2xcos(x^2+y)-y=y′(x-cos(x^2+ y))
y′={2xcos(x^2+y)-y}/(x-cos(x^2+ y))
则dy\dx= y′={2xcos(x^2+y)-y}/(x-cos(x^2+ y))
sin(x²+y)=xy
两边对x求导得cos(x²+y)*(2x+y')=y+x*y'
所以(cos(x²+y)-x)y'=y-2xcos(x²+y)
所以y'=[y-2xcos(x²+y)]/[cos(x²+y)-x]
即dy/dx=[y-2xcos(x²+y)]/[cos(x²+y)-x]