设f(u)具有二阶连续导数,而Z=f(e^xsiny),满足δ눀Z⼀δx눀+δ눀Z⼀δy눀=Ze^2x 求f(u).

2025-02-26 15:01:32
推荐回答(2个)
回答1:

解:
令:
u=(e^x)siny,根据链式法则,对函数Z=f[(e^x)siny]求关于x的偏导,则:
∂z/∂x
=(dz/du)·(∂u/∂x)
=f'(u)·[(e^x)siny]
=uf'(u)
∂²z/∂x²
=∂[uf'(u)]/∂x
={d[uf'(u)]/du}·(∂u/∂x)
=[u'f'(u)+uf''(u)]·u
=[f'(u)+uf''(u)]·u
=uf'(u)+u²f''(u)
后面类似,相信你已经会了

回答2:

简单计算一下即可,答案如图所示