Question

大一高数求解

Answer 1

z=[cos(x+y)]^(siny)
注：因为求的是z'x，故把y看做常数，从而siny也是常数
【相当于z=u^a，而u=g(x,y)，故z'x=u^a · ln|u|· u'x】
故z'x=[cos(x+y)]^(siny) · ln|cos(x+y)| · [-sin(x+y)]·1
=-sin(x+y)[cos(x+y)]^(siny) ln|cos(x+y)|
故z'x (x，π/2)=-sin(x+π/2)[cos(x+π/2)]^[sin(π/2)] ln|cos(x+π/2)|
=-cosx (-sinx)^1 · ln|-sinx|
=sinx cosx ln|sinx|

Answer 2

请点击输入图片描述

Answer 3

两边取对数，然后求导，lnz=siny*ln(cos(x+y))
z'/z=siny*-sin(x+y)/cos(x+y)
带入y=π/2
得出：z'/z=-sin(x+π/2)/cos(x+π/2)
z'=z*cos(x)/sin(x)=z*cot(x)