Question

高数谢谢(*°∀°)22

Answer 1

1、Z'x=cos(x²-y²) · 2x
Z''x=2cos(x²-y²)-2x sin(x²-y²) ·2x
=2cos(x²-y²)-4x² sin(x²-y²)
2、Z'x=（2x+y）^y · ln|2x+y| · 2
=2（2x+y）^y · ln|2x+y|
故Z'x(0,1)=2(0+1) ln(0+1)=0

Answer 2

(1)
z=sin(x^2-y^2)
∂z/∂x
= cos(x^2-y^2) . ∂/∂x (x^2-y^2)
=2xcos(x^2-y^2)
∂^2z/∂x^2
=2[ cos(x^2-y^2) + x ∂/∂x cos(x^2-y^2) ]
=2[ cos(x^2-y^2) - 2x^2.sin(x^2-y^2) ]
(2)
z=(2x+y)^y
lnz = yln(2x+y)
(1/z)∂z/∂x = 2y/(2x+y)
∂z/∂x = [2y/(2x+y)].(2x+y)^y
∂z/∂x|(0,1) =2