1 f(x.y)可微,F(X,X^2)=2X^2,f(x,x^2)对x的偏导数等于2xf(x.y)可微,F(X,X^2)=2X^2,f(x,x^2)对x的偏导数等于2x,则f(x,x^2)对Y的偏导数?答案为什么是1
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/11 02:30:54
1f(x.y)可微,F(X,X^2)=2X^2,f(x,x^2)对x的偏导数等于2xf(x.y)可微,F(X,X^2)=2X^2,f(x,x^2)对x的偏导数等于2x,则f(x,x^2)对Y的偏导数?
1 f(x.y)可微,F(X,X^2)=2X^2,f(x,x^2)对x的偏导数等于2xf(x.y)可微,F(X,X^2)=2X^2,f(x,x^2)对x的偏导数等于2x,则f(x,x^2)对Y的偏导数?答案为什么是1
1 f(x.y)可微,F(X,X^2)=2X^2,f(x,x^2)对x的偏导数等于2x
f(x.y)可微,F(X,X^2)=2X^2,f(x,x^2)对x的偏导数等于2x,则f(x,x^2)对Y的偏导数?
答案为什么是1
1 f(x.y)可微,F(X,X^2)=2X^2,f(x,x^2)对x的偏导数等于2xf(x.y)可微,F(X,X^2)=2X^2,f(x,x^2)对x的偏导数等于2x,则f(x,x^2)对Y的偏导数?答案为什么是1
1 f(x.y)可微,F(X,X^2)=2X^2,f(x,x^2)对x的偏导数等于2xf(x.y)可微,F(X,X^2)=2X^2,f(x,x^2)对x的偏导数等于2x,则f(x,x^2)对Y的偏导数?答案为什么是1
设f(x)可微.y=f(lnx)+f(sin^2*x),求dy
设f(x)可微,y=f(e^x)/e^[f(x)],y '=
f(x)连续且可导,并且f(x+y)=[f(x)+f(y)]/[1-f(x)f(y)],求f(x)
大一 多元函数微分学设函数f(x,y)可微,且f(x,x^2)=1 (1)若f(x,x^2)对x的偏导数=x,求f(x,x^2)对y的偏导数(2)若f(x,y)对y的偏导数=x^2+2y,求f(x,y)
已知f(x)=3^x,求证:(1)f(x)·f(y)=f(x+y);(2)f(x)/f(y)=f(x-y).
设f(x)=e的y次方,证明:(1),f(x)f(y)=f(x+y) ,(2),f (x)/f(y)=f(x-y)
设y=f[(sinx)^2]+f[(cosx)^2],f(x)可微,求dy
导数:f(x+y)=f(x)f(y),且f'(o)=1,求f'(x)f(x+y)=f(x)f(y),且f'(o)=1,求f'(x)f(x+y)=f(x)+f(y)+2xy,且f'(o)存在,求f'(x) f(1+x)=af(x),且f'(0)=b,求f'(1)
f(x+y)+f(xy-1)=f(x)f(y)+2f(n)表达式
高数题:y^2f(x)+xf(y)=x^2(^代表平方),f(x)可微,求dy.
设函数f可微,z=f(ye^x,x/(y^2)) 求z/x,z/y
设f(x)可微,2=f(x*x-y*y),求一阶偏导eZ/eX,eZ/eY.
设f(x)可导,且f'(0=1,又y=f(x^2+sin^2x)+f(arctanx),求dy/dx /x=0
若f(x)在(-∞,+∞)内处处可道,且f'(0)=1,此外,对任意实数x,y恒有f(x+y)=f(x)+f(y)+2xy,则f(x)=?
如果函数F(x)在R上处处可导F(0)'=1对于任意x,y恒有F(x+y)=F(x)+F(y)+2xy,求F(x)'?
求y=f(x)=x^2的导函数.【f(x)可导】
f(x+y)=f(x)+f(y)+2xy