1樓:可琪理瑞
∵z=f(2x-y,ysinx)∴?
?xz=??x
f(2x-y,ysinx)
=f1′??x
(2x-y)+f2'??x
(ysinx)
=2f1′+ycosxf2'
?2z?x?y=?
?y(2f1′+ycosxf2')=2?
?yf1′+cosx??y
(yf2')
因為:?
?yf1′=f11″??y
(2x-y)+f12″??y
(ysinx)
=-f11″+sinxf12″??y
(yf2')=f2'+y??y
f2'=f2'+y[f21″??y
(2x-y)+f22″??y
(ysinx)]
=f2'+y[-f21″+sinxf22″]
=f2'-yf21″+ysinxf22″
所以:?2z
?x?y=2?
?yf1′+cosx??y
(yf2')
=2(-f11″+sinxf12″)+cosx(f2'-yf21″+ysinxf22″)
=-2f11″+2sinxf12″+cosxf2'-ycosf21″+ysinxcosxf22″
又因為函式f具有連續二階導數,所以其二階混合偏導數相等,即:
f12″=f21″
所以:?2z
?x?y
=-2f11″+2sinxf12″+cosxf2'-ycosf21″+ysinxcosxf22″
=-2f11″+(2sinx-ycosx)f12″+cosxf2'+ysinxcosxf22″
故?2z
?x?y
的值為:
-2f11″+(2sinx-ycosx)f12″+cosxf2'+ysinxcosxf22″
2樓:茹翊神諭者
簡單計算一下即可,答案如圖所示
設z=f(2x-y,ysinx),其中f(u,v)具有連續的二階偏導數,求?2z?x?y
3樓:氣象天使丶
∵z=f(2x-y,ysinx)
∴??x
z=??x
f(2x-y,ysinx)
=f1′?
?x(2x-y)+f2'?
?x(ysinx)
=2f1′+ycosxf2'?z
?x?y
=??y
(2f1′+ycosxf2')
=2??y
f1′+cosx?
?y(yf2')
因為:?
?yf1′=f11″?
?y(2x-y)+f12″?
?y(ysinx)
=-f11″+sinxf12″??y
(yf2')=f2'+y?
?yf2'
=f2'+y[f21″?
?y(2x-y)+f22″?
?y(ysinx)]
=f2'+y[-f21″+sinxf22″]
=f2'-yf21″+ysinxf22″
所以:?
z?x?y
=2??y
f1′+cosx?
?y(yf2')
=2(-f11″+sinxf12″)+cosx(f2'-yf21″+ysinxf22″)
=-2f11″+2sinxf12″+cosxf2'-ycosf21″+ysinxcosxf22″
又因為函式f具有連續二階導數,所以其二階混合偏導數相等,即:
f12″=f21″
所以:?
z?x?y
=-2f11″+2sinxf12″+cosxf2'-ycosf21″+ysinxcosxf22″
=-2f11″+(2sinx-ycosx)f12″+cosxf2'+ysinxcosxf22″故?z
?x?y
的值為:
-2f11″+(2sinx-ycosx)f12″+cosxf2'+ysinxcosxf22″
4樓:茹翊神諭者
簡單計算一下即可答案如圖所示
5樓:郜飆操宛暢
zx=f1*2+f2
ycosx
=2f1+ycosxf2
zxy=-2f11+2sinxf12+cosxf2+ycosx(-f21+sinxf22)
=-2f11+2sinxf12+cosxf2-ycosxf21+ysinxcosxf22
設z=f(xy,x²+2y),其中f(u,v) 具有二階連續偏導數,求az/ax,az/ay,a²z/axay。
6樓:雀玉花弓綢
令xy=u,下面在xy為中間變數且要求導時用u表示,即f(x,xy)=f(x,u),ðu/ðx=y,ðu/ðy=x.
求z對x的一階偏導數ðz/ðx=2xf(x,xy)+x²[ðf(x,u)/ðx+ðf(x,u)/ðu*x)
f對y的一階偏導數後面要用到
ðf(x,u)/ðy=ðf(x,u)/ðu*x=xðf(x,u)/ðu
∴ð²z/ðxðy=ð(ðz/ðx)/ðy
=ð(2xf(x,xy)+x²*ðf(x,u)/ðx+x3*ðf(x,u)/ðu)/ðy
=2xðf(x,u)/ðu*x+x²*ð²f(x,u)/ðxðu*x+x立方ð²f(x,u)/ðu²*x
=2x²ðf(x,u)/ðu*x+x立方*ð²f(x,u)/ðxðu+x四次方ð²f(x,u)/ðu²
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