cos(xy)=x求隐函数的导数dy/dx 详细 求sin(xy)=x的隐函数的导数dy/dx
cos(xy)=x\uff0dy\u6240\u786e\u5b9a\u7684\u9690\u51fd\u6570y=y(x)\u7684\u5bfc\u6570dy/dxcos(xy)=x-y,\u9690\u51fd\u6570,\u4e24\u8fb9\u6c42\u5bfc
-sin(xy)*(xy)'=1-y'
-sin(xy)*(y+xy')=1-y'
-ysin(xy)-xcos(xy)*y'=1-y'
y'[1-xsin(xy)]=1+ysin(xy)
y'=[1+ysin(xy)]/[1-xsin(xy)]
\u4e5f\u53ef\u7528\u8bbe\u4e8c\u5143\u51fd\u6570f(x,y)=cos(xy)-x+y
\u7528\u9690\u51fd\u6570\u6c42\u5bfc\u6cd5:f'x(x,y)+f,y(x,y)*y'=0
f'x(x,y)=-sin(xy)*(xy)'-1
=-ysin(xy)-1
f'y(x,y)=-sin(xy)*(xy)'+1
=-xsin(xy)+1
\u2234[-ysin(xy)-1]+[-xsin(xy)+1]*y'=0
y'=-[ysin(xy)-1]/[-xsin(xy)+1]
y'=[1+ysin(xy)]/[1-xsin(xy)]
\u4e24\u8fb9\u540c\u65f6\u5bf9x\u6c42\u5bfc\u5f97
cos(xy)(y+xy')=1
\u89e3\u51fay'\u5373\u5f97
dy/dx=1/xcos(xy)-y/x
cos(xy)=x
两边对x求导：-sin(xy)[y+xy']=1
y+xy'=-1/sin(xy)
xy'=-y-(1/sin(xy))
y'=[-y-(1/sin(xy))]/x
隐函数导数的求解一般可以采用以下方法：
方法①：先把隐函数转化成显函数，再利用显函数求导的方法求导；
方法②：隐函数左右两边对x求导（但要注意把y看作x的函数）；
方法③：利用一阶微分形式不变的性质分别对x和y求导，再通过移项求得的值；
方法④：把n元隐函数看作(n+1)元函数，通过多元函数的偏导数的商求得n元隐函数的导数。
解：cos(xy)=x.关于x求导：[-sin(xy)]×(y+xy′)=1.===>y+xy′=-1/sin(xy).===>xy'=-y-[1/sin(xy)].===>y'={-y-[1/sin(xy)]}/x.∴(dy)/(dx)=-{y+[1/sin(xy)]}/x
cos(xy)=x
两边对x求导：-sin(xy)[y+xy']=1
y+xy'=-1/sin(xy)
xy'=-y-(1/sin(xy))
y'=[-y-(1/sin(xy))]/x
cos(xy)=x
dcos(xy)/d(xy)*d(xy)/dx=dx/dx
-sin(xy)*[y*dx/dx+x*dy/dx]=1
y+x*dy/dx=-csc(xy)
x*dy/dx=-csc(xy)-y
dy/dx=-[y+csc(xy)]/x
cosxy(y+ xy')=1 解的y'=(1-ycosxy)/ xcosxy
绛旓細cos(xy)=x 涓よ竟瀵x姹傚锛-sin(xy)路(xy)'=1 -sin(xy)(y+xy')=1 y'(-x)路sin(xy)=1+y路sin(xy)y'=-[1+y路sin(xy)]/[x路sin(xy)]
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绛旓細瑙ｆ瀽锛氶殣鍑芥暟姹傚 cos(xy)=x -sin(xy)鈼(y+xy')=1 y'=[-1/sin(xy)-y]/y cos(x+y)=y -sin(x+y)鈼(1+y')=y'y'=-sin(x+y)/[1+sin(x+y)]
绛旓細闅愬嚱鏁扮殑瀵兼暟鍙互杩欐牱姹傝В銆傞亣鍒拌嚜鍙橀噺x锛岀洿鎺ヨ繍鐢ㄥ熀鏈嚱鏁扮殑瀵兼暟鍏紡姹傚锛涢亣鍒板洜鍙橀噺y锛屽厛杩愮敤鍩烘湰鍑芥暟鐨勫鏁板叕寮忔眰瀵硷紝鐒跺悗闄勫姞瀵兼暟鍒嗗彿y'锛涙渶鍚庢暣鐞嗘眰寰梱'.瑙ｏ細鏂圭▼涓よ竟姹傚锛屾湁 (xlny)'+(ycos(xy))'=(x)'lny+x/yy'+cos(xy)y'-y²sin(xy)-xysin(xy)y'=1 {(x/y+cos(xy)-xy...
绛旓細鎴戜滑鍙互浣跨敤闅愬嚱鏁版眰瀵兼硶鏉ユ眰瑙ｈ繖涓棶棰樸傞鍏堬紝瀵规柟绋 x = cos(xy) 涓よ竟鍚屾椂瀵 x 姹傚锛屽埄鐢ㄩ摼寮忔硶鍒欙紝寰楀埌锛1 = (-sin(xy))(y + xy') + cos(xy)(y'x + y)鎺ヤ笅鏉ワ紝灏 y' 绉婚」锛屾暣鐞嗗緱鍒 y' 鐨勮〃杈惧紡锛歽' = (1 - cos(xy)y) / (x + sin(xy)y)杩欏氨鏄 y 瀵兼暟鐨勮〃杈惧紡锛...
绛旓細1銆cos(xy)=x+y涓よ竟姹傚寰-sin(xy)[y+xy']=1+y',y'=-[1+ysin(xy)]/[1+xsin(xy)]2銆亂=tan(x+y) 涓よ竟姹傚寰梱'=(sec(x+y))^2(1+y'),y'=(sec(x+y))^2/[1-(sec(x+y))^2]
绛旓細-sin(xy)*(y+xy')=1-y'-ysin(xy)-xcos(xy)*y'=1-y'y'[1-xsin(xy)]=1+ysin(xy)y'=[1+ysin(xy)]/[1-xsin(xy)]涔熷彲鐢ㄨ浜屽厓鍑芥暟f(x,y)=cos(xy)-x+y 鐢闅愬嚱鏁版眰瀵娉:f'x(x,y)+f,y(x,y)*y'=0 f'x(x,y)=-sin(xy)*(xy)'-1 =-ysin(xy)-1 f'y(x,y...
绛旓細-sin(xy)*(y+xy')=1-y'-ysin(xy)-xcos(xy)*y'=1-y'y'[1-xsin(xy)]=1+ysin(xy)y'=[1+ysin(xy)]/[1-xsin(xy)]涔熷彲鐢ㄨ浜屽厓鍑芥暟f(x,y)=cos(xy)-x+y 鐢闅愬嚱鏁版眰瀵娉:f'x(x,y)+f,y(x,y)*y'=0 f'x(x,y)=-sin(xy)*(xy)'-1 =-ysin(xy)-1 f'y(x,y...
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