y=cos(xy)-x,求dy-搜答案-搜搜做题作业网

y=(1+x^4)^cos x,求dy

(1+X4)COSx先求导数再乘以dx就行了

设y=cos^2(x+1),求dy

楼上的少写了“-”和“dx”吧dy=2cos(x+1)•[-sin(x+1)]dx=-sin2(x+1)dx

y= (cos x)^x,求dy/dx

最好的办法是求对数：lny=xlncosx,两边求导数得：y'/y=lncosx-x(sinx/cosx)=lncosx-xtanx所以：y'=y(lncosx-xtanx)=(cosx)^x(lnc

x^2+xy+y^3=1,求dy/dx

解析2xdx+ydx+xdy+3y²dy=0(2x+y)dx+(x+3y²)dy=0(2x+y)dx=-(x+3y²)dydy/dx=(2x+y)/-(x+3y²

ysinx=cos(xy)两边分别求导y'sinx+ycosx=-sin(xy)(y+xy')y'=-y(sin(xy)+cosx)/(sinx+xsin(xy))

y/x=ln(xy) 求详 dy/dx

方法一（微分法）d(y/x)=d(ln(xy))(xdy-ydx)/x²=1/xy*d(xy)即（xdy-ydx)/x²=(ydx+xdy)/xy∴dy/dx=(xy+y²

y/x=ln(xy) 求dy/dx

两边求导(y'x-y)/x^2=(y+xy')/xyxy+x^2y'=xyy'+y^2y'=(xy-y^2)/(xy+x^2)

xy=e^(x+y),求dx/dy

ydx/dy+x=(e^x)(e^y)dx/dy+(e^x)(e^y)dx/dy=[(e^x)(e^y)-x]/[y-(e^x)(e^y)]dx/dy=(xy-x)/(y-xy)dx/dy=x(y-1

f(x,y)=e^(x+y)+cos(xy)=0 //: 利用隐函数存在定理：f 'x(x,y)=e^

cos(x+y)(1+y')=y+xy'dy/dx=y'=[y-cos(x+y)]/[cos(x+y)-x]

e∧x+y=xy,求dy/dx

cos(xy)=x两边对x求导：-sin(xy)[y+xy']=1y+xy'=-1/sin(xy)xy'=-y-(1/sin(xy))y'=[-y-(1/sin(xy))]/x

应用复合函数求导方法,y′sinx+ycosx+（1+y′）sin（x+y）=0,（sinx+sin（x+y））y′+ycosx+sin（x+y）=0,y′=-（ycosx+sin（x+y））/（si

cos(x+y)+y=1 求dy/dx

我算的结果和你的一样,也是y'=sin(x+y)/1-sin(x+y)应该是书上写错了.在说xsin(x+y)中的x从何而来?找不到它的来源啊.不管是对cos(),还是对y求导都不会出现xsin()这

cos(xy)=x-y,隐函数,两边求导-sin(xy)*(xy)'=1-y'-sin(xy)*(y+xy')=1-y'-ysin(xy)-xcos(xy)*y'=1-y'y'[1-xsin(xy)]

x=0时,代入方程得：1+1=y,得：y=2对x求导：(y+xy')e^xy-sin(xy)*(y+xy')=y'将x=0,y=2代入得：2=y'故dy(0)=2dx

两边同时求导x+x(dy/dx)+1*(dy/dx)/y+1/x=0合并同类项dy/dx=-y/x

dy/dx=-2cosxsinx-5x的4次方所以dy=(-sin2x-5x的4次方)dx

∵x=cos(y/x)==>1=-sin(y/x)*(xy'-y)/x^2(等式两端对x求导)==>xy'-y=-x^2/sin(y/x)==>xy'=y-x^2/sin(y/x)==>y'=y/x-

求dy/dx+y/x=e^(xy)

令e^(xy)=u,y=lnu/xDy/dx=[(x/u)*(du/dx)-lnu]/x²,∴(1/ux)*(du/dx)-lnu/x²+lnu/x²=u即du/u