y=sin(x y),求y及dy
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![y=sin(x y),求y及dy](/uploads/image/f/910856-56-6.jpg?t=y%3Dsin%28x+y%29%2C%E6%B1%82y%E5%8F%8Ady)
将y看成是关于x的函数即y=f(x)我们在求导的同时要记得y也要对x求导即dy/dx我们两边分别对x求导得e^x+e^y*dy/dx=cos(xy)*(y+x*dy/dx)移项e^x-y*cos(xy
y=sinx/(1+x^2)先求导:y'=cosx*(1+x^2)-sinx*(2x)/(1+x^2)^2那么,dy=y'dx=[cosx*(1+x^2)-sinx*(2x)]dx/(1+x^2)^2
e^(xy)+sin(xy)=y(y+xy')e^(xy)+(y+xy')cos(xy)=y'y'=(ye^(xy)+ycos(xy))/(1-xe^(xy)-xcos(xy))
这个题目要利用隐函数的求导法则.则sin(x^2+y)=xy(两边同时求导,还要结合复合函数的求导法则)cos(x^2+y)*(2x+y′)=y+xy′2xcos(x^2+y)-y=xy′-y′cos
将原方程两边微分得d[xe^y+sin(xy)]=0→e^ydx+xe^ydy+cos(xy)(ydx+xdy)=0→移项[xe^y+xcos(xy)]dy=-[e^y+ycos(xy)]dx整理→d
sin(x^2+y^2)+e^x-xy^2=0左右微分得到cos(x^2+y^2)*(2xdx+2ydy)+(e^x)dx-(y^2)dx-2xydy=0余下的求出dy就可以了
cos(x+y)(1+y')=y+xy'dy/dx=y'=[y-cos(x+y)]/[cos(x+y)-x]
y=x*sin(lnx)y'=sin(lnx)+x*cos(lnx)*(lnx)'=sin(lnx)+x*cos(lnx)*1/x=sin(lnx)+cos(lnx)dy=[sin(lnx)+cos(
y'=1/(1+x²)×反正切就是括号里的手机打不出来dy=y'dx手机答好麻烦.给分.
三种方法1式中同时对x求导-(y+xy‘)cosxy+2yy'=0解出y’2式中同时取微分d{sin(xy)+y^2-e^2}=dsin(xy)+dy^2-de^2=-cosxydxy+2ydy=-c
就是把这dydx转为求导前的式子,然后再求导一遍验证一下对错.再问:就是算到最后有个积分搞不出来。求过程。
等式两边对x求导:cos(xy)*(y+x*y')-(2x*2y+x^2*2*y'=0解出y'即为所求
y+xy'-cos(πy²)2πyy'=0y=[2πycos(πy²)-x]y'y'=y/[2πycos(πy²)-x]即:dy/dx=y/[2πycos(πy²
∵siny+e^x-xy^2=0,∴(dy/dx)cosy+e^x-[y^2+2xy(dy/dx)]=0,∴(cosy-2xy)(dy/dx)=y^2-e^x,∴dy/dx=(y^2-e^x)/(co
y=sin(x+y)dy=cos(x+y)(dx+dy)dy=cos(x+y)dx+cos(x+y)dydy/dx=cos(x+y)/(1-cos(x+y))
第一题问得不清楚,看不懂.第二题,两边求导,得e^x+y'-(x'y+xy')=0整理得,dy=(e^x-y)*dx/(x-1)
dy/dx=-fx/fy,你自己可以算吧
化为:e^(ylnx)-e^y=sin(xy)两边对x求导:e^(ylnx)(y'lnx+y/x)-y'e^y=cos(xy)(y+xy')y'[lnxe^(ylnx)-e^y-xcos(xy)]=[