y=sin(x y),求dy dx-搜答案-大师作文网

方程两边求关x的导数ddx(xy)＝(y+xdydx)； ddxex+y＝ex+y(1+dydx)；所以有（y+xdy

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

两边求导得：cos(xy)*(y+xy')+1+y'=0y'[xcos(xy)+1]=-ycos(xy)-1所以,y'=-[ycos(xy)+1]/[xcos(xy)+1]

再答：隐函数高阶求导。再答：

方程两边对x求导得2x+y′x2+y＝3x2y+x3y′+cosxy′＝2x−(x2+y)(3x2y+cosx)x5+x3y−1由原方程知，x=0时y=1，代入上式得y′|x＝0＝dydx|x＝0＝1

e^(xy)+sin(xy)=y(y+xy')e^(xy)+(y+xy')cos(xy)=y'y'=(ye^(xy)+ycos(xy))/(1-xe^(xy)-xcos(xy))

满意请采纳,谢谢

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]

这是一阶线性微分方程，其中P（x）=1，Q（x）=e-x∴通解y＝e−∫dx(∫e−x•e∫dxdx+C)＝e−x(∫e−x•exdx+C)＝e−x(x+C)．

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

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

limsin(xy)/x(x.y)->(0.2)=lim{[sin(xy)/xy]*y}=im[sin(xy)/xy]*(limy)(x.y)->(0.2)=1*2=2这里把(xy)看作一个整体,当(

y+xy'-cos(πy²)2πyy'=0y=[2πycos(πy²)-x]y'y'=y/[2πycos(πy²)-x]即：dy/dx=y/[2πycos(πy²

sin(xy)=x 求dx/dy

x/[sec(xy)-y]dx/dy.

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

sin(xy)-ln((x+1)/y)+1=0对x求导有：（y+xy')cos(xy)-y/(x+1)·[y-(x+1)y']/y^2-y/(x+1)·(x+1)(-1/y^2)y'=0x=0代入有：

在方程ex+y+cos（xy）=0左右两边同时对x求导，得：ex+y（1+y′）-sin（xy）•（y+xy′）=0，化简求得：y′＝dydx＝ysin(xy)−ex+yex+y−xsin(xy)．

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)]=[