y=sin（3x^2+5）求dy/dx

设y=sin(3x^[2]+[5]x+1),求dy/dx.dy/dx=(6x+5)*cos(3x^2+5x+1)

x=sin(y/x)+e^2求dy/dxx=sin(y/x)+e^2求dy/dxd(x)=d(sin(y/x)+e^2)dx=dsin(y/x)+de^2dx=cos(y/x)d(y/x)dx=cos(y/x)(xdy-ydx)/x^2x^

y=sin²（x²）求dy/dxdy=2sin(x²)dsin(x²)=2sin(x²)cos(x²)dx²=2sin(2x²)dx所以dy/dx=2sin(2

3X^2+4X^3y-5y^2+sin(y)=6求dy/dx等式两边求导6x+12x^2y+4x^3dy/dx-10ydy/dx+cosydy/dx=0(4x^3-10y+cosy)dy/dx=-6x-12x^2ydy/dx=(6x+12x

数学题求dy/dx设f'(x)=sin√x定义（x>0）,又y=f[e^(2x)]求dy/dx=f'(e^2x)*e^2x*2=sin(根号e^2x)*e^2x*2=2sin(e^x)*e^(2X)

设sin（xy）-x^2y=0,求dy／dx,求导什么的都要疯了!等式两边对x求导：cos(xy)*(y+x*y')-(2x*2y+x^2*2*y'=0解出y'即为所求大一知识全忘了，不过看起来好简单的样子

求dy/dx=(x-y+5)/(x+y-2)dy/dx=(x-y+5)/(x+y-2)=[(x+3/2)-(y-7/2)]/[(x+3/2)+(y-7/2)]令v=y-7/2,u=x+3/2,原方程化为dv/du=(u-v)/(u+v)变为

设sin(x+y)=xy,求dy/dx.cos(x+y)(1+y')=y+xy'dy/dx=y'=[y-cos(x+y)]/[cos(x+y)-x]dx/dy=y/[cos(x+y)-x]

求通解dy/dx=sin(x-y)∵dy/dx=sin(x-y)==>dy=sin(x-y)dx==>dx-dy=dx-sin(x-y)dx==>d(x-y)=(1-sin(x-y))dx==>d(x-y)/(1-sin(x-y))=dx=

y=sin[sin(x^2)]则dy/dx=?求详解dy/dx相当于对x进行求导：dy/dx=y'=2x*cos[sin(x^2)]*cos(x^2)由于：sinx=cosx,sin(x^2)=2x*cos(x^2)y=sin[sin(x^

求∫dy/(2sin(y/2))=-∫sin(x/2)dxps：原来的问题是求解微分方程dy/dx+cos(x-y)/2=cos(x+y)/2,求解微分方程dy/dx+cos((x-y)/2)=(cos(x+y)/2),原来那个少打了两个括

y=sin(x^2),dy/dx=?公式

dy/dx=（sin根号下x）/（y根号下x）求通解dy/dx=sin(x^1/2)/y*x^(1/2)1/ydy=sin(x^1/2)/x^(1/2)dx=2d(sin(x^1/2))lny=2sin(x^1/2)+cy=c1*e^(2s

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

y=(2x-7)^5求dy/dx=?设2x-7=ay=a^5所以dy/dx=（dy/da）×（da/dx）dy/da=5a^4da/dx=2dy/dx=10a^4=10(2x-7)^4这其实用一个chainrule就很容易了令2x-7=ty

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

y=[sin(x^4)]^2,则dy/dx=?,dy^2/dx^2=?,dy/d(x^2)=?dy/dx=2sin(x^4)cos(x^4)*4x^3复合函数求导dy^2/dx^2=[8x^3sin(x^4)cos(x^4)]^2dy/d(

高等数学求微分方程的通解dy/dx=y/x+e^(y/x)x^2*dy/dx+2xy=5y^3y''''-2y'''+5y''=01,dy/dx=y/x+e^(y/x)为齐次微分方程,令u=y/x,则y=xu,原方程化为u+xdu/dx=u

y=（x^2·e^x）+㏑3,求dy/dx.y=（x^2﹚﹙e^x）+㏑3,求dy/dx.解y'=dy/dx=(x²e^x)'+(ln3)'=2xe^x+x²e^x+0=2xe^x+x²e^x

设函数y=sin(2x²+1),求dy/dx,d²y/dx²