t=sin(2x 1),dy=
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function z=krsx(t,y) &nb
sin(x^2+y^2)=x两边同时求导,得(x^2+y^2)'cos(x^2+y^2)=dx(2xdx+2ydy)cos(x^2+y^2)=dx2xdx+2ydy=dx/cos(x^2+y^2)2y
y'=2xsin4x-x²cos4x·4所以dy=(2xsin4x-4x²cos4x)dxy=ln√4+t²=1/2ln(4+t²)y'=1/2·1/(4+t&
e'表示对自然对数e求导,e'=0但是在dy/dx的过程中由于分子和分母都有e',可以约掉,所以不用急着把分子分母都等于0,这样就做不出来了.dy/dx=(dy/dt)/(dx/dt)dy/dt=(e
dx/dt=4(cost)^3*(cost)'dy/dt=4(sint)^3*(sint)'而(cost)'=-sint(sint)'=cost于是dy/dx=(dy/dt)/(dx/dt)=4(co
d/dx∫(1/x→√x)sin(t²)dt=d(√x)/dx·sin(√x²)-d(1/x)/dx·sin(1/x²)=1/(2√x)·sin|x|-(-1/x
dy/dx=(3t+1)sin(t²)/(6t+2)=1/2sin(t²)dt²y/dx²=d[1/2sin(t^2)]/dx=t*cos(t²)*d
x=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^
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)
加油啊,不然大学考不起啊!
dy=2sin[x(x+1)]cos[x(x+1)](2x+1)
此微分方程没有显式解,建议用数值解法 function dyy=xielei(t,y) %&nbs
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(x^2)=2sin(x^4)cos(x^4)*2x^2
y+xy'-cos(πy²)2πyy'=0y=[2πycos(πy²)-x]y'y'=y/[2πycos(πy²)-x]即:dy/dx=y/[2πycos(πy²
请看图:请看图:请看图:再答:
原式y=sinx^2+2xdy/dx=2x·cosx^2+2
dy/d(x^3)=(dy/dx)/(d(x^3)/dx)=cosx/3(x^2)
dy/dx=[sin(-x)²](-x)'=-sin(x²)