x=2sin(t),y=cos(t+pie),求dy/dx,和d^2y/dx^2,在点（0,-1）

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/11/09 09:43:37

x=2sin(t),y=cos(t+pie),求dy/dx,和d^2y/dx^2,在点（0,-1）
我求出了dy/dx=sin(t)/2cos(t),在点（0,-1）时,dy/dt=0
哪个高手帮我求一下d^2y/dx^2,

x=2sin(t),y=cos(t+pie
x=2sint
y=-cost
x/2)^2+y^2=1
x^2+4y^2-4=0
则：2x+8ydy/dx=0
dy/dx=-x/4y=-2sint/(-4cost)=sint/(2cost)
2x+8ydy/dx=0
1+4[(dy/dx*dy/dx)+yd^2y/dx^2]=0
代入dy/dx=-x/4y
1+4[（-x/4y)^2+yd^2y/dx^2]=0
1+4(x^2/(16y^2)+ 4yd^2y/dx^2=0
4y^2+x^2+16y^3 d^2y/dx^2=0
d^2ydx^2=-(4y^2+x^2)/(16y^3)
代入x=2sint
y=-cost
d^2ydx^2=-(4(cost)^2+4(sint)^2)/(-16(cost)^3)=((cost)^2+(sint)^2)/(4(cost)^3)=1/(4(cost)^3)

x=2sin(t),y=cos(t+pie),求dy/dx,和d^2y/dx^2,在点（0,-1）设函数y=y(x)由x=1-e^t和y=t+e^-t确定,求dy/dx和d^2y/dx^2 dx/dt=6t+2,dy/dt=（3t+1）sin（t^2）,求d^2y/dx^2 设x=1+t^2、y=cost 求 dy/dx 和d^2y/dx^2 sint-tcost/4t^3 和 sint-tc 急设x=2t^(2)-1,y=根号(1+t^2).求dy/dx和d^2y/dx^2 设x=e'sin t,y=e'cos t,求dy/dx. 求导：x=cos^4*t,y=sin^4*t,求dy/dx x=2(cot t),y=2(sin²t),用t表示dy/dx 着急!作变换t=tanx,将微分方程cos^4x(d^2y/dx^2)+2cos^2x(1-sinxcosx)dy/dx y＝∫（x 1）sin（t∧2）dt,求dy/dx y＝∫（1 -x）sin（t∧2）dt,求dy/dx x=t^2+t y=ln(1+t) 求dy/dx