x=a(θ-sinθ) y=a(1-cosθ)x=a(θ-sinθ),y=a(1-cosθ).求d^2y/dx^2.dy/dx=cotθ/2,d^2y/dx^2=d/dxdy/dx=d（dy/dx)/dθdθ/dx=d(dy/dx)/dθ1/dx/dθ= 我想问哈d^2y/dx^2那后面d（dy/dx)/dθdθ/dx=d(dy/dx)/dθ*1/dx/dθz这两步是如何化解的.

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x=a(θ-sinθ)y=a(1-cosθ)x=a(θ-sinθ),y=a(1-cosθ).求d^2y/dx^2.dy/dx=cotθ/2,d^2y/dx^2=d/dx*dy/dx=d（dy/dx)/

x=a(θ-sinθ) y=a(1-cosθ)x=a(θ-sinθ),y=a(1-cosθ).求d^2y/dx^2.dy/dx=cotθ/2,d^2y/dx^2=d/dx*dy/dx=d（dy/dx)/dθ*dθ/dx=d(dy/dx)/dθ*1/dx/dθ= 我想问哈d^2y/dx^2那后面d（dy/dx)/dθ*dθ/dx=d(dy/dx)/dθ*1/dx/dθz这两步是如何化解的.
x=a(θ-sinθ) y=a(1-cosθ)
x=a(θ-sinθ),y=a(1-cosθ).求d^2y/dx^2.dy/dx=cotθ/2,d^2y/dx^2=d/dx*dy/dx=d（dy/dx)/dθ*dθ/dx=d(dy/dx)/dθ*1/dx/dθ= 我想问哈d^2y/dx^2那后面d（dy/dx)/dθ*dθ/dx=d(dy/dx)/dθ*1/dx/dθz这两步是如何化解的.化解的目的在哪里,这样化解是为了得到什么.

x=a(θ-sinθ) y=a(1-cosθ)x=a(θ-sinθ),y=a(1-cosθ).求d^2y/dx^2.dy/dx=cotθ/2,d^2y/dx^2=d/dx*dy/dx=d（dy/dx)/dθ*dθ/dx=d(dy/dx)/dθ*1/dx/dθ= 我想问哈d^2y/dx^2那后面d（dy/dx)/dθ*dθ/dx=d(dy/dx)/dθ*1/dx/dθz这两步是如何化解的.
高等数学上册(同济版）,第二章里有追问：童鞋 .我上面写了步骤的.就是问那样化解撒意思.回答：θ自变量换成是X,就必须要求dx／dθ,因为二阶导数是y对x的,不是对θ,所以要代换掉

y=sin(2x+a)(0 y=sin(2x+a)（0 x=1+2cosθ,(a≠0) y=sinθ 消去参数已知(x/a)cosθ+(y/b)sinθ=1,(x/a)sinθ-(y/b)cosθ=1,求证(x^2/a^2)+(y^2/b^2)=2如题示. matlab公式“y=a.*(1-sin(x))”怎么画 x=a(θ－sinθ) y=a(1-cosθ) 参数方程求导,答案是sinθ/1-cosθ 曲线x=1+cosθ,y=2sinθ 经过点(3/2,a),则a= C语言求三角函数f(x,y,z)=sin(x)/(sin(x-y)*sin(x-z))+sin(y)/(sin(y-x)*sin(y-z))+sin(z)/(sin(z-x)*sin(z-y))代码#include #include double fun(double x,double y,double z){double a,b,c,sum;a=sin(x)/(sin(x-y)*sin(x-z));b=sin(y)/(sin(y-x)*sin(y-z));c 函数Y=2sin（3X+A）（|A| 笛卡尔坐标系//请问 r=a(1-cosθ)或r=a(1+cosθ) 水平方向：r=a(1-cosθ)或r=a(1+cosθ) (a>0) 或垂直方向：r=a(1-sinθ)或r=a(1+sinθ) (a>0) 平面直角坐标系表达式分别为x^2+y^2+a*x=a*sqrt(x^2+y^2)和x^2+y^2-a*x=a*sqrt(x^2+y^ 已知x+y=根号2sin(a+π/4),x-y=根号2sin(a-π/4),求证x^2+y^2=1 已知x+y=根号2sin(a+π/4),x-y=根号2sin(a-π/4),求证x^2+y^2=1 x=sin a+cos a 平方后减去y=1+sin 2a,得到x^2=y 为什么? （1）x=a/cosθ y=btanθ (θ)为参数 2、x=sinθ y=cos2θ 将其化为普通方程在下列函数中,其最小值为2的函数是（）A.y=x+1/x B.y=sinθ+secθ (0 若函数y=sin(x+a)(0 函数Y=sin(2x+a),0 函数Y=sin(2x+a),0