如果dy/dx = 1/(1+4t^2),y = 写错了，是dy/dt

如果dy/dx = 1/(1+4t^2),y = 写错了，是dy/dt 设x=4t y=t^2+1 求dy/dx d(t(dy/dt))/dx为什么等于t² d²y/dt²+t dy/dt作变量代换x=lnt简化方程d^2y/dx^2-dy/dx+e^2x*y=0x=lntdx/dt=1/tdy/dx=(dy/dt)/(dx/dt)=t dy/dtd²y/dx²=[d/dt(dy/dx)]/(dx/dt)=t² d²y/dt²+t dy/dt代入d^2y/dx^2- x = t - ln(1+t) y = t^3 + t^2 求dy/dx 参数方程x=(t-1)e^t,y=1-t^4,求dy/dx x=1+t^2 y=t+t^3 求dy/dx x=t^2+t y=ln(1+t) 求dy/dx 1-dy/dx+1/2cosy*dy/dx=0变形成dy/dx=2/2-cosy x=ln(1+t^2)+1,y=2arctant-t.求dy/dx. x=t^2/2 y=1-t 求dy/dx 已知x=1+t^2,y=t^2,则dy/dx y= ∫[0,x](t-1)^3(t-2)dt,dy/dx(x=0) dx/dt=6t+2,dy/dt=（3t+1）sin（t^2）,求d^2y/dx^2d^2y/dx^2不是直接对dy/dx求导吗 设参数函数x=ln(1+t^2),y=t-arctant.求(d^2y)/(dx^2).另外,如果求dy/dx^2,d^2y/dx又怎样算? 解方程 x(dy/dx)^3=(1+dy/dx) 设x=1+t^2、y=cost 求 dy/dx 和d^2y/dx^2 sint-tcost/4t^3 和 sint-tcost/4t^2 哪个对?设x=1+t^2、y=cost 求 dy/dx 和d^2y/dx^2sint-tcost/4t^3 和 sint-tcost/4t^2 哪个对? x=t^3+t-1 y=3-2t^2 dy/dx|t-1 等于多少 设x=t+arctan t+1,y=t的立方+6t－2,求dy／dx