x=sint,y=2t的导数
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![x=sint,y=2t的导数](/uploads/image/f/892824-24-4.jpg?t=x%3Dsint%2Cy%3D2t%E7%9A%84%E5%AF%BC%E6%95%B0)
再问:果然是大神呀。。
再问:太感谢了!!!
dx/dt=(e^t)sint+(e^t)cost=(e^t)(sint+cost)dy/dt=(e^t)cost-(e^t)sint=(e^t)(cost-sint)dy/dx=(dy/dt)/(d
x=a(t-sint)dx/dt=a(1-cost)y=a(1-cost)dy/dt=asintdy/dx=dy/dt.(dt/dx)=sint/(1-cost)d^2y/dx^2=d/dt(dy/d
lnz=y*lnx=tant*lnsint两边同时求导:dz/z=sec^2t*lnsintdt+tant*cost/sintdtdz=z(sec^2t*lnsint+tan^2t)dt.dz=(si
∵x=1+t²,y=cost==>dx/dt=2t,dy/dt=-sint∴d²y/dx²=d(dy/dx)/dx=(d((dy/dt)/(dx/dt))/dt)/(dx
dx/dt=-e^(-t)sint+e^(-t)cost=e^(-t)(cost-sint)dy/dt=e^tcost+e^t(-sint)=e^t(cost-sint)dy/dx=(dy/dt)/(
由对称性,S=4∫(0→a)ydx=4∫(π/2→0)a(sint)^3d[a(cost)^3]=12a^2×∫(0→π/2)(sint)^4×(cost)^2dt=12a^2×∫(0→π/2)[(s
x=e^t*sinty=e^t*cost所以dx/dt=e^t*(sint+cost),dy/dt=e^t*(cost-sint)故dy/dx=(dy/dt)/(dx/dt)=(cost-sint)/
x^2=9sin^ty^2=16sin^tz^2=25cos^t三式相加可得一般方程x^2+y^2+z^2=25
t=arccos(1-y)x=arccos(1-y)-sin[arccos(1-y)]【sin(arccosx)=√(1-x²)】=arccos(1-y)-√[1-(1-y)²]=
x't=costy't=-2sin2tdy/dx=y't/x't=-2sin2t/cost=-4sintcost/cost=-4sint再问:y't为什么等于-2sin2t?再问:哦!我懂了!这是复合
需要注意的是有个隐藏条件:(sint)^2+(cost)^2=1即(sint+cost)^2-2sint*cost=1将x=cost+sint,y=sint*cost代入得x^2-2y=1,即y=(x
解dy/dx=(1-sint)'/(t²+cost)'=(-cost)/(2t-sint)
z=e^(x-2y)dz=e^(x-2y)(dx-2dy)(1)x=sintdx=costdt(2)y=t^2dy=2tdt(3)将(2),(3)代入(1)得dz=e^(x-2y)(cost-4t)d
dy/dt=costdx/dt=4tdy/dx=cost/4t
显然dx/dt=a(1-cost)dy/dt=a*sint那么dy/dx=sint/(1-cost)继续求二阶导就得到d(dy/dx)/dt*dt/dx=[(sint)'*(1-cost)-sint*
解答过程如下:再问:抄错了,题目是计算下列函数的导数:y=∫(sint/t)dt,t从1到x²+1再答:只要你题目对了就好办,解答过程如下:
dy/dt=e^t(cost+sint)dx/dt=e^t(cost-sint)所以dy/dx=(dy/dt)/(dx/dt)=(cost+sint)/(cost-sint)=1/)cos²