Question

sin(x+t)=xt,e^ty=y+t+1求dy/dx

Answer 1

sin(x+t)=xt
两边对t求导，cos(x+t)*(dx/dt+1)=x+tdx/dt
[cos(x+t)-t](dx/dt)=x-cos(x+t)
dx/dt=[x-cos(x+t)]/[cos(x+t)-t]
e^(ty)=y+t+1
两边对t求导，e^(ty)*(y+tdy/dt)=dy/dt+1
[te^(ty)-1](dy/dt)=1-ye^(ty)
dy/dt=[1-ye^(ty)]/[te^(ty)-1]
所以dy/dx=(dy/dt)/(dx/dt)
={[1-ye^(ty)]/[te^(ty)-1]}*{[cos(x+t)-t]/[x-cos(x+t)]}
=[1-ye^(ty)][cos(x+t)-t]/[te^(ty)-1][x-cos(x+t)]