y = f(e^t/(e^t+1))
dy/dt = dy/dx * dx/dt chain rule
= dy/dx * d /dt [ e^t/(e^t+1) ]
= dy/dx [ (e^t+1)e^t - e^2t ]/(e^t+1)^2
now what is x(1-x) ?
[e^t/(e^t+1)][1-e^t/(e^t+1)]
= [ e^t(e^t+1)- e^2t]/(e^t+1)^2
The same !!!
if y is a function of x and x=e^t/(e^t+1) show that dy/dt=x(1-x)dy/dx
help me plz
2 answers
x=e^t/(e^t+1)
dx/dt= -[e^t*e^t)/(e^t+1)^2 + e^t/(e^t+1)
= e^t/(e^t+1) *(-e^t/(e^t+1)+1)
= x * (1-x)
so
dy/dt=dx/dt*dy/dx
and you have it.
dx/dt= -[e^t*e^t)/(e^t+1)^2 + e^t/(e^t+1)
= e^t/(e^t+1) *(-e^t/(e^t+1)+1)
= x * (1-x)
so
dy/dt=dx/dt*dy/dx
and you have it.