We are given that #y# is a function of #x#.
Also, #x=e^t/(e^t+1) :. x" is a function of "t#.
Altogether, #y# is a fun. of #x#, and #x# of #t#.
Hence, by the Chain Rule, #dy/dt=dy/dx*dx/dt......(square)#,
Here, #x=e^t/(e^t+1)#,
Applying the Quotient Rule, we have,
# dx/dt={(e^t+1)*d/dt(e^t)-e^t*d/dt(e^t+1)}/(e^t+1)^2#,
#={(e^t+1)e^t-e^t*e^t}/(e^t+1)^2#,
# rArr dx/dt=e^t/(e^t+1)^2#,
#:. dx/dt=e^t/(e^t+1)^2*e^t/e^t={e^t/(e^t+1)}^2*1/e^t...(square^1)#.
#"Now, "x=e^t/(e^t+1)rArr 1/x=(e^t+1)/e^t=e^t/e^t+1/e^t=1+1/e^t#.
#:.1/x-1=1/e^t, or, 1/e^t=1/x-1=(1-x)/x#.
Utilising this in #(square^1), dx/dt={e^t/(e^t+1)}^2*1/e^t#,
#=x^2*(1-x)/x#.
# rArr dx/dt=x(1-x)#.
Hence, by #(square), dy/dt=x(1-x)dy/dx#, as desired!
Enjoy Maths.!
