It depends on what derivative you want. If you want #dy/(dt) and (dx)/dt#, simply differentiate the x and y functions with respect to t, the same way you would differentiate a function f(x) with respect to x.
#x(t)= e^(1-t) -> dx/dt = -e^(1-t)#
#y(t) = tan^2(t) = (tan t)(tan t) -> dy/dt = 2tan(t)d/dt(tan t) = (2 tan t)/(sec^2(t) #
If you specifically want #dy/dx#, we must use the chain rule .
#dy/dx = dy/dt dt/dx#
This last term is simply one over #dx/dt#...
#dy/dx = (2tant)/(sec^2(t)) * 1/(-e^(1-t)#
Recalling that #e^(a-b) = e^a/e^b...#
#dy/dx = (2tant)/(sec^2(t)) * -1/((e/e^t)) = (2tant)/(sec^2(t)) * -e^t/e = (2tant)/(sec^2(t)) * -e^(t-1) = ((-2tant)(e^(t-1)))/(sec^2(t))#
