It depends on what derivative you want. If you want dy/(dt) and (dx)/dtdydtanddxdt, 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)x(t)=e1−t→dxdt=−e1−t
y(t) = tan^2(t) = (tan t)(tan t) -> dy/dt = 2tan(t)d/dt(tan t) = (2 tan t)/(sec^2(t) y(t)=tan2(t)=(tant)(tant)→dydt=2tan(t)ddt(tant)=2tantsec2(t)
If you specifically want dy/dxdydx, we must use the chain rule .
dy/dx = dy/dt dt/dxdydx=dydtdtdx
This last term is simply one over dx/dtdxdt...
dy/dx = (2tant)/(sec^2(t)) * 1/(-e^(1-t)dydx=2tantsec2(t)⋅1−e1−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))
