Good evening.
I am so confused, I am trying to find the derivative of $x=e^t$ and $y=t^2e^{-t}$ to find the tangent of a parametric function. The only thing is, is that when I derive $dy/dt$ I get $-(t-2)te^{-t}$ and $dx/dt = e^t$; thus, $dy/dx = (2-t)t/e^{2t}$. However, when I simplify the whole expression $y/x = t^2e^{-t}/e^t$ and derive it I get $-2(t-1)te^{-2t}$. Aren't these answers suppose to be the same? Please help me.
Gabrielle
Your second method does not get you to $dy/dx$. You’re differentiating the quotient of two functions, so you’re actually computing (by the quotient rule) $$\frac d{dt}{y(t)\over x(t)}={\dot y(t)x(t)-y(t)\dot x(t)\over x(t)^2}.$$
The two answers are not the same.
In the first you have used the chain rule (correctly) in the form $\dfrac {dy}{dx}= {\dfrac {dy}{dt}} \div {\dfrac {dx}{dt}}$
In the second you are finding $\dfrac {d}{dx}\left (\dfrac yx \right)$, but this is not equal to $\dfrac {dy}{dx}$