What is the derivative of f(t) = (1-te^t, t^2+t ) f(t)=(1tet,t2+t)?

1 Answer
Jun 24, 2017

f'(t) = -(2t+1)/((t+1)e^t)

Explanation:

We have:

f(t) = (1-te^t, t^2+t)

Which we can write as

f(t) = (x(t), y(t))

Where:

x(t) = 1-te^t
y(t) = t^2+t

Then differentiating wrt t (and applying the product rule) we have:

dx/(dt) = (-t)(e^t)+(-1)(e^t) = -te^t-e^t
dy/(dt) = 2t+1

So then by the chain rule we have:

dy/dx = (dy//dt)/(dx//dt)

" " = (2t+1)/(-te^t-e^t)

" " = -(2t+1)/((t+1)e^t)