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