What is the derivative of $f (t) = (1 - t e^{t}, t^{2} + t)$ $f(t) = (1-te^t, t^2+t )$ ?

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Answer 1 · 2017-06-24T22:24:17

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

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:

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

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

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

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