What is the derivative of #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) #