What is the derivative of #f(t) = (e^(t^2-1)-t, -t^3-4t ) #?

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JMicheli
Jul 12, 2016

#(df)/dt = (2t e^(t^2 - 1) - 1, -3t^2 - 4)#

Explanation:

Since we are taking the derivative of a vector-valued function parameterized over a single variable, #t#, we need only to take the individual derivative of each component:
#(d/dt f(t) = d/dt(e^(t^2-1)-t, -t^3-4t) = (d/dt(e^(t^2-1)-t), d/dt(-t^3-4t))#

So ultimately this is simply two very simple derivatives:
#d/dt(e^(t^2-1)-t) = 2t e^(t^2 - 1) - 1#
(by applying the chain rule and standard polynomial differentiation)
#d/dt(-t^3-4t)) = -3t^2 - 4#
(by standard polynomial differentiation)

These are combined into the final answer:
#(df)/dt = (2t e^(t^2 - 1) - 1, -3t^2 - 4)#

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