Can you take out a common factor when taking a derivative of a vector function? For example, if you have a tangent unit vector T(t) = ⟨(sqrt2)/(e^(7t)+e^(-7t)), (e^(7t))/(e^(7t)+e^(−7t)), (−e^(−7t))/(e^(7t)+e^(−7t))⟩ and you want to find its derivative, could you factor out (1)/(e^(7t)+e^(−7t)) to get T(t) = (1)/(e^(7t)+e^(−7t))⟨sqrt2, e^(7t), −e^(−7t)⟩

Licinilg

Licinilg

Open question

2022-08-22

For example, if you have a tangent unit vector
T(t) = ⟨ 2 e 7 t + e 7 t , e 7 t e 7 t + e 7 t , e 7 t e 7 t + e 7 t
and you want to find its derivative, could you factor out 1 e 7 t + e 7 t to get
T(t) = 1 e 7 t + e 7 t 2 , e 7 t , e 7 t
and then take the derivative of the interior term by term and the factor to get
T'(t) = 1 7 e 7 t 7 e 7 t ⟨0, 7 e 7 t , 7 e 7 t ⟩?
Any advice as to take a derivative of a vector such as this in an easier way would be welcome.
...or is brute forcing a derivative of these kinds of vectors simply inevitable?

Answer & Explanation

Bogalskilk

Bogalskilk

Beginner2022-08-23Added 9 answers

Yes, you can pull out the factor. But when you do, because the factor depends on t, you'll have to use the (usual) product rule to compute the derivative:
d d t [ f ( t ) v ( t ) ] = d d t f ( t ) v ( t ) + f ( t ) d d t v ( t )
In this case,
d d t [ 1 e 7 t + e 7 t 2 , e 7 t , e 7 t ] = d d t [ 1 e 7 t + e 7 t ] 2 , e 7 t , e 7 t + 1 e 7 t + e 7 t d d t 2 , e 7 t , e 7 t
= 1 7 e 7 t 7 e 7 t 2 , e 7 t , e 7 t + 1 e 7 t + e 7 t 0 , 7 e 7 t , 7 e 7 t

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