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2 answers:
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We can consider each unique 
as the as the 
-th unit vector. So your set 
can be considered as the vectors
Then check for independence your favorite way. In this case, I'll see if the linear map A of the new basis vectors doesn't map to a subspace via the determinant not being zero:
![det(A) = det \left ( \left [ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & 1 & 4 \end{array}\right ] \right ) = 1(4-3) = 1](https://tex.z-dn.net/?f=%20det%28A%29%20%3D%20det%20%5Cleft%20%28%20%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bccc%7D%201%20%26%200%20%26%200%20%5C%5C%200%20%26%201%20%26%203%20%5C%5C%200%20%26%201%20%26%204%20%5Cend%7Barray%7D%5Cright%20%5D%20%5Cright%20%29%20%3D%201%284-3%29%20%3D%201)
So they are linear independent.
Then check for independence your favorite way. In this case, I'll see if the linear map A of the new basis vectors doesn't map to a subspace via the determinant not being zero:
So they are linear independent.
Answer:
Step-by-step explanation:
We are given the following in the question:
We have to find the derivative of the given function.
Formula:
Derivative of a constant is zero.
Derivation takes place as: