Answer
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Hint: As we know that $ i $ is an imaginary part of the complex number. It is known as iota. We know that a complex number is a number which can be expressed in the $ a + bi $ form, where $ a $ and $ b $ are real numbers and $ i $ is the imaginary number. It means it consists of both real and imaginary parts. We can find the value of the imaginary unit number which is a negative number inside the square root. It is given by $ \sqrt { - 1} $
Complete step-by-step answer:
As per the given we have to find the derivative of iota i.e. $ i $ .
From the above we can see that the value of iota is i.e. $ i = \sqrt { - 1} $ . We can see that it is a constant.
We know that the derivative of any constant number is always zero. Here the value is also constant though imaginary.
Therefore, $ \dfrac{d}{{dx}}C = 0 $
$\Rightarrow \dfrac{d}{{dx}}i = 0 $
Hence we can say that the derivative of $ i $ is $ 0 $ .
So, the correct answer is “0”.
Note: We should know the constant rule which is , Let $ C $ be the constant. If $ f(x) = C, $ then $ f'(x) = 0 $ or we can write that $ \dfrac{d}{{dx}}C = 0 $ . The constant rule says that the derivative of any constant function is always $ 0 $ . We should be careful while calculating the values and in the square of the imaginary part we should note that the square of any negative number is always positive, the negative sign changes.
Complete step-by-step answer:
As per the given we have to find the derivative of iota i.e. $ i $ .
From the above we can see that the value of iota is i.e. $ i = \sqrt { - 1} $ . We can see that it is a constant.
We know that the derivative of any constant number is always zero. Here the value is also constant though imaginary.
Therefore, $ \dfrac{d}{{dx}}C = 0 $
$\Rightarrow \dfrac{d}{{dx}}i = 0 $
Hence we can say that the derivative of $ i $ is $ 0 $ .
So, the correct answer is “0”.
Note: We should know the constant rule which is , Let $ C $ be the constant. If $ f(x) = C, $ then $ f'(x) = 0 $ or we can write that $ \dfrac{d}{{dx}}C = 0 $ . The constant rule says that the derivative of any constant function is always $ 0 $ . We should be careful while calculating the values and in the square of the imaginary part we should note that the square of any negative number is always positive, the negative sign changes.
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