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What is the derivative of $2{{x}^{3}}$ ?

Answer
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Hint: We know that the differentiation of ${{x}^{n}}$ is $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$, where n is a constant value. Also, we are very well aware that when a constant c is multiplied by a function then its differentiation is given as $\dfrac{d}{dx}\left( c\cdot f\left( x \right) \right)=c\cdot \dfrac{d}{dx}\left( f\left( x \right) \right)$. Using these two identities, we can find the derivative of $2{{x}^{3}}$.

Complete step-by-step answer:
In our question, we need to find the derivative of $2{{x}^{3}}$.
Here, since nothing is specified, we should assume that the differentiation is to be done with respect to the variable x.
This implies that we need to find $\dfrac{d\left( 2{{x}^{3}} \right)}{dx}$.
We all are very well aware that, when a constant c is multiplied by a function then its derivative is equal to the product of constant c and the derivative of that function.
We can write this property mathematically as, $\dfrac{d}{dx}\left( c\cdot f\left( x \right) \right)=c\cdot \dfrac{d}{dx}\left( f\left( x \right) \right)$.
So, we can write $\dfrac{d\left( 2{{x}^{3}} \right)}{dx}=2\dfrac{d\left( {{x}^{3}} \right)}{dx}...\left( i \right)$
Also, we know that, the differentiation of ${{x}^{n}}$ is given by,
$\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$
So, by using this identity, we can write
$\dfrac{d\left( {{x}^{3}} \right)}{dx}=3{{x}^{2}}...\left( ii \right)$
Using the value from equation (ii) into the right hand side (RHS) of equation (i), we get
$\dfrac{d\left( 2{{x}^{3}} \right)}{dx}=2\times 3{{x}^{2}}$
Thus, we get
$\dfrac{d\left( 2{{x}^{3}} \right)}{dx}=6{{x}^{2}}$
Hence, the derivative of $2{{x}^{3}}$ is $6{{x}^{2}}$.

Note: We can always verify our answer for questions of differentiation, by integrating the result and verifying whether we get the expression given in question or not. Here, in this question, we can see that by integrating $6{{x}^{2}}$ we get $2{{x}^{3}}+c$, with c = 0 in this case.
We must also remember all formulae of derivatives by heart, as without them, we will not be able to solve the problems based on the similar concepts.