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**Hint**: In the question they have clearly mentioned to use the first principle of differentiation which is stated as follows. Given a function $y = f(x)$, its derivative or the rate of change of $f(x)$ with respect to $x$ is defined as $\dfrac{d}{{dx}}f(x) = f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$, Where $h$ is an infinitesimally small positive number.

**:**

__Complete step-by-step answer__Let \[f\left( x \right)\] be a real function in its domain. A function defined such that $\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$. if it exists is said to be derivative of the function \[f\left( x \right)\]. This is known as the first principle of the derivative. The first principle of a derivative is also called the Delta Method.

Let's consider the given function.

\[ \Rightarrow \,\,\,\,f\left( x \right) = {x^3} - 27\]

Let us differentiate $f(x)$ with respect to $x$ by using the formula $\dfrac{d}{{dx}}f(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$

For finding $f(x + h)$ we replace $x$ by $x + h$ in the given function.

$ \Rightarrow \dfrac{d}{{dx}}\left( {{x^3} - 27} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {{{\left( {x + h} \right)}^3} - 27} \right) - \left( {{x^3} - 27} \right)}}{h}$

Now, by using a algebraic identity \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}\].

Here, \[a = x\] and \[b = h\], then we have

$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {{x^3} + {h^3} + 3{x^2}h + 3x{h^2} - 27} \right) - \left( {{x^3} - 27} \right)}}{h}$

$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{x^3} + {h^3} + 3{x^2}h + 3x{h^2} - 27 - {x^3} + 27}}{h}$

On simplification, we get

\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{h^3} + 3{x^2}h + 3x{h^2}}}{h}\]

Taking h as common in numerator, then

\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left( {{h^2} + 3{x^2} + 3xh} \right)}}{h}\]

On cancelling the like terms i.e., h on both numerator and denominator, we have

\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left( {{h^2} + 3{x^2} + 3xh} \right)\]

On applying a limit h tends to 0 \[\left( {h \to 0} \right)\] to the function, we have

\[ \Rightarrow f'\left( x \right) = {\left( 0 \right)^2} + 3{x^2} + 3x\left( 0 \right)\]

\[ \Rightarrow f'\left( x \right) = 0 + 3{x^2} + 0\]

On simplification, we get

\[ \Rightarrow f'\left( x \right) = 3{x^2}\]

Therefore, the derivative of \[{x^3} - 27\] is \[3{x^2}\].

**So, the correct answer is “ \[3{x^2}\]”.**

**Note**: In the question if they do not mention using first principle, we can use a direct method to differentiate the function by using a standard differentiation formula, which is easier than first principle. When differentiate using a first principle we must know the formula and know the product and quotient properties of the limit functions.

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