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Find the derivative of the following functions from first principle \[{x^3} - 27\].

Last updated date: 16th Mar 2023
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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.