Question

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

Answer 1

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.