Find the derivative using the first principle of differentiation: ${x^3}$ ?
Answer
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Hint: In the given question we have to evaluate the derivative of a given function using the first principle of derivative. The first principle of differentiation involves the concepts of limits, derivatives, continuity and differentiability. It helps us to determine the derivative of the function using concepts of limits.
Complete answer:
Given the function ${x^3}$ and we have to calculate the derivative of the function using the first principle of derivatives.
Using formula of first principle of derivatives, we know that derivative of a function is calculated as:
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\]
So, derivative of the given function$f(x) = {x^3}$,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {x + h} \right)}^3} - {x^3}}}{h}\]
Expanding the whole cube term using the algebraic identity ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$, we get,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {{x^3} + {h^3} + 3xh\left( {x + h} \right)} \right) - {x^3}}}{h}\]
Opening the brackets, we get,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{x^3} + {h^3} + 3{x^2}h + 3x{h^2} - {x^3}}}{h}\]
Cancelling the like terms with opposite signs,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{h^3} + 3{x^2}h + 3x{h^2}}}{h}\]
Cancelling the common terms in numerator and denominator,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left[ {{h^2} + 3{x^2} + 3xh} \right]}}{h}\]
\[ \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \left[ {{h^2} + 3{x^2} + 3xh} \right]\]
Now, putting in the value of h in the limit, we get,
\[ \Rightarrow f'(x) = \left[ {{{\left( 0 \right)}^2} + 3{x^2} + 3x\left( 0 \right)} \right]\]
Simplifying the expression, we get,
\[ \Rightarrow f'(x) = 3{x^2}\]
Therefore, we get the derivative of ${x^3}$ as \[3{x^2}\] using the first principle of differentiation.
Note: The calculation of derivative using the first principle of derivative is the most basic and core method to find the derivative of a specific function. Another method of finding derivatives of a function can be simply differentiating that particular function with respect to the variable or unknown using some basic differentiation rules like addition rule, subtraction rule, product rule, and quotient rule.
So, we can apply the power rule of differentiation as $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$.
So, we get the derivative of ${x^3}$ as $\dfrac{d}{{dx}}\left( {{x^3}} \right) = 3{x^{3 - 1}} = 3{x^2}$.
Complete answer:
Given the function ${x^3}$ and we have to calculate the derivative of the function using the first principle of derivatives.
Using formula of first principle of derivatives, we know that derivative of a function is calculated as:
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\]
So, derivative of the given function$f(x) = {x^3}$,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {x + h} \right)}^3} - {x^3}}}{h}\]
Expanding the whole cube term using the algebraic identity ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$, we get,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {{x^3} + {h^3} + 3xh\left( {x + h} \right)} \right) - {x^3}}}{h}\]
Opening the brackets, we get,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{x^3} + {h^3} + 3{x^2}h + 3x{h^2} - {x^3}}}{h}\]
Cancelling the like terms with opposite signs,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{h^3} + 3{x^2}h + 3x{h^2}}}{h}\]
Cancelling the common terms in numerator and denominator,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left[ {{h^2} + 3{x^2} + 3xh} \right]}}{h}\]
\[ \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \left[ {{h^2} + 3{x^2} + 3xh} \right]\]
Now, putting in the value of h in the limit, we get,
\[ \Rightarrow f'(x) = \left[ {{{\left( 0 \right)}^2} + 3{x^2} + 3x\left( 0 \right)} \right]\]
Simplifying the expression, we get,
\[ \Rightarrow f'(x) = 3{x^2}\]
Therefore, we get the derivative of ${x^3}$ as \[3{x^2}\] using the first principle of differentiation.
Note: The calculation of derivative using the first principle of derivative is the most basic and core method to find the derivative of a specific function. Another method of finding derivatives of a function can be simply differentiating that particular function with respect to the variable or unknown using some basic differentiation rules like addition rule, subtraction rule, product rule, and quotient rule.
So, we can apply the power rule of differentiation as $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$.
So, we get the derivative of ${x^3}$ as $\dfrac{d}{{dx}}\left( {{x^3}} \right) = 3{x^{3 - 1}} = 3{x^2}$.
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