Question

How do you find the derivative of ${x^4}$ ?

Answer 1

Hint:The given question requires us to find the derivative of a function. The first principle of differentiation helps us evaluate the derivative of a function using limits.Calculating the derivative of a function using the first principle of differentiation may be a tedious task. We may employ algebraic identities and tricks to calculate the limits and evaluate the required derivative.

Complete step by step answer:
We have to evaluate the derivative of ${x^4}$ with respect to x using the first principle of differentiation.So, let us consider $f\left( x \right) = {x^4}$. According to the first principle of differentiation, the derivative of a function can be evaluated by calculating the limit as,
\[f'\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f{\text{(x)}}}}{h}\] .

So, the derivative of the function ${x^4}$ can be calculated by the first rule of differentiation as:
\[f'\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {x + h} \right)}^4} - {x^4}}}{h}\]
Taking the LCM of the fractions, we get,
\[ \Rightarrow f'\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {x + h} \right)}^4} - {x^4}}}{h}\]

Using the algebraic identity $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$, we get,
\[ \Rightarrow f'\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {{{\left( {x + h} \right)}^2} - {x^2}} \right]\left[ {{{\left( {x + h} \right)}^2} + {x^2}} \right]}}{h}\]
Opening the whole square terms using the algebraic identities ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$, we get,
\[ \Rightarrow f'\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {{x^2} + 2hx + {h^2} - {x^2}} \right]\left[ {{x^2} + 2hx + {h^2} + {x^2}} \right]}}{h}\]
Cancelling the numerator and denominator and simplifying the limit further,
\[ \Rightarrow f'\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {2hx + {h^2}} \right]\left[ {2{x^2} + 2hx + {h^2}} \right]}}{h}\]

Taking $h$ common from the first bracket and cancelling the common terms in numerator and denominator, we get,
\[ \Rightarrow f'\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \dfrac{{h\left[ {2x + h} \right]\left[ {2{x^2} + 2hx + {h^2}} \right]}}{h}\]
\[ \Rightarrow f'\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {2x + h} \right]\left[ {2{x^2} + 2hx + {h^2}} \right]\]
Now, putting the limit of the variable and getting to the final answer, we get,
\[ \Rightarrow f'\left( x \right){\text{ = }}\left[ {2x + 0} \right]\left[ {2{x^2} + 0 + {0^2}} \right]\]
Simplifying the calculations, we get,
\[ \therefore f'\left( x \right){\text{ = }}4{x^3}\]

Therefore, the derivative of the function $f\left( x \right) = {x^4}$ with respect to x is $f'(x){\text{ = }}4{x^3}$.

Note:The derivative of the given function can also be calculated by using the power rule of differentiation. According to the power rule of differentiation, the derivative of ${x^n}$ is $n{x^{n - 1}}$. So, going by the power rule of differentiation, the derivative of $f(x) = {x^4}$ is $4{x^3}$. So, we get the same answer from both the methods.