Solveeit Logo
Login

Question

Question: How do you find the derivative of \({x^4}\) ?...

How do you find the derivative of x4{x^4} ?

Explanation

Solution

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 x4{x^4} with respect to x using the first principle of differentiation.So, let us consider f(x)=x4f\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′(x) = lim⁡h→0f(x+h)−f(x)hf'\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f{\text{(x)}}}}{h} .

So, the derivative of the function x4{x^4} can be calculated by the first rule of differentiation as:
f′(x) = lim⁡h→0(x+h)4−x4hf'\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,
⇒f′(x) = lim⁡h→0(x+h)4−x4h\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 (a2−b2)=(a−b)(a+b)\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right), we get,
⇒f′(x) = lim⁡h→0[(x+h)2−x2][(x+h)2+x2]h\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 (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}, we get,
⇒f′(x) = lim⁡h→0[x2+2hx+h2−x2][x2+2hx+h2+x2]h\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,
⇒f′(x) = lim⁡h→0[2hx+h2][2x2+2hx+h2]h\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 hh common from the first bracket and cancelling the common terms in numerator and denominator, we get,
⇒f′(x) = lim⁡h→0h[2x+h][2x2+2hx+h2]h\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}
⇒f′(x) = lim⁡h→0[2x+h][2x2+2hx+h2]\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,
⇒f′(x) = [2x+0][2x2+0+02]\Rightarrow f'\left( x \right){\text{ = }}\left[ {2x + 0} \right]\left[ {2{x^2} + 0 + {0^2}} \right]
Simplifying the calculations, we get,
∴f′(x) = 4x3\therefore f'\left( x \right){\text{ = }}4{x^3}

Therefore, the derivative of the function f(x)=x4f\left( x \right) = {x^4} with respect to x is f′(x) = 4x3f'(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 xn{x^n} is nxn−1n{x^{n - 1}}. So, going by the power rule of differentiation, the derivative of f(x)=x4f(x) = {x^4} is 4x34{x^3}. So, we get the same answer from both the methods.