Question

If $f\left( x \right) = {x^n}$, then find the value of $f\left( 1 \right) + \dfrac{{{f^1}\left( 1 \right)}}{{1!}} + \dfrac{{{f^2}\left( 1 \right)}}{{2!}} + ...... + \dfrac{{{f^n}\left( 1 \right)}}{{n!}}$, where ${f^r}\left( x \right)\;$ denotes the ${r^{th}}$ derivative of $f\left( x \right)\;$ w.r.t. $x$.

Answer 1

The equation $F(x,y) = 0$are sequential (n times) differentiation when nth order derivatives of an implicit function are found. At each step, after appropriate substitutions and transformations, we will obtain an explicit expression for the derivative, which depends only on the variables x and y, i.e. the derivatives in the form of $y\prime = {f_1}(x,y),y\prime \prime = {f_2}(x,y), \ldots ,{y^{(n)}} = {f_n}(x,y)$
This process can continue but notice that we will start getting zero for all derivatives after this point.
Sometimes, we are able to see a pattern in successive derivatives of a given function. In such kinds of problems, we need to consider those patterns to solve them.

Complete step-by-step answer:
The function provided in the question is $f\left( x \right) = {x^n}$
And we need to calculate the value of $f\left( 1 \right) + \dfrac{{{f^1}\left( 1 \right)}}{{1!}} + \dfrac{{{f^2}\left( 1 \right)}}{{2!}} + ...... + \dfrac{{{f^n}\left( 1 \right)}}{{n!}}$
So $f\left( x \right)$, and all derivatives of $f\left( x \right)$has to be calculated with the value of $x = 1$.
So, if $f\left( x \right) = {x^n}$ then $f\left( 1 \right) = {1^n} = 1$
We calculate several first derivatives:
So, first derivative of the equation is equal to ${f^1}\left( x \right) = n{x^{n - 1}}$ then ${f^1}\left( 1 \right) = n{1^{n - 1}} = n$
So, second derivative of the equation is equal to ${f^2}\left( x \right) = n(n - 1){x^{n - 2}}$ then ${f^2}\left( 1 \right) = n(n - 1){1^{n - 2}} = n(n - 1)$
So, third derivative of the equation is equal to ${f^3}\left( x \right) = n(n - 1)(n - 2){x^{n - 3}}$ then ${f^3}\left( 1 \right) = n(n - 1)(n - 2){1^{n - 3}} = n(n - 1)(n - 2)$
Each derivative gives us a pattern.
Hence it becomes easy to establish a general expression for the nth order derivative by looking at the pattern:
${f^n}\left( x \right) = n(n - 1)(n - 2).........................\left[ {n - \left( {n - 1} \right)} \right]{x^{n - n}}$ then
${f^n}\left( 1 \right) = n(n - 1)(n - 2).........................\left[ {n - \left( {n - 1} \right)} \right]{1^{n - n}} = n(n - 1)(n - 2).............1$
Now we will calculate the value of $f\left( 1 \right) + \dfrac{{{f^1}\left( 1 \right)}}{{1!}} + \dfrac{{{f^2}\left( 1 \right)}}{{2!}} + ...... + \dfrac{{{f^n}\left( 1 \right)}}{{n!}}$by substituting the value of $f\left( 1 \right)$, ${f^1}\left( 1 \right)$, ${f^2}\left( 1 \right)$…………………${f^n}\left( 1 \right)$.
So, the equation is converted to $f\left( 1 \right) + \dfrac{{{f^1}\left( 1 \right)}}{{1!}} + \dfrac{{{f^2}\left( 1 \right)}}{{2!}} + ...... + \dfrac{{{f^n}\left( 1 \right)}}{{n!}}$
$= 1 + \dfrac{n}{{1!}} + \dfrac{{n(n - 1)}}{{2!}} + ...... + \dfrac{{n(n - 1)(n - 2).............1}}{{n!}}$
The formula of combination is ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$, So replacing these in the above equation.
$= 1 + {}^n{C_1} + {}^n{C_2} + ...... + {}^n{C_n}$
$= {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ...... + {}^n{C_n}$
We also know the sum of all possible combinations of n distinct things is
${}^n{C_0} + {}^n{C_1} + {\text{ }}{}^n{C_2} + .{\text{ }}.{\text{ }}. + {}^n{C_n} = {\text{ }}{2^n}$. Hence, we calculate the above equation sum as
$= {2^n}$

If $f\left( x \right) = {x^n}$, then the value of $$f\left( 1 \right) + \dfrac{{{f^1}\left( 1 \right)}}{{1!}} + \dfrac{{{f^2}\left( 1 \right)}}{{2!}} + ...... + \dfrac{{{f^n}\left( 1 \right)}}{{n!}}$$$ = {2^n}$.

Note: In general, to find out the n-th derivative of function $y = f\left( x \right)$ we need to find all derivatives of previous orders starting from 1 to nth. But sometimes n-th order derivative is possible to obtain as it depends on n and doesn't contain previous derivatives.
In order to calculate higher-order derivatives, we can use Leibniz's formula:

$${(uv)^n} = \sum\limits_{k = 0}^n {C_n^k{u^{(n - k)}}{v^{(k)}}} \\\ C_n^k = \dfrac{{n!}}{{(n - k)!k!}} \\\$$