Question

If $f\left( x \right)=1+x+{{x}^{2}}+.................+{{x}^{100}}$ then find ${f}'\left( 1 \right)$ .

Answer 1

Hint: For solving this question first we will see two important results like $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ and the sum of the first $n$ natural numbers will be equal to the value of $\dfrac{n\left( n+1 \right)}{2}$. After that, we will differentiate the given function with respect to $x$ and then substitute the value of $x=1$ in the expression of ${f}'\left( x \right)$ to get the value of ${f}'\left( 1 \right)$.

Complete step-by-step answer:
Given:
We have the following function $f\left( x \right)$ in terms of $x$ :
$f\left( x \right)=1+x+{{x}^{2}}+.................+{{x}^{100}}$
And in the question, we have to find the value of ${f}'\left( 1 \right)$.
Now, before we proceed we should know the following two important formulas:
1. If $y={{x}^{n}}$ then, $\dfrac{dy}{dx}=n{{x}^{n-1}}$. For example: if $y={{x}^{23}}$ then, $\dfrac{dy}{dx}=23{{x}^{22}}$.
2. Sum of the first $n$ natural numbers will be equal to the value of $\dfrac{n\left( n+1 \right)}{2}$ . For example: the sum of the first 5 natural numbers will be $1+2+3+4+5=\dfrac{5\left( 5+1 \right)}{2}=15$.
Now, we will be using the above results for solving this question.
Now, as it is given that $f\left( x \right)=1+x+{{x}^{2}}+.................+{{x}^{100}}$ so, we can differentiate $f\left( x \right)$ with respect to $x$ with the help of the formula mentioned in the first point. Then,
$\begin{align}
  & f\left( x \right)=1+x+{{x}^{2}}+.................+{{x}^{100}} \\
 & \Rightarrow {f}'\left( x \right)=0+1+2x+3{{x}^{2}}+4{{x}^{3}}+5{{x}^{4}}+.....................+98{{x}^{97}}+99{{x}^{98}}+100{{x}^{99}} \\
\end{align}$
Now, as we have to find the value of ${f}'\left( 1 \right)$ so, put $x=1$ in the above expression of ${f}'\left( x \right)$. Then,
$\begin{align}
  & {f}'\left( x \right)=0+1+2x+3{{x}^{2}}+4{{x}^{3}}+5{{x}^{4}}+.....................+98{{x}^{97}}+99{{x}^{98}}+100{{x}^{99}} \\
 & \Rightarrow {f}'\left( 1 \right)=1+2+3\left( {{1}^{2}} \right)+4\left( {{1}^{3}} \right)+5\left( {{1}^{4}} \right)+................+98\left( {{1}^{97}} \right)+99\left( {{1}^{98}} \right)+100\left( {{1}^{99}} \right) \\
 & \Rightarrow {f}'\left( 1 \right)=1+2+3+4+5+...................................+98+99+100 \\
\end{align}$
Now, after analysing the result of the above calculation it is evident that the value of ${f}'\left( 1 \right)$ will be equal to the sum of the first 100 natural numbers. So, we can find it with the help of formula for the sum of first $n$ natural numbers and put the value of $n=100$ in the formula $\dfrac{n\left( n+1 \right)}{2}$. Then,
$\begin{align}
  & {f}'\left( 1 \right)=1+2+3+4+5+...................................+98+99+100 \\
 & \Rightarrow {f}'\left( 1 \right)=\dfrac{100\times \left( 100+1 \right)}{2} \\
 & \Rightarrow {f}'\left( 1 \right)=\dfrac{100\times 101}{2} \\
 & \Rightarrow {f}'\left( 1 \right)=5050 \\
\end{align}$
Now, from the above result, we can say that if $f\left( x \right)=1+x+{{x}^{2}}+.................+{{x}^{100}}$ then, the value of ${f}'\left( 1 \right)=5050$.
Thus, the value of ${f}'\left( 1 \right)=5050$ for the given function.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. Moreover, though the question is very easy, we should be careful while differentiating the given function and proceed in a stepwise manner to avoid mistakes. And substitute the value of the variable $x$ correctly in the expression of ${f}'\left( x \right)$ and analyse the result to get the final answer without any error.