Question

Find the derivative of $f\left( x \right) = 1 + x + {x^2} + ... + {x^{50}}$ at $x = 1$.

Answer 1

Hint: Write the given function in summation form. Apply the differentiation formulas \[{\left( {f\left( x \right) + g\left( x \right) + h\left( x \right)} \right)^\prime } = f'\left( x \right) + g'\left( x \right) + h'\left( x \right)\] and $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ to find the differentiation of $f\left( x \right)$. Then put $x = 1$ and apply the formula of sum of first $n$ natural numbers, i.e. $\dfrac{{n\left( {n + 1} \right)}}{2}$ to get the final answer.

Complete step-by-step answer:
According to the question, we have been given a function $f\left( x \right) = 1 + x + {x^2} + ... + {x^{50}}$.
This function can be written as:
$ \Rightarrow f\left( x \right) = \sum\limits_{i = 0}^{50} {{x^i}} $
We have to find the derivative of $f\left( x \right)$ at $x = 1$.
So, differentiating the function, we’ll get:
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {\sum\limits_{i = 0}^{50} {{x^i}} } \right)$
Now, we know that the differentiation of sum of two or more functions is the sum of the differentiations of the respective functions, i.e.
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {f\left( x \right) + g\left( x \right) + h\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right) + \dfrac{d}{{dx}}g\left( x \right) + \dfrac{d}{{dx}}h\left( x \right)\]
Applying this rule for the above differentiation, we’ll get:
$ \Rightarrow f'\left( x \right) = \sum\limits_{i = 0}^{50} {\dfrac{d}{{dx}}{x^i}} $
Further, we know that the differentiation of ${x^n}$ is given as:
$ \Rightarrow \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
Applying this formula for the above differentiation, we’ll get:
$ \Rightarrow f'\left( x \right) = \sum\limits_{i = 0}^{50} {i{x^{i - 1}}} $
Putting $x = 1$ in above expression, we’ll get:
$
   \Rightarrow f'\left( 1 \right) = \sum\limits_{i = 0}^{50} {i{{\left( 1 \right)}^{i - 1}}} \\
   \Rightarrow f'\left( 1 \right) = \sum\limits_{i = 0}^{50} i \\
$
Expanding the summation by putting the values of $i$, we have:
$ \Rightarrow f'\left( 1 \right) = 0 + 1 + 2 + 3 + ...... + 50$
Thus, the differentiation of $f\left( x \right)$ at $x = 1$ is the sum of first 50 natural numbers.
We know that the sum of first $n$ natural numbers is given by the formula $\dfrac{{n\left( {n + 1} \right)}}{2}$. Putting $n = 50$ for finding the value of $f'\left( 1 \right)$, we’ll get:
$
   \Rightarrow f'\left( 1 \right) = \dfrac{{50 \times 51}}{2} \\
   \Rightarrow f'\left( 1 \right) = 25 \times 51 \\
   \Rightarrow f'\left( 1 \right) = 1275 \\
$

Thus the derivative of the given function at $x = 1$ is 1275.

Additional Information:
The sum of first $n$ natural numbers is given by the formula:
$ \Rightarrow \sum\limits_{k = 1}^n {k = \dfrac{{n\left( {n + 1} \right)}}{2}} $
Similarly, the sum of the squares first $n$ natural numbers is given by the formula:
$ \Rightarrow \sum\limits_{k = 1}^n {{k^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} $
And the sum of the cubes first $n$ natural numbers is given by the formula:
$ \Rightarrow \sum\limits_{k = 1}^n {{k^3} = {{\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]}^2}} $

Note: The formula for the differentiation of ${x^n}$, as used above, is given as:
$ \Rightarrow \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$, where $n \ne - 1$.
Thus the above formula is not applicable for $n \ne - 1$. So, if we have to find the differentiation of ${x^{ - 1}}$(i.e. when $n = - 1$), its formula is given as:
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) = {\log _e}x = \ln x$