Question

Find the derivative of $f(x) = 1 + x + {x^2} + {x^3} + ......{x^{50}}$ at x=1

Answer 1

Hint: We’ll first differentiate the function with-respect-to x using the properties of differentiation which are $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$and $\dfrac{{d(A + B)}}{{dx}} = \dfrac{{d(A)}}{{dx}} + \dfrac{{d(B)}}{{dx}}$.
Then on substituting x=1, we’ll see a well-known series coming as the result, we’ll then solve that series using the sum of the first n natural numbers which is $1 + 2 + ....... + n = \dfrac{{n(n + 1)}}{2}$ , to approach our answer.

Complete step-by-step answer:
Given data: $f(x) = 1 + x + {x^2} + {x^3} + ......{x^{50}}$
We that the derivative of ${x^n}$ is given by $n{x^{n - 1}}$
i.e. $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$
and the addition rule of derivatives
i.e. $\dfrac{{d(A + B)}}{{dx}} = \dfrac{{d(A)}}{{dx}} + \dfrac{{d(B)}}{{dx}}$
Now, on differentiating the given function with-respect-to x, we get,
$ \Rightarrow f'(x) = \dfrac{{d(1)}}{{dx}} + \dfrac{{d(x)}}{{dx}} + \dfrac{{d({x^2})}}{{dx}} + ....... + \dfrac{{d({x^{50}})}}{{dx}}$
Now, on simplifying each term, we get,
$ \Rightarrow f'(x) = 0 + 1{x^{1 - 1}} + 2{x^{2 - 1}} + ....... + 50{x^{50 - 1}}$
On solving the exponents of each term, we get,
$ \Rightarrow f'(x) = 0 + 1{x^0} + 2{x^1} + ....... + 50{x^{49}}$
As we need to find the derivative of the function at x=1
Substituting x=1,
$ \Rightarrow f'(1) = 1 + 2(1) + ....... + 50{(1)^{49}}$
Now, we know that ${(1)^n} = 1$, where n is a natural number
\[ \Rightarrow f'(1) = 1 + 2 + ....... + 50\]
Now, using the formula of the sum of first natural numbers
i.e.$1 + 2 + ....... + n = \dfrac{{n(n + 1)}}{2}$,
As \[n = 50\], we get,
$ \Rightarrow f'(1) = \dfrac{{50(50 + 1)}}{2}$
On Simplifying the brackets, we get,
$ \Rightarrow f'(1) = \dfrac{{50(51)}}{2}$
On Dividing numerator and denominator by 2, we get,
$ \Rightarrow f'(1) = 25(51)$
$\therefore f'(1) = 1275$
Therefore, derivative of function at x=1 is 1275.

Note: As here sum of series is included on should use the result, $1 + 2 + ....... + n = \dfrac{{n(n + 1)}}{2}$, rather than manually calculating the numbers uptil 50, as that could lead to calculation errors.
Alternative method for this question can be
From the function given, we can say that the function is a G.P. where the first term is 1 and the common ratio is x having 51 terms
We know that sum of first n terms of a G.P. when a is the first term and r is the common ratio i.e. ${S_n}$ is given by
${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$
$ \Rightarrow f(x) = \dfrac{{1({x^{51}} - 1)}}{{x - 1}}$
$\therefore f(x) = \dfrac{{({x^{51}} - 1)}}{{x - 1}}$
On differentiating with-respect-to x
Using division rule of derivative i.e. $d\left( {\dfrac{u}{v}} \right) = \dfrac{{vdu - udv}}{{{v^2}}}$
\[ \Rightarrow f'(x) = \dfrac{{51{x^{50}}(x - 1) - ({x^{51}} - 1)}}{{{{\left( {x - 1} \right)}^2}}}\]
On Simplifying the brackets, we get,
\[ \Rightarrow f'(x) = \dfrac{{51{x^{51}} - 51{x^{50}} - {x^{51}} + 1}}{{{{\left( {x - 1} \right)}^2}}}\]
On Simplifying the like terms, we get,
\[ \Rightarrow f'(x) = \dfrac{{50{x^{51}} - 51{x^{50}} + 1}}{{{{\left( {x - 1} \right)}^2}}}\]
Substituting x=1
The left-hand side is forming indeterminate form i.e. $\dfrac{0}{0}$
Therefore using l’ hospital rule, i.e. differentiating both numerator and denominator, we get,
\[ = \dfrac{{(50)(51){x^{50}} - (51)50{x^{49}}}}{{2\left( {x - 1} \right)}}\]
Now taking \[(51)50{x^{49}}\]common from both terms, we get,
\[ = \dfrac{{(51)50{x^{49}}(x - 1)}}{{2\left( {x - 1} \right)}}\]
On further simplifying substituting x=1,
\[ = \dfrac{{(51)50{{(1)}^{49}}}}{2}\]
\[ = 1275\], which is a similar answer as of the above solution.