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

Find the derivative of f(x)=1+x+x2+...+x50f\left( x \right) = 1 + x + {x^2} + ... + {x^{50}} at x=1x = 1.

Explanation

Solution

Write the given function in summation form. Apply the differentiation formulas (f(x)+g(x)+h(x))′=f′(x)+g′(x)+h′(x){\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 ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} to find the differentiation of f(x)f\left( x \right). Then put x=1x = 1 and apply the formula of sum of first nn natural numbers, i.e. n(n+1)2\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(x)=1+x+x2+...+x50f\left( x \right) = 1 + x + {x^2} + ... + {x^{50}}.
This function can be written as:
⇒f(x)=∑i=050xi\Rightarrow f\left( x \right) = \sum\limits_{i = 0}^{50} {{x^i}}
We have to find the derivative of f(x)f\left( x \right) at x=1x = 1.
So, differentiating the function, we’ll get:
⇒f′(x)=ddx(∑i=050xi)\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.
⇒ddx[f(x)+g(x)+h(x)]=ddxf(x)+ddxg(x)+ddxh(x)\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:
⇒f′(x)=∑i=050ddxxi\Rightarrow f'\left( x \right) = \sum\limits_{i = 0}^{50} {\dfrac{d}{{dx}}{x^i}}
Further, we know that the differentiation of xn{x^n} is given as:
⇒ddxxn=nxn−1\Rightarrow \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}
Applying this formula for the above differentiation, we’ll get:
⇒f′(x)=∑i=050ixi−1\Rightarrow f'\left( x \right) = \sum\limits_{i = 0}^{50} {i{x^{i - 1}}}
Putting x=1x = 1 in above expression, we’ll get:
⇒f′(1)=∑i=050i(1)i−1 ⇒f′(1)=∑i=050i  \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 ii, we have:
⇒f′(1)=0+1+2+3+......+50\Rightarrow f'\left( 1 \right) = 0 + 1 + 2 + 3 + ...... + 50
Thus, the differentiation of f(x)f\left( x \right) at x=1x = 1 is the sum of first 50 natural numbers.
We know that the sum of first nn natural numbers is given by the formula n(n+1)2\dfrac{{n\left( {n + 1} \right)}}{2}. Putting n=50n = 50 for finding the value of f′(1)f'\left( 1 \right), we’ll get:
⇒f′(1)=50×512 ⇒f′(1)=25×51 ⇒f′(1)=1275  \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=1x = 1 is 1275.

Additional Information:
The sum of first nn natural numbers is given by the formula:
⇒∑k=1nk=n(n+1)2\Rightarrow \sum\limits_{k = 1}^n {k = \dfrac{{n\left( {n + 1} \right)}}{2}}
Similarly, the sum of the squares first nn natural numbers is given by the formula:
⇒∑k=1nk2=n(n+1)(2n+1)6\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 nn natural numbers is given by the formula:
⇒∑k=1nk3=[n(n+1)2]2\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 xn{x^n}, as used above, is given as:
⇒ddxxn=nxn−1\Rightarrow \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}, where n≠−1n \ne - 1.
Thus the above formula is not applicable for n≠−1n \ne - 1. So, if we have to find the differentiation of x−1{x^{ - 1}}(i.e. when n=−1n = - 1), its formula is given as:
⇒ddx(1x)=log⁡ex=ln⁡x\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) = {\log _e}x = \ln x