Question

If \[f\left( x \right) = \dfrac{{\left( {x - 1} \right)}}{4} + \dfrac{{{{\left( {x - 1} \right)}^3}}}{{12}} + \dfrac{{{{\left( {x - 1} \right)}^5}}}{{20}} + \dfrac{{{{\left( {x - 1} \right)}^7}}}{{28}} + ...\] where \[0 < x < 4\], then find the value of \[f'\left( x \right)\].
A. \[\dfrac{1}{{4x\left( {2 - x} \right)}}\]
B. \[\dfrac{1}{{4{{\left( {x - 2} \right)}^2}}}\]
C. \[\dfrac{1}{{\left( {2 - x} \right)}}\]
D. \[\dfrac{1}{{\left( {2 + x} \right)}}\]

Answer 1

Hint In the given question the sum of infinite terms is given . We will find the derivative of the equation \[f\left( x \right)\] with respect to \[x\]. The infinite terms of the derivative equation \[f'\left( x \right)\] are in geometric progression. So, by using the formula of sum of infinite terms in geometric progression, we will find the value of \[f'\left( x \right)\].

Formula used
Power rule of derivative: \[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]
The sum of infinite terms in GP: \[{S_\infty } = \dfrac{a}{{\left( {1 - r} \right)}}\]
where \[a\] is the first term and \[r\] is the common ratio, where \[r \ne 0\] and \[r < 1\] .

Complete step by step solution:
The given equation is \[f\left( x \right) = \dfrac{{\left( {x - 1} \right)}}{4} + \dfrac{{{{\left( {x - 1} \right)}^3}}}{{12}} + \dfrac{{{{\left( {x - 1} \right)}^5}}}{{20}} + \dfrac{{{{\left( {x - 1} \right)}^7}}}{{28}} + ...\] where \[0 < x < 4\].
Let’s calculate the derivative of the above equation with respect to \[x\].

 \[f'\left( x \right) = \dfrac{1}{4} + \dfrac{{3{{\left( {x - 1} \right)}^2}}}{{12}} + \dfrac{{5{{\left( {x - 1} \right)}^4}}}{{20}} + \dfrac{{7{{\left( {x - 1} \right)}^6}}}{{28}} + ...\]
Simplify the above equation
\(f'\left( x \right) = \dfrac{1}{4} + \dfrac{{{{\left( {x - 1} \right)}^2}}}{4} + \dfrac{{{{\left( {x - 1} \right)}^4}}}{4} + \dfrac{{{{\left( {x - 1} \right)}^6}}}{4} + ...\)
_{Simplify further the above equation}
\[ \Rightarrow \] \[f'\left( x \right) = \dfrac{1}{4}\left[ {1 + {{\left( {x - 1} \right)}^2} + {{\left( {x - 1} \right)}^4} + {{\left( {x - 1} \right)}^6} + ...} \right]\]
The terms present in the square bracket are in geometric progression where the first term is \(a = 1\) and the common ratio is \[r = \frac{{\frac{{{{\left( {x - 1} \right)}^2}}}{4}}}{{\frac{1}{4}}} = {\left( {x - 1} \right)^2}\].
Apply the formula of the sum of infinite terms in geometric progression in the square bracket.
\[f'\left( x \right) = \dfrac{1}{4}\left[ {\dfrac{1}{{\left( {1 - {{\left( {x - 1} \right)}^2}} \right)}}} \right]\]
Simplify the above equation
\[f'\left( x \right) = \dfrac{1}{4}\left[ {\dfrac{1}{{1 - \left( {{x^2} - 2x + 1} \right)}}} \right]\]
\[ \Rightarrow \]\[f'\left( x \right) = \dfrac{1}{4}\left[ {\dfrac{1}{{2x - {x^2}}}} \right]\]
\[ \Rightarrow \]\[f'\left( x \right) = \dfrac{1}{{4x\left( {2 - x} \right)}}\]
Hence the correct option is option A.

Note: Students are often confused with the formula \[\dfrac{d}{{dx}}{\left( {ax + b} \right)^n} = n{\left( {ax + b} \right)^{n - 1}}\] and \[\dfrac{d}{{dx}}{\left( {ax + b} \right)^n} = n{\left( {ax + b} \right)^{n - 1}} \times a\] . But the correct formula is \[\dfrac{d}{{dx}}{\left( {ax + b} \right)^n} = n{\left( {ax + b} \right)^{n - 1}} \times a\]. Because we also have to calculate the derivative of the terms present in the bracket.