If $f(x) = \frac{x - 1}{4} + \frac{(x - 1)^{3}}{12} + \frac{(x - 1)^{5}}{20} + \frac{(x - 1)^{7}}{28} + \ldots,$ where $0\

Question

If $f(x) = \frac{x - 1}{4} + \frac{(x - 1)^{3}}{12} + \frac{(x - 1)^{5}}{20} + \frac{(x - 1)^{7}}{28} + \ldots,$ where $0\ < \ x\ < \ 2$, then $f^{'}(x)$ is equal to

Dr. Abubacker · Accepted Answer

(a)
Given,
$f(x) = \frac{1}{4}\ \left\lbrack \frac{x - 1}{1} + \frac{(x - 1)^{3}}{3} + \frac{(x - 1)^{5}}{5} + \frac{(x - 1)^{7}}{7} + \ldots ightbrack\$
$\Rightarrow \ \ f(x) = \frac{1}{4}\ \left\lbrack \frac{1}{2}\log{(\frac{1 + (x - 1)}{1 - (x - 1)}\ } ightbrack = \frac{1}{8}\log\left( \frac{x}{2 - x} ight)$
$\Rightarrow \ f^{'}(x) = \frac{1}{8} 	imes \frac{1}{\left( \frac{x}{2 - x} ight)}\left\lbrack \frac{(2 - x)1 - x( - 1)}{(2 - x)^{2}} ightbrack = \frac{1}{4x(2 - x)}$

Q. If $f(x) = \frac{x - 1}{4} + \frac{(x - 1)^{3}}{12} + \frac{(x - 1)^{5}}{20} + \frac{(x - 1)^{7}}{28} + \ldots,$ where $0\ < \ x\ < \ 2$, then $f^{'}(x)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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