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Question Number 75392 by wo1lxjwjdb last updated on 10/Dec/19

w h a t i s t h e g e n e r a l f o r m u l a r

f o r t h e \frac{d^{n}}{{d x}^{n}} (\frac{x}{e^{x} - 1})

Answered by mind is power last updated on 10/Dec/19

\frac{x}{e^{x} - 1} = x . \frac{1}{e^{x} - 1}

\frac{d^{n}}{{dx}^{n}} . (\frac{x}{e^{x} - 1}) = \underset{k = 0}{\sum^{n}} C_{n}^{k} . {(x)}^{(k)} . {(\frac{1}{e^{x} - 1})}^{(n - k)}

= x . {(\frac{1}{e^{x} - 1})}^{(n)} + n . {(\frac{1}{e^{x} - 1})}^{(n - 1)}

\frac{d}{{dx}^{n}} (\frac{1}{e^{x} - 1})

\frac{d}{dx} (\frac{1}{e^{x} - 1}) = - \frac{e^{x}}{{(e^{x} - 1)}^{2}} = - \frac{1}{e^{x} - 1} - \frac{1}{{(e^{x} - 1)}^{2}}

f (x) = \frac{1}{e^{x} - 1}

\Rightarrow f^{'} (x) = - f (x) - f^{2} (x) = P (f (x))

P (X) = - X - X^{2}

let P_{n} (X) defind like

P_{n} (f (x)) = \frac{d^{n}}{{dx}^{n}} f (x)

\Rightarrow P_{n + 1} (f (x)) = \frac{d^{n + 1}}{{dx}^{n + 1}} f (x) = \frac{d}{dx} (P_{n} (f (x))

= f^{'} (x) . P_{n}^{^{'}} (f (x))

= (- f (x) - f^{2} (x)) . P_{n}^{'} (f (x)) = P_{n + 1} (f (x))

\Rightarrow P_{n + 1} (X) = - (X + X^{2}) P_{n}^{^{'}} (X)

P_{0} = X

\Rightarrow \frac{d^{n}}{{dx}^{n}} (f (x)) = P_{n} (f (x))

\Rightarrow \frac{d}{{dx}^{n}} (\frac{x}{e^{x} - 1}) = {xp}_{n} (f (x)) + {nP}_{n - 1} (f (x))

{\begin{cases} P_{0} (X) = X \\ P_{n + 1} (X) = - (X^{2} + X) P_{n}^{'} (X) \end{cases}