f-x-1-x-2-x-2018-f-0-

Question Number 147402 by vvvv last updated on 20/Jul/21

\boldsymbol f (\boldsymbol x + 1) (\boldsymbol x + 2) \dots . (\boldsymbol x + 2018)

\boldsymbol f^{'} (0) = ?

Answered by SEKRET last updated on 21/Jul/21

\boldsymbol y = (\boldsymbol x + 1) (\boldsymbol x + 2) \cdot \dots . (\boldsymbol x + 2018)

\boldsymbol \ln (\boldsymbol y) = \boldsymbol \ln (\boldsymbol x + 1) + \boldsymbol \ln (\boldsymbol x + 2) + . . + \boldsymbol \ln (\boldsymbol x + 2018)

\frac{\boldsymbol y^{'}}{\boldsymbol y} = \frac{1}{\boldsymbol x + 1} + \frac{1}{\boldsymbol x + 2} + \dots + \frac{1}{\boldsymbol x + 2018}

\boldsymbol y^{'} = (\boldsymbol x + 1) (\boldsymbol x + 2) \cdot . . (\boldsymbol x + 2018) \cdot (\frac{1}{\boldsymbol x + 1} + \frac{1}{\boldsymbol x + 2} + \dots + \frac{1}{\boldsymbol x + 2018})

\boldsymbol y^{'} (0) = 2018! \cdot (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + . . + \frac{1}{2018})

\boldsymbol ABDULAZIZ \boldsymbol ABDUVALIYEV

Answered by Olaf_Thorendsen last updated on 20/Jul/21

I suppose you mean

f (x) = (x + 1) (x + 2) \dots (x + 2018)

f (x) = \underset{k = 1}{\prod^{2018}} (x + k)

\ln f (x) = \underset{k = 1}{\sum^{2018}} \ln (x + k)

\frac{f^{'} (x)}{f (x)} = \underset{k = 1}{\sum^{2018}} \frac{1}{x + k}

\frac{f^{'} (0)}{f (0)} = \underset{k = 1}{\sum^{2018}} \frac{1}{k}

\frac{f^{'} (0)}{2018!} = H_{2018}

f^{'} (0) = 2018! \times H_{2018}