f-x-x-1-x-2-x-2021-f-2021-

Question Number 151884 by mathdanisur last updated on 23/Aug/21

f (x) = (x - 1) (x - 2) \dots (x - 2021)

f^{^{'}} (2021) = ?

Answered by mr W last updated on 24/Aug/21

\ln y = \ln (x - 1) + \ln (x - 2) + \dots + \ln (x - 2021)

\frac{y^{'}}{y} = \frac{1}{x - 1} + \frac{1}{x - 2} + \dots + \frac{1}{x - 2021}

y^{'} = (x - 1) (x - 2) \dots (x - 2021) [\frac{1}{x - 1} + \frac{1}{x - 2} + \dots + \frac{1}{x - 2021}]

y^{'} ∣_{x = 2021} = 2020 \times 2019 \times \dots \times 2 \times 1 = 2020!

Commented by mathdanisur last updated on 24/Aug/21

Thank you S er

Answered by Olaf_Thorendsen last updated on 23/Aug/21

f (x) = \underset{k = 1}{\prod^{2021}} (x - k)

f (x) = (x - 2021) \underset{k = 1}{\prod^{2020}} (x - k)

Let g (x) = \underset{k = 1}{\prod^{2020}} (x - k)

f (x) = (x - 2021) g (x)

f^{'} (x) = (x - 2021) g^{'} (x) + g (x)

f^{'} (2021) = g (2021) = \underset{k = 1}{\prod^{2020}} (2021 - k)

f^{'} (2021) = \underset{k = 1}{\prod^{2020}} k = 2020!

Commented by mathdanisur last updated on 23/Aug/21

Thank You S er