find-the-3-rd-derivative-of-x-5-ln-2x-using-the-Leibniz-theorem-

Question Number 83189 by john santu last updated on 28/Feb/20

find the 3^{rd} derivative of

x^{5} \ln (2 x) using the Leibniz theorem

Commented by jagoll last updated on 28/Feb/20

y = x^{5} \ln (2 x)

y^{'} (x) = {5 x}^{4} \ln (2 x) + \frac{{2 x}^{5}}{2 x}

y^{'} (x) = {5 x}^{4} \ln (2 x) + x^{4}

y^{″} (x) = {20 x}^{3} \ln (2 x) + \frac{{10 x}^{4}}{2 x} + {4 x}^{3}

y^{″} (x) = {20 x}^{3} \ln (2 x) + {9 x}^{3}

y^{‴} (x) = {60 x}^{2} \ln (2 x) + \frac{{40 x}^{3}}{2 x} + {27 x}^{2}

y^{‴} (x) = {60 x}^{2} \ln (2 x) + {47 x}^{2}

Commented by mr W last updated on 28/Feb/20

{(x^{5} \ln (2 x))}^{(3)}

= x^{5} (\frac{2}{x^{3}}) + 3 (5 x^{4}) (- \frac{1}{x^{2}}) + 3 (20 x^{3}) (\frac{1}{x}) + (60 x^{2}) \ln (2 x)

= 2 x^{2} - 15 x^{2} + 60 x^{2} + 60 x^{2} \ln (2 x)

= x^{2} (47 + 60 \ln (2 x))

Commented by mathmax by abdo last updated on 28/Feb/20

d i r e c t c a l c u l u s f (x) = x^{5} l n (2 x) \Rightarrow

f^{(1)} (x) = 5 x^{4} l n (2 x) + x^{5} \times \frac{1}{x} = 5 x^{4} l n (2 x) + x^{4}

f^{(2)} (x) = 20 x^{3} l n (2 x) + 5 x^{4} \times \frac{1}{x} + 4 x^{3} = 20 x^{3} l n (2 x) + 5 x^{3} + 4 x^{3}

= 20 x^{3} l n (2 x) + 9 x^{3} \Rightarrow f^{(3)} (x) = 60 x^{2} l n (2 x) + 20 x^{3} \times \frac{1}{x} + 27 x^{2}

= 60 x^{2} l n (2 x) + 47 x^{2} = x^{2} {60 l n (2 x) + 47}

Commented by jagoll last updated on 28/Feb/20

what generally Leibniz theorem sir

Commented by jagoll last updated on 28/Feb/20

yes sir . thank you

Commented by mathmax by abdo last updated on 29/Feb/20

i f t h e f u n c t i o n s f a n d g a r e C^{n} o n I \subset R (o r C) w e h a v e

{(f . g)}^{(n)} = \sum_{k = 0}^{n} C_{n}^{k} f^{(k)} g^{(n - k)} (l e i b n i z f o r m u l a f o r d e r i v a t i o n)