Question

Successive Differentiation

please help

Answer 1

hi

Answer 2

\[\tt Find~nth~derivative~:~\frac{ x^2 }{ (x-1)(x-20)(x-3) }\]

Answer 3

don't we have a smaller way for this?

Answer 4

we can simplify by using partial fractions,right?
btw hello :)

Answer 5

\[ f(x) = \frac{ x^2 }{ (x-1)(x-20)(x-3) }
\\ ~\\ f ' (x) = -\frac{x(x^3-83x+120) } {(x-1)^2(x-20)^2(x-3)^2 }
\\ ~\\ f '' (x) = -\frac{2(x^6-249x^4+2412x^3-4320x^2+3600) } {(x-1)^3(x-20)^3(x-3)^3 }
\]

Answer 6

thats a great idea :)

Answer 7

yuppp

Answer 8

so now what will be the values for A B and C

Answer 9

\[ \frac{x^2}{(x-1)(x-20)(x-3)} = \frac 1 {38(x-1)}+\frac {400}{323(x-20)}-\frac{9}{34(x-3)}\]

Answer 10

yay
i got the same
yayayayay

Answer 11

you are fast :)

Answer 12

lol
thank you so much
furthermore \[\rm \frac{ (-1)^n n!}{ (x-1)^{1+n} }\]

Answer 13

nice expression

Answer 14

i wrote for the first term
is it correct?

Answer 15

\[ f(x) = \frac{x^2}{(x-1)(x-20)(x-3)}
\\~\\ f(x) = \frac{1}{38} \cdot \frac{1} {(x-1)}+\frac {400}{323} \cdot \frac{1}{(x-20)}-\frac{9}{34} \cdot \frac{1}{(x-3)}
\\~\\
\\ f '(x) =
\]yes you just need the constant 1/38 in front

Answer 16

yupp...
i got it SIr :)

Answer 17

\[ f(x) = \frac{x^2}{(x-1)(x-20)(x-3)}
\\~\\ f(x) = \frac{1}{38} \cdot \frac{1} {(x-1)}+\frac {400}{323} \cdot \frac{1}{(x-20)}-\frac{9}{34} \cdot \frac{1}{(x-3)}
\\~\\
\\ f ^{(n)}(x) = \frac{1}{38} \cdot \frac{ (-1)^n n!}{ (x-1)^{n+1} }
+ \frac {400}{323} \cdot \frac{ (-1)^n n!}{ (x-20)^{n+1} }
-\frac{9}{34} \cdot \frac{ (-1)^n n!}{ (x-3)^{n+1} }

\]
now we can factor

Answer 18

i think this is fine

Answer 19

no need of further simplification

Answer 20

:) nice problem

Answer 21

thank you @jayzdd