Question

successive differentiation

please helppp

Answer 1

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

Answer 2

do you know how to do this?

Answer 3

im very much new to this :(
please help me

Answer 4

ok i got you

Answer 5

hmmm

Answer 6

find the first few derivatives as a first step

Answer 7

it would be too big

Answer 8

we are going to use that short list to help come up with the nth derivative

Answer 9

is thelre any other way?

Answer 10

like if i use partial fractions then???

Answer 11

sure you can try partial fractions before the differentiation to see if things come out "less big"

Answer 12

\[\rm \frac{ 1 }{ (3x-2)(x-3)^2 }=\frac{ A }{ (3x-2) }+\frac{ B }{ (x-3) }+\frac{ C }{ (x-3)^2 }\]
correct?

Answer 13

???

Answer 14

yes

Answer 15

now how do i find the values of A B and C respectively?

Answer 16

combine the fractions

Answer 17

both sides will have the same denominator when doing that
so you will just have to compare tops

Answer 18

im getting the value of A to be : 9/49

Answer 19

but after this im not able to get the values for B & C :(

Answer 20

so you must have gotten the numerator for the right hand side since you were able to find A...
\[1=A(x-3)^2+B(x-3)(3x-2)+C(3x-2) \\ x=\frac{2}{3} \text{ gives } \\ 1=A(\frac{2}{3}-3)^2 \\ A=\frac{1}{(\frac{2}{3}-3)^2}=\frac{1}{(\frac{-7}{3})^2}=\frac{1}{\frac{49}{9}}=\frac{9}{49} \\ \text{ try inputting } x=3\]

Answer 21

ohh i got the value of A correct :)
give me a sec
let me try putting 3

Answer 22

i get a equation

Answer 23

what equation?

Answer 24

there is a reason I selected 3
I see it would give a couple of terms 0 which should give you a pretty easy equation

Answer 25

sorry
c = 1/7

Answer 26

yes
1=C(3*3-2)
1=C(9-2)
1=C(7)
C=1/7 is right
great job

Answer 27

we are out of see picks that will make it easy on us
so I guess we can just choose 0 or something so we can solve for B
you must remember to replace A with 9/49 and C with 1/7 though

Answer 28

easy picks* not see picks

Answer 29

B : -3/49 ?

Answer 30

sounds great
you are so fast

Answer 31

:)

Answer 32

\[y=\frac{9}{49} \cdot \frac{1}{3x-2}-\frac{3}{49} \cdot \frac{1}{x-3}+\frac{1}{7} \cdot \frac{1}{(x-3)^2}\]

Answer 33

now we can just find the nth derivative of each term

Answer 34

yup---
what about the last term
how do i find its nth derivative
?

Answer 35

the only way I know to do this is to find the first few derivatives (like let's see if we can stop after taking the 4th)

Answer 36

I would probably not choose quotient rule for this
I would write everything with negative exponents and using chain rule

Answer 37

let's look at one term as an example
\[y=\frac{1}{7} (x-3)^{-2} \\ y^{(1)}=\frac{1}{7}(-2)(x-3)^{-3} \\ y^{(2)}=\frac{1}{7}(-2)(-3)(x-3)^{-4} \\ y^{(3)}=\frac{1}{7}(-2)(-3)(-4)(x-3)^{-5}\]

Answer 38

our nth derivative will definitely have the 1/7 constant multiple
we have a negative/positive alternating thingy
like odd derivative are negative and even derivatives are positive
we also have a bit of factorial business going on
also the power of (x-3) is decreasing consecutively

Answer 39

\[\frac{ (-1)^n(1+n)! }{ 7(x-3)^{2+n} }\]

Answer 40

that looks great

Answer 41

fr the last one?

Answer 42

and I didn't go to the 4th derivative because I think it was pretty obvious at the 3rd what the pattern was

Answer 43

yes
if
\[y=\frac{1}{7} \cdot \frac{1}{(x-3)^2} \\ \text{ then } y^{(n)}=\frac{(-1)^n (1+n)!}{7(x-3)^{2+n}}\]

Answer 44

now we have to look at the other two terms

Answer 45

you think you are able to do this?

Answer 46

yepp
os was lagging for me
sorry for late reply sir
i need help with more sums too.....


thank you so much for helping me

Answer 47

np
I will be back on a little later
I must put food in the belly