Question

Please help, I really need it: Compute the 8th derivative of...

Answer 1

\[\large f(x)=\frac{\cos(3x^3)-1}{x^{4}}\]

Answer 2

@amistre64

Answer 3

a hint was to use the MacLaurin series for f(x)

Answer 4

if you do the MacLaurin series for the cos(3x^3) you get:\[\Large \sum_{n=0}^{\infty} {\frac{(3x^{3})^{2n}}{(2n)!}}(-1)^{n}\]

Answer 5

@ash2326 @estudier

Answer 6

and then if you minus 1 (as per f(x)) and divide each term by x^4, then, when n=4 (the 8th derivative) i get:\[\Large \frac{\frac{3^{8}x^{24}}{8!}}{x^{4}}\]

Answer 7

which would imply that\[f^{8}(0)=3^{8}\]

Answer 8

(the question is to calculate the 8th derivative at x=0:\[f^{8}(0)=.....\])

Answer 9

@remnant What's your f(x) in terms of maclaurin series of cos 3x^3?

Answer 10

\[f(x) = \frac{\Large \sum_{n=0}^{\infty} {\frac{(3x^{3})^{2n}}{(2n)!}}(-1)^{n}-1}{x^4}\]

Answer 11

Expand it and write the terms

Answer 12

i did...and my fourth term (which has the 8th derivative in it) was:\[\Large \frac{\frac{3^{8}x^{24}}{8!}}{x^{4}}\]

Answer 13

Nope, second term
n=2
\[\frac{(3x^3)^4}{4!} \div x^4\]
\[\frac{3^4 x^8}{4!}\]
all terms on the left will get to 0 by 8th derivative, if you take 8th derivative of this term you'd get
\[\frac{3^4 \times 8!}{4!}\]
if you put x=0, only this will be left
\[f_8(0)=\frac{3^4 \times 8!}{4!}\]

Answer 14

surely according to my MacLaurin series for cos(3x^3), the eight derivative will happen when n=4 as cos skips one per one done

Answer 15

@remnant
1) first step is to simplify f(x). You could write in series form, first two terms maybe
2) Next step is to notice what happens on derivative, start in steps of 1

first term is x^2 and second term is x^8, rest are higher powe terms
obviously x^2 diminishes to 0 in 3rd derivative
x^8 becomes 1 if you take 8 times derivative
higher order terms will have x
if you put x=0, only one term is left.

Answer 16

I'll be back in 20 minutes, to explain you.
Till then you try to check

Answer 17

thanks, will do

Answer 18

okay so i get the right answer now and I kinda get how/why i'm doing what i'm doing....when n=2 i get\[\frac{81x^{12}}{24x^{4}}\]and so then i get\[\frac{27}{8}x^8\] which if you take 8 derivatives of, you get\[(8!)\frac{27}{8} = 136080\]

Answer 19

BUT....i have no idea why i would just randomly stop when n=2? ... is it cuz before then i would get 0 and after then i would still have some x's lying around

Answer 20

when n=2, I get:\[\frac{81x^{12}-1}{24x^{4}}\]

Answer 21

nah, the first term of the series is 1 so the -1 cancels that out

Answer 22

ahhh :)

Answer 23

i wish i had more time to focus on this, but trying to do some job related stuff.

Answer 24

\[f(x)=\frac{\cos (3x^3)-1}{x^4}\]
rewriting \(\cos (3x^3) \) in terms of its MacLaurin series
\[\Large f(x)=\frac{(1-\frac {(3x^3)^2}{2!}+\frac {(3x^3)^4}{4!}-\frac {(3x^3)^6}{6!}....)-1}{x^4}\]

\[\Large f(x)=\frac{(\cancel 1-\frac {(3x^3)^2}{2!}+\frac {(3x^3)^4}{4!}-\frac {(3x^3)^6}{6!}....)-\cancel 1}{x^4}\]

\[\Large f(x)=\frac{(\frac {(3x^3)^2}{2!}+\frac {(3x^3)^4}{4!}-\frac {(3x^3)^6}{6!}....)}{x^4}\]

\[\Large f(x)=\frac{\frac {3^2x^6}{2!}+\frac {3^4x^{12}}{4!}-\frac {3^6x^{18}}{6!}....}{x^4}\]
\[\Large f(x)=\frac {3^2x^2}{2!}+\frac {3^4x^{8}}{4!}-\frac {3^6x^{14}}{6!}....\]
Do you get this ?

Answer 25

yip :) (thanks for writing that out :) )

Answer 26

Now let's understand when a x term will become zero on differentiation
Example:
\[f(x)=x\]
\[f'(x)=1\]
\[f''(x)=0\]

\[f(x)=x^2\]
\[f'(x)=2x\]
\[f''(x)=2\]
\[f'''(x)=0\]
so if the power of x is n, then on (n+1) th derivative the term will become zero
Do you get this?

Answer 27

yip :)

Answer 28

\[\Large f(x)=\frac {3^2x^2}{2!}+\frac {3^4x^{8}}{4!}-\frac {3^6x^{14}}{6!}....\]
here first term will become zero on 8th derivative
second will become zero only on 9 th,
higher power terms will have x term on 8th derivative
\[f''''''''(x)=0+\frac{3^4 \times 8\times 7 \times 6\times 5\times 4 \times 3 \times 2\times 1}{4!}-\frac {3^6 \times 14\times 13\times ....7\times x^6}{6!}....\]
now put x=0

Answer 29

Only second term will be there, rest all will vanish
\[f''''''''(x)=0+\frac{3^4 \times 8!}{4!}+0\]

Answer 30

Sorry
\[f''''''''(0)=0+\frac{3^4 \times 8!}{4!}+0\]

Answer 31

aaaaaaah :)

Answer 32

@remnant do you understand this?

Answer 33

fully with you....just dont think i would ever "see" that

Answer 34

you got it now?

Answer 35

thanks for spending so much time going through the answer....was really really helpful :)

Answer 36

yeah i got it :)

Answer 37

Glad to help you:D