Question

find the given derivative by finding the first few derivatives and observing the pattern that occurs : D^35 xsin pi x

Answer 1

I am not sure what \(\large\color{black}{ \displaystyle {\rm D}^{35} }\) is, does that come to denote 35th derivative?

Answer 2

In any case,

\(\large\color{black}{ \displaystyle f(x)=x\sin\left(\pi x\right) }\)
\(\large\color{black}{ \displaystyle f'(x)=\sin\left(\pi x\right)+\pi x\cos\left(\pi x\right) }\)

\(\color{black}{ \displaystyle f''(x)=\pi\cos\left(\pi x\right)+\pi \cos\left(\pi x\right)-\pi^2 x\sin\left(\pi x\right) = 2\pi \cos\left(\pi x\right)-\pi^2 x\sin\left(\pi x\right) }\)

Answer 3

Yes I think that it denotes the 35th derivative, I understand since the derivative resets every 4th derivative, the answer would be the 3rd derivative of the equation. (since 4 goes into 32 evenly)

Answer 4

with the x in front of the sine, the derivatives don't reset.

Answer 5

Oh
So in this case it would just be the second derivative ?

Answer 6

What do you mean?

Answer 7

Because you said that it doesn't reset if there is an x in front

Answer 8

well, not reset to the very first function, but I think the pattern is present

Answer 9

I think I have got the pattern.

Answer 10

How do you determine the pattern ?

Answer 11

I was just taking derivatives and simplifying them.... it might be a pain in the neck, but this is how you do it.

Answer 12

Are my notations making sense?

Answer 13

I understand what you just did, but would that change the final answer for the original questions I asked?

Answer 14

you have to observe the pattern for the derivatives.
Every 4th derivative...

I would observe every pattern for
(4k)th
(4k+1)th
(4k+2)th
(4k+3)th

(and (4k+4)th is again restarting the cycle of (4k)th )

Answer 15

35(sin(pix) + pi x cos (pi x)

Answer 16

you need 35th, or that would be,
(4k+3)th derivative. and percisely: (4×8+3)th

Answer 17

which would make it the 3rd derivative of the equation ?

Answer 18

this is what I got:

My function is: \(\large\color{black}{ \displaystyle f^{(0)}(x)=x\sin\left(\pi x\right) }\)


\(\large\color{blue}{ \displaystyle f^{(4k)}(x)=\pi^{4k-1}\left(\pi x \sin \pi x~-4k\cos\pi x\right) }\)

\(\large\color{blue}{ \displaystyle f^{(4k+1)}(x)=\pi^{4k}\left((4k+1) \sin \pi x+\pi x \cos\pi x\right) }\)

\(\large\color{blue}{ \displaystyle f^{(4k+2)}(x)=\pi^{4k+1}\left((4k+2) \cos \pi x-\pi x \sin \pi x\right) }\)

\(\large\color{blue}{ \displaystyle f^{(4k+3)}(x)=-\pi^{4k+2}\left((4k+3) \sin \pi x+\pi x \cos\pi x\right) }\)

Answer 19

these are the patterns I got.... took a while to type and check them, apologize.

Answer 20

Try finding 1st, 2nd, 3rd, 4th, and so forth derivative.... (I would recommend factoring like I did), you will notice the pattern repeats every 4th time (for 4k, 4k+1, 4k+2, and 4k+3)

Answer 21

Thank you so much for your help !!

Answer 22

Patterns do take pain, but all you need is work... you can do it:)
you welcome