Question

Find the given derivative by finding the first few derivatives and observing the pattern that occurs.

Answer 1

\[\frac{ d ^{35} }{ dx ^{35} }(xsinx)\]

Answer 2

Another example from the book:
\[\frac{ d ^{99} }{ dx ^{99} }\]
\[\frac{ d ^{1} }{ dx ^{1} }(sinx)=cosx\]
\[\frac{ d ^{2} }{ dx ^{2} }(sinx)=-sinx\]
\[\frac{ d ^{3} }{ dx ^{3} }(sinx)=-cosx\]
\[\frac{ d ^{4} }{ dx ^{4} }(sinx)=sinx\]
The derivatives of sinx occur in a cycle of four. Since 99 = 4(24) + 3, we have
\[\frac{ d ^{99} }{ dx ^{99} }(sinx) = \frac{ d ^{3} }{ d x^{3} }(sinx) = -cosx\]

Answer 3

Ya, your problem will work out similarly I think.
But you have product rule, so uhhhh try a few,
see what it looks like :)

Answer 4

\[y' = (x)(cosx) + (1)(sinx) = xcosx + sinx\]

Answer 5

So you chain ruled, what happened to the sinx on the far right though? Hmm

Answer 6

are the derivatives of the first two correct?

Answer 7

first derivative is good.

Answer 8

\[y'' = (x)(-sinx) + (1)(cosx) + cosx = -xsinx + 2cosx\]

Answer 9

Is the 2nd deriv. correct?

Answer 10

Ok that looks better :)

Answer 11

xsinx is what we started with, right?

Answer 12

yes

Answer 13

\(\large\rm y''=-y+2cos x\)
I'm not sure if this helps us or not, just something to think about.

Answer 14

I was trying something else sneaky here :)
\(\large\rm y'''=-y'-2sin x\)
\(\large\rm y^{(4)}=-y''-2cos x=-(-y+2cos x)-2cos x\)

Answer 15

So every `two derivatives`,
we gain 2 more cosines, and the sign changes on everything.

Answer 16

If we don't want to deal with the sign changes,
we can count by 4's instead maybe.
So um,
\(\large\rm y^{(4)}=y-4cos x\)
Then,
\(\large\rm y^{(32)}=y-32cos x\)
somethingggggg like that maybe get us on the right track?

Answer 17

so, the 32th derivative would get us back to xsinx?

Answer 18

\(\large\rm y^{(32)}=x\sin x-32\cos x\)
Assuming I didn't make a boo boo somewhere,
ya it seems that way.

Answer 19

the -32cosx shouldn't be there, I think

Answer 20

should be just only xsinx

Answer 21

Were the 2cosx cancelling out each time?
Maybe I messed up a sign earlier, lemme check.
Cause I thought we were GAINING 2cosx each two derivatives.

Answer 22

It could be some trig. identity somewhere in the middle

Answer 23

but then, just an assumption

Answer 24

\(\large\rm y=x sin x\)
\(\large\rm y'=x cos x+sin x\)
\(\large\rm y''=-x sin x+2cos x=-y+2cos x\)
\(\large\rm y'''=-y'-2sin x\)
\(\rm y^{(4)}=-y''-2cos x=-(-x sin x+2cos x)-2cos x=x sin x-4cos x=y-4cos x\)

Answer 25

Oh, you are actually correct

Answer 26

y'' = -(y-2cosx)
y^(4) = +(y-4cosx)

Do you see how the sign is changing every two derivatives,
and our cosines are increasing at the same rate as the derivative number?

Answer 27

Seems like, every even derivative there is a cosine at the end.
For every odd derivative there is a sine at the end

Answer 28

I'm not sure if we should even worry about the odd derivatives :)
If we look for a 5th derivative, maybe we can count by 5's....

But I'm thinking, using what we have here, we can get from the 4th to the 32nd very easily, and then from the 32nd to the 34th using what we know about the change that happens every 2 derivatives.

And then take 1 derivative from there.

Answer 29

\[\frac{ d }{ dx }=xcosx+sinx\]
\[\frac{ d^{2} }{ dx^{2} }=-xsinx+2cosx\]
\[\frac{ d^{3} }{ dx^{3} }=-xcosx-3sinx\]
\[\frac{ d^{4} }{ dx^{4} }=xsinx-4cosx\]
\[\frac{ d^{5} }{ dx^{5} }=xcosx+5sinx\]
\[\frac{ d^{6} }{ dx^{6} }=-xsinx+6cosx\]

Answer 30

The first 4 terms seem to be of the following choices: xcosx, -xsinx, -xcosx, xsinx
and the last 4 terms: sinx, cosx, -sinx, -cosx

Answer 31

\[\frac{ d^{35} }{ dx^{35} }=-xcosx-35sinx \] is the answer

Answer 32

Is the xcosx positive maybe?
Hmm sec I gotta check the pattern again :) lol

Answer 33

yea, we got the math down, it is just analyzing the pattern

Answer 34

nono I think you're right :)
I was approaching it a little different though:
y4 = y-4cosx
So then
y32 = y-32cosx
and 2 more derivatives,
y34 = -(y-34cosx)
and doing the hard work taking another derivative,
y35 = -(y'+34sinx)
and then a little more work.

Yah your way seems better,
just looking at the pattern completely :D hehe

Answer 35

One second.

Answer 36

https://www.wolframalpha.com/input/?i=35th+derivative+of+xsinx

Answer 37

\[\frac{ d^{35} }{ dx^{35} }(xsinx) => 35 = 4(8) + 3 = 35\]

Answer 38

Since, this one has 4 patterns (the same as the example)

Answer 39

ooo neato :D

Answer 40

so, it will look the same as \[\frac{ d^{3} }{ dx^{3} }\]

Answer 41

-xcosx - sinx

Answer 42

yah, very cool c:

Answer 43

then, you just plug in the numbers

Answer 44

-xcosx-35sinx

Answer 45

:)

Answer 46

That small "formula" helps, so we don't have to manually count