Question

Find the derivative with respect to x of the integral from 2 to x squared of the cosine of the quantity t squared plus 1, dt

Answer 1

https://gyazo.com/34ac1d12d431fd3ca6ce517b4733a203

Answer 2

@jim_thompson5910

Answer 3

Why have you posted a problem with no work shown? How far have you gotten? Give us SOMETHING to go on?

Answer 4

you'll need to use the fundamental theorem of calculus

http://aleph0.clarku.edu/~ma120/FTC.jpg

the second part of it. Don't forget to use the chain rule



this page may help when it comes to the chain rule:
http://www.sosmath.com/calculus/integ/integ03/integ03.html


Look at the part that has
\[\LARGE F(x) = \int_{a}^{u(x)}f(t)dt\]


notice just under that is
\[\LARGE F \ '(x) = f{\Huge(}u(x){\Huge )}u \ '(x)\]

Answer 5

@tkhunny I wasn't sure where to start with this one
@jim_thompson5910 I'll give it a go

Answer 6

Insufficient response. You have a problem section or assignment because you have the required information. You CAN'T have NO idea. That's just not right. Start SOMEWHERE - ANYWHERE. You won't break it. It's just a problem statement. We like to help. Help us do that. :-)

Answer 7

@tkhunny do my thoughts count? ;p

Answer 8

@jim_thompson5910 would I integrate the function, find the the value then the domain in which the value exists?

Answer 9

that sounds like too much work

Answer 10

the fundamental theorem ties together derivatives and integrals

in a way, the two cancel each other out (sorta)

Answer 11

did what I posted above make sense at all?

Answer 12

the two links posted

Answer 13

I got part of it. Currently trying to see how I would manipulate it for \[\int\limits_{2}^{x^2}\cos(t^2+1)\]
Since they have it simpler here https://gyazo.com/9f4530e3a1e2dc1c478a40f8f353166e

Answer 14

let's start with something a bit simpler

let's say we had

\[\Large \int \cos(x)dx\]

what is the value of that integral?

Answer 15

sin(x)

Answer 16

+C

Answer 17

yep, now let's throw in endpoints

what is the value of this

\[\Large \int_{0}^{\pi}\cos(x)dx\]

don't simplify or evaluate with calculator, just do a substitution/replacement

Answer 18

\[\sin (\pi) \]

Answer 19

I am looking for sin(pi) - sin(0)

Answer 20

- sin(0)

Answer 21

yes

Answer 22

The equation writer glitched :/

Answer 23

so imagine we are now integrating with respect to t

and we now have this

\[\Large \int_{0}^{x}\cos(t)dt\]

what is the value of this?

Answer 24

Sin(x)-sin(0)?

Answer 25

yes

Answer 26

now if we apply the derivative to sin(x)-sin(0), we will get back to cos(x), agreed?

Answer 27

Ok

Answer 28

agreed? or no?

Answer 29

Yes since we would just undo it

Answer 30

correct

Answer 31

let's replace the x with x^2

what is the value of this integral?

\[\Large \int_{0}^{x^2}\cos(t)dt\]

Answer 32

sin(x^2)-sin(0)
I see what you're getting at.
so in the problem it would be sin(x^4+1)-sin(5)

Answer 33

yes, sin(x^2) - sin(0)

now if we apply the derivative to `sin(x^2) - sin(0)` what do we get?

Answer 34

cos(x^2) - cos(0)

Answer 35

close, but 2 things

a) the derivative of a constant is 0
b) you forgot the chain rule

Answer 36

sin(0) is a constant

Answer 37

Oh

Answer 38

I see

Answer 39

\[\sin(x)=u'\cos(u)\]

Answer 40

So \[2x \cos(x^2)\]

Answer 41

if you really want to be technical, you apply the chain rule every time even if it's simply x (and not x^2)


y = sin(x)
dy/dx = cos(x)*d/dx[x]
dy/dx = cos(x)*1
dy/dx = cos(x)

Answer 42

yes, if y = sin(x^2) - sin(0), then dy/dx = 2x*cos(x^2)

Answer 43

so that's effectively how that second link I posted is able to say

\[\LARGE F(x) = \int_{a}^{u(x)}f(t)dt\]

\[\LARGE F \ '(x) = f{\Huge (}u(x){\Huge )}u \ '(x)\]

Answer 44

Thoughts? Absolutely! Talk to us! That will work. :-)

Answer 45

Ok so I have

Answer 46

\[2x^2 \sin(x^4+1) - 2x^2\sin(5)\] @jim_thompson5910 would I just take the derivative?

Answer 47

You don't have to tag in every post. Anyone contributing to the thread is likely to get a notification anytime you post.

Answer 48

Tag is to show who it was directed at

Answer 49

@jim_thompson5910 I got D for the final answer. Mind checking?

Answer 50

In this case, the u(x) function is x^2

what is u ' (x) equal to?

Answer 51

I think you took the derivative of x^4 when you shouldn't have done so

Answer 52

2x

Answer 53

I took the derivative of t^2 which is 2t then plugged in x^2

Answer 54

Unless you're talking about the second part.

Answer 55

So,

\[\LARGE F(x) = \int_{a}^{u(x)}f(t)dt\]

\[\LARGE F(x) = \int_{2}^{x^2}\cos(t^2+1)dt\]

we have a = 2, u(x) = x^2 and f(t) = cos(t^2+1)



\[\LARGE F \ '(x) = f{\Huge (}u(x){\Huge )}u \ '(x)\]

\[\LARGE F \ '(x) = f{\Huge (}x^2{\Huge )}*2x\]

\[\LARGE F \ '(x) = 2x*f{\Huge (}x^2{\Huge )}\]

\[\LARGE F \ '(x) = 2x*\cos\left((x^2)^2+1\right)\]

\[\LARGE F \ '(x) = 2x*\cos\left(x^4+1\right)\]

Answer 56

Where I did take the derivative of x^4

Answer 57

when you said it was D

Answer 58

Yeah, so i should've taken it of x^2 instead. I see my mistake. Thanks for pointing it out.

Answer 59

no problem