anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
https://gyazo.com/34ac1d12d431fd3ca6ce517b4733a203
anonymous
  • anonymous
@jim_thompson5910
tkhunny
  • tkhunny
Why have you posted a problem with no work shown? How far have you gotten? Give us SOMETHING to go on?

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jim_thompson5910
  • jim_thompson5910
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 \[\LARGE F \ '(x) = f{\Huge(}u(x){\Huge )}u \ '(x)\]
anonymous
  • anonymous
@tkhunny I wasn't sure where to start with this one @jim_thompson5910 I'll give it a go
tkhunny
  • tkhunny
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. :-)
anonymous
  • anonymous
@tkhunny do my thoughts count? ;p
anonymous
  • anonymous
@jim_thompson5910 would I integrate the function, find the the value then the domain in which the value exists?
jim_thompson5910
  • jim_thompson5910
that sounds like too much work
jim_thompson5910
  • jim_thompson5910
the fundamental theorem ties together derivatives and integrals in a way, the two cancel each other out (sorta)
jim_thompson5910
  • jim_thompson5910
did what I posted above make sense at all?
jim_thompson5910
  • jim_thompson5910
the two links posted
anonymous
  • anonymous
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
jim_thompson5910
  • jim_thompson5910
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?
anonymous
  • anonymous
sin(x)
anonymous
  • anonymous
+C
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
\[\sin (\pi) \]
jim_thompson5910
  • jim_thompson5910
I am looking for sin(pi) - sin(0)
anonymous
  • anonymous
- sin(0)
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
The equation writer glitched :/
jim_thompson5910
  • jim_thompson5910
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?
anonymous
  • anonymous
Sin(x)-sin(0)?
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
now if we apply the derivative to sin(x)-sin(0), we will get back to cos(x), agreed?
anonymous
  • anonymous
Ok
jim_thompson5910
  • jim_thompson5910
agreed? or no?
anonymous
  • anonymous
Yes since we would just undo it
jim_thompson5910
  • jim_thompson5910
correct
jim_thompson5910
  • jim_thompson5910
let's replace the x with x^2 what is the value of this integral? \[\Large \int_{0}^{x^2}\cos(t)dt\]
anonymous
  • anonymous
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)
jim_thompson5910
  • jim_thompson5910
yes, sin(x^2) - sin(0) now if we apply the derivative to `sin(x^2) - sin(0)` what do we get?
anonymous
  • anonymous
cos(x^2) - cos(0)
jim_thompson5910
  • jim_thompson5910
close, but 2 things a) the derivative of a constant is 0 b) you forgot the chain rule
jim_thompson5910
  • jim_thompson5910
sin(0) is a constant
anonymous
  • anonymous
Oh
anonymous
  • anonymous
I see
anonymous
  • anonymous
\[\sin(x)=u'\cos(u)\]
anonymous
  • anonymous
So \[2x \cos(x^2)\]
jim_thompson5910
  • jim_thompson5910
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)
jim_thompson5910
  • jim_thompson5910
yes, if y = sin(x^2) - sin(0), then dy/dx = 2x*cos(x^2)
jim_thompson5910
  • jim_thompson5910
so that's effectively how that second link I posted is able to say \[\LARGE F(x) = \int_{a}^{u(x)}f(t)dt\] \[\LARGE F \ '(x) = f{\Huge (}u(x){\Huge )}u \ '(x)\]
tkhunny
  • tkhunny
Thoughts? Absolutely! Talk to us! That will work. :-)
anonymous
  • anonymous
Ok so I have
anonymous
  • anonymous
\[2x^2 \sin(x^4+1) - 2x^2\sin(5)\] @jim_thompson5910 would I just take the derivative?
tkhunny
  • tkhunny
You don't have to tag in every post. Anyone contributing to the thread is likely to get a notification anytime you post.
anonymous
  • anonymous
Tag is to show who it was directed at
anonymous
  • anonymous
@jim_thompson5910 I got D for the final answer. Mind checking?
jim_thompson5910
  • jim_thompson5910
In this case, the u(x) function is x^2 what is u ' (x) equal to?
jim_thompson5910
  • jim_thompson5910
I think you took the derivative of x^4 when you shouldn't have done so
anonymous
  • anonymous
2x
anonymous
  • anonymous
I took the derivative of t^2 which is 2t then plugged in x^2
anonymous
  • anonymous
Unless you're talking about the second part.
jim_thompson5910
  • jim_thompson5910
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) \[\LARGE F \ '(x) = f{\Huge (}u(x){\Huge )}u \ '(x)\] \[\LARGE F \ '(x) = f{\Huge (}x^2{\Huge )}*2x\] \[\LARGE F \ '(x) = 2x*f{\Huge (}x^2{\Huge )}\] \[\LARGE F \ '(x) = 2x*\cos\left((x^2)^2+1\right)\] \[\LARGE F \ '(x) = 2x*\cos\left(x^4+1\right)\]
anonymous
  • anonymous
Where I did take the derivative of x^4
jim_thompson5910
  • jim_thompson5910
when you said it was D
anonymous
  • anonymous
Yeah, so i should've taken it of x^2 instead. I see my mistake. Thanks for pointing it out.
jim_thompson5910
  • jim_thompson5910
no problem

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