Question

Find F'(x) when F(x)=sin(-x^2) intergral from 0 to x^2 of e^(t)^(2)dt.

Answer 1

Ooo interesting problem! :)

Answer 2

\[\Large\rm F(x)=\sin(-x^2)\int\limits_0^{x^2} e^{t^{2}}~dt\]
Find \(\Large\rm F'(x)\).

This is the problem? That look correct?

Answer 3

Yes.

Answer 4

I'm gonna use some shorthand notation for a sec,
\[\Large\rm F(x)=\sin(-x^2)\cdot Int\]So to find the derivative, we apply product rule, yes?\[\Large\rm F'(x)=\color{royalblue}{\left[\sin(-x^2)\right]'}\cdot Int+\sin(-x^2)\cdot \color{royalblue}{\left[Int\right]'}\]

Answer 5

So we need the derivative of these blue parts.
Understand how to deal with the first one?

Answer 6

Okay, but where do I go from there? So far in calculus we have only learned the fundamental theorem of calculus. So they expect us to do it that way.

Answer 7

You apply the Fundamental Theorem of Calculus, Part 1, to take the derivative of that integral.

We can go over that in a sec if you need.
But first deal with the sin(-x^2) derivative.
You just apply normal rules to deal with that.
Do you remember how?

Answer 8

Yeah, that would be -2cos(x^2), right?

Answer 9

\[\Large\rm (\sin(-x^2))'=\cos(-x^2)(-2x)\]If you want to absorb the negative inside the cosine function, (since it's an odd function) I guess you can do that...\[\Large\rm =-2x \cos(x^2)\] but it's probably better to leave the negative inside alone.\[\Large\rm -2x \cos(-x^2)\]

Answer 10

Yours looks a little different, do you understand where the x came from?
Derivative of x^2 is 2x, right?

Answer 11

Yeah, I just wrote it down wrong when taking the derivative. Sorry about that. I just redid it and got the same answer as you.

Answer 12

cool c:
so for the derivative of the integral, it's a little weird!
We'll apply the FTC, part 1:\[\Large\rm \frac{d}{dx}\int\limits_0^x f(t)~dt=f(x)\]

Answer 13

So that would be \[e ^{x ^{2}}\].

Answer 14

It's a bit more complicated than that since our upper limit isn't just "x".
Lemme try to explain this in a way that will make sense...

Answer 15

Okay.

Answer 16

\[\huge\rm \int\limits\limits_0^{x^2} e^{\color{orangered}{t}^{2}}~dt\quad=\quad \int\limits\limits_0^{x^2} f(\color{orangered}{t})~dt\]So you correctly identified the f(t) that we're integrating.

Answer 17

Integrating will give us "something", let's call it F(t) the anti-derivative.

This is the part that is a little hard for some students to understand:
We're not actually looking for an anti-derivative, we're assuming one exists and writing it as F(t). We'll plug in our boundaries:\[\huge\rm F(\color{orangered}{t})|_0^{x^2}\quad=\quad F(\color{orangered}{x^2})-F(\color{orangered}{0})\]

Answer 18

Then if we want to take the derivative, we end up back at our original f function,\[\huge f(\color{orangered}{x^2})\cdot 2x-0\]

Answer 19

Since the inner function (x^2) was more than just "x", I had to apply the chain rule.

Answer 20

I know I know it's a weird process!!
~You anti-differentiate
~plug in boundaries
~differentiate

So you end up with the same thing you started with, but it's got a little baggage along with it now.

Answer 21

Isn't that the second step of the fundamental theorem of calculus though?

Answer 22

True I guess it is! :)
I'm using both steps of the fundamental theorem to show what's going on.

Answer 23

Oh, okay!

Answer 24

F(0) is some anti-derivative of e^{t^2} with 0 plugged in.
So the whole thing ends up being a costant value.

When you take the derivative of a constant, you get zero right?
That's why F(0) is giving us zero when we take it's derivative with respect to x.

Answer 25

\[\huge =f(\color{orangered}{x^2})\cdot 2x\]\[\huge\rm =e^{(\color{orangered}{x^2})^2}\cdot 2x\]

Answer 26

You have to plug the entire x^2 into the t.
See how we get an extra square because of the square that was already on the t?

Answer 27

Oh, that makes a lot more sense now.

Answer 28

Yah this is a tricky problem!
If you have a good grasp on the FTC, Part 2, it really helps here.
When you're able to setup your F(b)-F(a), you can see where the pieces are coming from, like that 2x.

Answer 29

So would you have to do anything else with the derivative of sin(-x^2)?

Answer 30

cos(-x) = cos(x)
because cosine is an even function* blah I said odd before.. mistake.
So you can write your \[\Large\rm \cos(-x^2)\]as\[\Large\rm \cos(x^2)\]if you want.
But that doesn't seem necessary.

Answer 31

\[\Large\rm F'(x)=\color{royalblue}{\left[\sin(-x^2)\right]'}\cdot Int+\sin(-x^2)\cdot \color{royalblue}{\left[Int\right]'}\]
\[\Large\rm F'(x)=-2x\cos(-x^2)\cdot Int+\sin(-x^2)\cdot 2x\cdot e^{x^4}\]Something like that yah? :o

Answer 32

With the integral part written out*
I'm too lazy though lol

Answer 33

Okay, thank you so much for all your help!