Homework Help: Calc 2 final tomorrow need help

1. Dec 10, 2009

calchelp

calc 2 final tomorrow need urgent help!!

Evaluate the derivative of the function

G(x) = ({integral} from x to x^2) of sin(-t^2)dt

this was on my review for the final and i just cannot get an answer. any help will be greatly appreciated.

2. Dec 10, 2009

calchelp

need help evaluating derivative of sin(x^2)

sorry i'm new to this site.
i tried substitution which did not get me anywhere.

next, i did taylor series but could not get an answer.

now i am stuck. i am almost positive i have to use a taylor series.

with the series i got

x^2 - (x^6)/3! + (x^10)/5! ...

i plugged in x^2 for x
and i plugged in x for x and subtracted the two and got another series...
now what? antidifferentiate the series and say that that is my answer or did i mess up somewhere?

3. Dec 10, 2009

tiny-tim

Welcome to PF!

Hi calchelp! Welcome to PF!

(have an integral: ∫ and try using the X2 tag just above the Reply box )
Hint: fundamental theorem of calculus.

4. Dec 10, 2009

calchelp

Re: calc 2 final tomorrow need urgent help!!

can i change the title of this thread??

5. Dec 10, 2009

calchelp

Re: calc 2 final tomorrow need urgent help!!

sin(-x^4)2x - sin(-x^3)

6. Dec 10, 2009

tiny-tim

Almost … where did your x3 come from?

7. Dec 10, 2009

calchelp

Re: calc 2 final tomorrow need urgent help!!

ok... so after about 20 minutes of looking at it i realized that i completely made up the x^3 i dont know why or where it came from...

sin(-x^4)2x - sin(-x^2)

please tell me that is right. i am about to go insane!!!

8. Dec 10, 2009

calchelp

Re: calc 2 final tomorrow need urgent help!!

9. Dec 10, 2009

Dick

Re: calc 2 final tomorrow need urgent help!!

That's right.

10. Dec 10, 2009

calchelp

Re: calc 2 final tomorrow need urgent help!!

thank you! dick and tiny-tim!!

11. Dec 10, 2009

andylu224

Re: calc 2 final tomorrow need urgent help!!

Essentially this is using the chain rule whereby when:

G(x) = (integral of f1(x) to f2(x) ) g(t) dt

G'(x) = g(f2(x))*f2'(x) - g(f1(x))*f1'(x)

so in your case, G'(x) = sin(-x^4)*2x - sin(-x^2)

12. Dec 11, 2009

tiny-tim

Hi calchelp!

(just got up :zzz: …)
You can tidy it up a bit by taking the minus out of the sin (because sin(-θ) = -sinθ) …

sin(x2) - 2x*sin(x4)