How do I evaluate the derivative of G(x) for my Calc 2 final tomorrow?

[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.

 

[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?

 

[tiny-tim]

Welcome to PF!

Hi calchelp! Welcome to PF!

(have an integral: ∫ and try using the X² tag just above the Reply box )

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.

Hint: fundamental theorem of calculus.

 

[calchelp]

can i change the title of this thread??

 

[calchelp]

is the answer:

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

 

[tiny-tim]

is the answer:

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

Almost … where did your x³ come from?

 

[calchelp]

ok... so after about 20 minutes of looking at it i realized that i completely made up the x^3 i don't know why or where it came from...

so is the answer:
sin(-x^4)2x - sin(-x^2)

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

 

[calchelp]

can anyone confirm this please?

 

[Dick]

ok... so after about 20 minutes of looking at it i realized that i completely made up the x^3 i don't know why or where it came from...

so is the answer:
sin(-x^4)2x - sin(-x^2)

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

That's right.

 

[calchelp]

thank you! dick and tiny-tim!

 

[andylu224]

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)

 

[tiny-tim]

Hi calchelp!

(just got up :zzz: …)

so is the answer:
sin(-x^4)2x - sin(-x^2)

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

sin(x²) - 2x*sin(x⁴)