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

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SUMMARY

The derivative of the function G(x) = ∫ from x to x² of sin(-t²) dt is evaluated using the Fundamental Theorem of Calculus and the chain rule. The correct derivative is G'(x) = sin(-x⁴) * 2x - sin(-x²). This expression can be simplified to G'(x) = -sin(x²) + 2x * sin(x⁴). The discussion emphasizes the importance of correctly applying differentiation techniques to integral functions.

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  • Proficiency in applying the chain rule in differentiation
  • Familiarity with Taylor series expansions
  • Knowledge of trigonometric identities, specifically sin(-θ) = -sin(θ)
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  • Practice differentiation using the chain rule with various functions
  • Explore Taylor series and their applications in calculus
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Students preparing for calculus exams, particularly those focusing on derivatives of integral functions, as well as educators seeking to clarify concepts related to the Fundamental Theorem of Calculus and differentiation techniques.

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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.
 
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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?
 
Welcome to PF!

Hi calchelp! Welcome to PF! :smile:

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


can i change the title of this thread??
 


is the answer:

sin(-x^4)2x - sin(-x^3)
 
calchelp said:
is the answer:

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

Almost … where did your x3 come from? :confused:
 


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!
 


can anyone confirm this please?
 


calchelp said:
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.
 
  • #10


thank you! dick and tiny-tim!
 
  • #11


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
Hi calchelp! :smile:

(just got up :zzz: …)
calchelp said:
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) :wink:
 

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