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anonymous
 5 years ago
Taking the derivative of an integral...
anonymous
 5 years ago
Taking the derivative of an integral...

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How do I take the derivative of the function \[g(x)=\int\limits_{x}^{x^2}(t^25)/(t^2+2)dt\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the derivative and an integral are inverse operations: thed cancel each other out and youd end up with: g'(x) = (t^25)/(t^2+2) ..... as far as I recollect

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How exactly does it cancel out?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm..... you might have a point, I failed to recognize the variables....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So.. how would you do it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0take the integral, and substitute for t with x...... then derive it back down..... but the actuall process eludes me :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0can we get the integrand into a sum of terms?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dunno if a trig substitution would be useful: t = sqrt(2) tan(a)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Idk.. I'm just totally lost on this question. So I substitute the t's for x's?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or do I substitute the t^2 in?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0my limited knowledge on this would suggest that I find g(x) from the integrand first.... an exact HOW, is beyind me at the moment :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lol ok. Thank you for your help though :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if I did it right in my thoguths here: the new integrand becomes: tan^2(a) 5/2 which looks easier to work with...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nicole did you finish most of the packet? i didn't...i do have a much geater understanding about integrals though :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No.. I think I'm gonna stay after school.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mother just told me she is picking me up. i have NO idea why. i want to stay too..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i substituted: t = sqrt(2) tan(a) ; dt = sqrt(2) sec^2 2tan^2(a) 5 sqrt(2)sec^2(a) da   2(tan^2(a) + 1) 1 2tan^2(a) 5 sqrt(2)sec^2(a) da   2sec^2(a) 1 2tan^2(a) 5 sqrt(2) da   2 1 2tan^2(a) 5 sqrt(2) da     2 2 1 sqrt(2) tan^2(a)  5sqrt(2)/2 da is what I get actually, but it might be easier to integrate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just stay.. find a ride home.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wow.. That's a whole lot..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or since the sqrt(2) is a constant, we can yank it out to the other side: sqrt(2) [S] tan^2(a)  5/2 da

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0myins here....now ill now what Im doing wrong ;)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0Let f(t)=(t^25)/(t^2+2) Then int(f(t), t=x..x^2)=F(x^2)F(x) To take derivative of this. We have to use chain rule. Therefore g'(x)=2xf(x^2)f(x)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0g'(x)=2x[(x^45)/(x^4+2)](x^25)/(x^2+2)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0we dont need to integrate to find the derivative of the integral

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0......thats what I thought :) but those x..x^2 things threw me for a loop

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0lol where did you get all of those fans? I'm so jealous

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0they were having a sale at pier1 ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0maybe, but the math remained lol

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0true im going to go play call of duty now peace
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