Taking the derivative of an integral...

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Taking the derivative of an integral...

Mathematics
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How do I take the derivative of the function \[g(x)=\int\limits_{x}^{x^2}(t^2-5)/(t^2+2)dt\]
the derivative and an integral are inverse operations: thed cancel each other out and youd end up with: g'(x) = (t^2-5)/(t^2+2) ..... as far as I recollect
How exactly does it cancel out?

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hmmm..... you might have a point, I failed to recognize the variables....
So.. how would you do it?
take the integral, and substitute for t with x...... then derive it back down..... but the actuall process eludes me :)
can we get the integrand into a sum of terms?
dunno if a trig substitution would be useful: t = sqrt(2) tan(a)
Idk.. I'm just totally lost on this question. So I substitute the t's for x's?
Or do I substitute the t^2 in?
my limited knowledge on this would suggest that I find g(x) from the integrand first.... an exact HOW, is beyind me at the moment :)
Lol ok. Thank you for your help though :)
if I did it right in my thoguths here: the new integrand becomes: tan^2(a) -5/2 which looks easier to work with...
nicole did you finish most of the packet? i didn't...i do have a much geater understanding about integrals though :)
No.. I think I'm gonna stay after school.
mother just told me she is picking me up. i have NO idea why. i want to stay too..
i 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
Just stay.. find a ride home.
Wow.. That's a whole lot..
or 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
myins here....now ill now what Im doing wrong ;)
Let f(t)=(t^2-5)/(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)
g'(x)=2x[(x^4-5)/(x^4+2)]-(x^2-5)/(x^2+2)
any questions?
we dont need to integrate to find the derivative of the integral
......thats what I thought :) but those x..x^2 things threw me for a loop
lol where did you get all of those fans? I'm so jealous
they were having a sale at pier1 ;)
damn i missed it :(
i think he lefted :(
maybe, but the math remained lol
true im going to go play call of duty now peace

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