Taking the derivative of an integral...

- anonymous

Taking the derivative of an integral...

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- anonymous

How do I take the derivative of the function
\[g(x)=\int\limits_{x}^{x^2}(t^2-5)/(t^2+2)dt\]

- amistre64

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

- anonymous

How exactly does it cancel out?

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## More answers

- amistre64

hmmm..... you might have a point, I failed to recognize the variables....

- anonymous

So.. how would you do it?

- amistre64

take the integral, and substitute for t with x...... then derive it back down..... but the actuall process eludes me :)

- amistre64

can we get the integrand into a sum of terms?

- amistre64

dunno if a trig substitution would be useful: t = sqrt(2) tan(a)

- anonymous

Idk.. I'm just totally lost on this question. So I substitute the t's for x's?

- anonymous

Or do I substitute the t^2 in?

- amistre64

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 :)

- anonymous

Lol ok. Thank you for your help though :)

- amistre64

if I did it right in my thoguths here:
the new integrand becomes:
tan^2(a) -5/2 which looks easier to work with...

- anonymous

nicole did you finish most of the packet? i didn't...i do have a much geater understanding about integrals though :)

- anonymous

No.. I think I'm gonna stay after school.

- anonymous

mother just told me she is picking me up. i have NO idea why. i want to stay too..

- amistre64

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

- anonymous

Just stay.. find a ride home.

- anonymous

Wow.. That's a whole lot..

- amistre64

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

- amistre64

myins here....now ill now what Im doing wrong ;)

- myininaya

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)

- myininaya

g'(x)=2x[(x^4-5)/(x^4+2)]-(x^2-5)/(x^2+2)

- myininaya

any questions?

- myininaya

we dont need to integrate to find the derivative of the integral

- amistre64

......thats what I thought :) but those x..x^2 things threw me for a loop

- myininaya

lol where did you get all of those fans? I'm so jealous

- amistre64

they were having a sale at pier1 ;)

- myininaya

damn i missed it :(

- myininaya

i think he lefted :(

- amistre64

maybe, but the math remained lol

- myininaya

true im going to go play call of duty now peace

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