Quote from Sagaris

Quote from chineserider

Basically, you have [integral]1/(1+sinx) dx
Then you separate that into [integral]1/1 + [integral]1/sinx (which is cscx)

Fractions don't work that way.

Integral of dx/(1+sinx) is the same as dx * 1/(1+sinx). I have never seen people prove it otherwise. You cannot have a dx on the denominator, but dx on the numerator is acceptable.

Edit: Sorry. Wasn't paying attention to the parenthesis there.

Nulla in mundo pax sincera
"Always go too far, because that's where you'll find the truth." - Albert Camus

Quote from Sagaris

Quote from chineserider

Basically, you have [integral]1/(1+sinx) dx
Then you separate that into [integral]1/1 + [integral]1/sinx (which is cscx)

Fractions don't work that way.

@chineserider - They don't. It's in the form of (1/(2+3)). You can't right that as (1/2+1/3).

@Mewman - Did you try rationalising it? Edit: Look below to sarah-eats-cupcakes' post.

There are times when you will miss what you never had. I wonder how you will find what you so desperately need.

Quote from MewMan

Compute this integral: ∫dx/(1+sinx)

my answer is tanx - 1/cosx + C
i don't understand why it's wrong 🤢

multiply the numerator and denominator by (1-sinx)
so you get (1-sinx)/(cos²x)
now seperate fractions, it becomes: (1/(cos²x))-(sinx/(cos²x))
so, sec²x-(sinx/(cos²x))
sec²x - (1/cosx)(sinx / cosx)
sec²x-(secx)(tanx)

∫(sec²x)dx-∫(secx*tanx) dx
tanx-secx+c

nothings wrong with your answer...i think 😐 you just have to write secx instead of 1/cosx...but its fine if you dont its still correct

thanks for all the help everyone
but the problem is so simple that i want to punch myself in the face
you see, for this problem they already gave C in the answer, so my C was unnecessary
sorry for the trouble 🤢

its cold down here fam ~

EDIT: Nvm. I googled it. lol

♪MONSTARR~ will eat all your cookies and steal your bishies~♪ Φ_Φ

So I'm not sure how to delete a post so I'm just gonna say that I figured out my homework so it's ok ^-^

High school calculus:

After 3 unsuccessful attempts, I still can't calculate the derivative of this function using the chain rule (with possibly, power rule?) correctly:

http://i.imgur.com/IWUIF.png

There's a second part for this question, but I appreciate any help to figure out the derivative only. I treated the function as if it was a composite function, but I just managed to puzzle myself...

Whether we fall by ambition, blood, or lust,
Like diamonds, we are cut with our own dust.

Quote from 15C08

High school calculus:

After 3 unsuccessful attempts, I still can't calculate the derivative of this function using the chain rule (with possibly, power rule?) correctly:

http://i.imgur.com/IWUIF.png

There's a second part for this question, but I appreciate any help to figure out the derivative only. I treated the function as if it was a composite function, but I just managed to puzzle myself...

I'm getting [(1/3)(((x^2)+5)^(1/2)) + (1/3x)(1/2)(x^2)+5)(2x))], if that is readable. Is that what you wanted?

There are times when you will miss what you never had. I wonder how you will find what you so desperately need.

Quote from Casey D. Geek

Quote from 15C08

High school calculus:

After 3 unsuccessful attempts, I still can't calculate the derivative of this function using the chain rule (with possibly, power rule?) correctly:

http://i.imgur.com/IWUIF.png

There's a second part for this question, but I appreciate any help to figure out the derivative only. I treated the function as if it was a composite function, but I just managed to puzzle myself...

I'm getting [(1/3)(((x^2)+5)^(1/2)) + (1/3x)(1/2)(x^2)+5)(2x))], if that is readable. Is that what you wanted?

Yes, thank you for the quick response. 🙂

Whether we fall by ambition, blood, or lust,
Like diamonds, we are cut with our own dust.

Quote from 15C08

Quote from Casey D. Geek

Quote from 15C08

High school calculus:

After 3 unsuccessful attempts, I still can't calculate the derivative of this function using the chain rule (with possibly, power rule?) correctly:

http://i.imgur.com/IWUIF.png

There's a second part for this question, but I appreciate any help to figure out the derivative only. I treated the function as if it was a composite function, but I just managed to puzzle myself...

I'm getting [(1/3)(((x^2)+5)^(1/2)) + (1/3x)(1/2)(x^2)+5)(2x))], if that is readable. Is that what you wanted?

Yes, thank you for the quick response. 🙂

You're welcome. I got the answer using [(uv)' = u'v + uv']; I don't remember what rule it is known as.

There are times when you will miss what you never had. I wonder how you will find what you so desperately need.

Quote from Casey D. Geek

Quote from 15C08

Quote from Casey D. Geek

[quote=15C08]High school calculus:

After 3 unsuccessful attempts, I still can't calculate the derivative of this function using the chain rule (with possibly, power rule?) correctly:

http://i.imgur.com/IWUIF.png

There's a second part for this question, but I appreciate any help to figure out the derivative only. I treated the function as if it was a composite function, but I just managed to puzzle myself...

I'm getting [(1/3)(((x^2)+5)^(1/2)) + (1/3x)(1/2)(x^2)+5)(2x))], if that is readable. Is that what you wanted?

Yes, thank you for the quick response. 🙂

You're welcome. I got the answer using [(uv)' = u'v + uv']; I don't remember what rule it is known as.[/quote]

It's called Integration by Parts. Calculus II curriculum.

Edit: Sorry, it's the product rule. I misread.

Nulla in mundo pax sincera
"Always go too far, because that's where you'll find the truth." - Albert Camus

Quote from chineserider

It's called Integration by Parts. Calculus II curriculum.

No, it's product rule.

Hi, I was wondering if someone could assist me on a project for my Economics class. We have to compare an aspect of medicare between the United States and another country, but the problem is that I can't find much info on the medicare systems of other countries beside Canada, which I'm not allowed to pick.

Could anyone help give some ideas to which aspect of medicare to use as comparison? Or sites where I can find such info on other countries?

Nulla in mundo pax sincera
"Always go too far, because that's where you'll find the truth." - Albert Camus

Cuba, France. Those two come up to mind.

Edit: Nvm, Not China.

[img]http://i1237.photobucket.com/albums/ff463/Asura007/Enishi-1.jpg[/img]

Quote from Sagaris

Quote from chineserider

It's called Integration by Parts. Calculus II curriculum.

No, it's product rule.

And it's a differential rule :/

There are times when you will miss what you never had. I wonder how you will find what you so desperately need.