# handful of derivatives using trig idents, have 3 hrs

• Oct 18th 2006, 06:41 AM
thedoge
handful of derivatives using trig idents, have 3 hrs
Hi. It's been some time since I've done any trig work. BTW, I'm sorry to doublepost in the calc forum, meant to post here.

Could anyone guide me through these particular problems?

I need the derivatives of the following:

1. (2 sin x)/(1+cos x)
2. (5 tan x)/(x)
3. (-8xsinxcosx)
4. (tan x - 3)/(sec x)

Also, can anyone briefly tell me what the following are equal to?

sin^2 x
cos^2 x
tan^2 x
sec^2 x

And then, how to use them in conjunction with each other.
• Oct 18th 2006, 06:56 AM
Jameson
#1) d/dx(2 sin x)/(1+cos x) = [(1+cosx)(2cosx)-2sinx(-sinx)] / [1+cosx]^2
• Oct 18th 2006, 07:05 AM
earboth
1rst problem only
Quote:

Originally Posted by thedoge
Hi. It's been some time since I've done any trig work. BTW, I'm sorry to doublepost in the calc forum, meant to post here.

Could anyone guide me through these particular problems?
I need the derivatives of the following:
1. (2 sin x)/(1+cos x)
...
Also, can anyone briefly tell me what the following are equal to?

sin^2 x
cos^2 x
...

Hi,

I've attached a diagram to show you what I've calculated:
• Oct 18th 2006, 07:25 AM
ThePerfectHacker
Quote:

Originally Posted by thedoge

sin^2 x
cos^2 x
tan^2 x
sec^2 x

And then, how to use them in conjunction with each other.

The 3 Pythagorean Identities are:
sin^2 x+cos^2 x=1
tan^2 x+1=sec^2 x
cot^2 x+1=csc^2 x

Thus,
Solving these equations for the other functions we have,
sin^2 x=1-cos^2 x
cos^2 x=1-sin^2x
tan^2 x=sec^2 x-1
cot^2 x=csc^2 x -1
• Oct 18th 2006, 07:46 AM
thedoge
Thanks a lot for the help so far, but I'm still stuck with the others.

Also, I didn't know how to remove my other post or else I would have.

Do you guys have an irc chat or something?
• Oct 18th 2006, 08:28 AM
thedoge
Well, can you at least tell me what I'm doing wrong here?

Prob (sqrt(x)-4)/(sqrt(x)+4)

I use the quotient rule.
gdf-fdg/g^2.
(sqrt(x)+4)(1/2sqrt(x))-(sqrt(x)-4)(1/2sqrt(x))/(sqrt(x)+4)^2
(4sqrt(x))/(sqrt(x)+4)^2
• Oct 18th 2006, 08:38 AM
topsquark
Quote:

Originally Posted by thedoge
Well, can you at least tell me what I'm doing wrong here?

Prob (sqrt(x)-4)/(sqrt(x)+4)

I use the quotient rule.
gdf-fdg/g^2.
(sqrt(x)+4)(1/2sqrt(x))-(sqrt(x)-4)(1/2sqrt(x))/(sqrt(x)+4)^2
(4sqrt(x))/(sqrt(x)+4)^2

Watch the parenthesis. Either you did the derivative wrong or you got lost on what was in the denominators.

d/dx[sqrt(x)] = (1/2) * 1/sqrt(x)

Look at it in terms of powers:
sqrt(x) = x^(1/2)

So
d/dx[x^(1/2)] = (1/2)*x^(1/2 - 1) = (1/2)*x^(-1/2)

-Dan
• Oct 18th 2006, 08:49 AM
thedoge
Well, where in what I did was the error? I'm so lost:(

To tell you the truth, the last HS math I took in full was geometry. I had a quick refresher in College Algebra and now I'm in a calculus course. I follow the calc fine so long as the stuff I skipped isn't brought up. I can follow general principles, but I have alot of holes in my background. I'm working to fix them, but failing this hw won't help me do that :/

I -really- want to learn everything I've missed and excel in math, but with nothing to really work with, hard to find help, and lack of proper examples, I can not figure this stuff out.

I know this stuff isn't difficult, but without knowing all the rules, that doesn't matter...
• Oct 18th 2006, 09:41 AM
earboth
Quote:

Originally Posted by thedoge
Well, can you at least tell me what I'm doing wrong here?
Prob (sqrt(x)-4)/(sqrt(x)+4)
I use the quotient rule.
gdf-fdg/g^2.
(sqrt(x)+4)(1/2sqrt(x))-(sqrt(x)-4)(1/2sqrt(x))/(sqrt(x)+4)^2
(4sqrt(x))/(sqrt(x)+4)^2

Hi,

you've forgotten to set the denominator into brackets.

(sqrt(x)+4)(1/(2sqrt(x)))-(sqrt(x)-4)(1/(2sqrt(x)))/(sqrt(x)+4)^2
• Oct 18th 2006, 12:09 PM
thedoge
Hrm, I've tried that answer, but it doesn't seem to work earboth. Thanks though.
• Oct 18th 2006, 12:14 PM
thedoge
Any help on the other 3?

2. (5 tan x)/(x)
3. (-8xsinxcosx)
4. (tan x - 3)/(sec x)