# Trig/Quotient Rule

Discussion in 'Undergraduate Math' started by James Stein, Oct 1, 2004.

1. ### James SteinGuest

Two point post, actually. The first is a question; does anyone know of a
good site covering basic trig, up to trig identities and trig functions?
I've always been bad with that, and I was hoping to get some review in (my
textbook doesn't cover it well.)

The second is the following:

f(x) = 5sin(-3x) / 1 + cosx.
Find f ' (3).

So, using quotient rule:
(1 + cos x) (5cos(-3x)) - (5sin(-3x))(-sinx)
-------------------------------------------
(1 + cosx)^2

I'm really not sure how to foil it, because I'm not sure what happens with
5cos(-3x) * cos x. Does it become 5cos^2(-3x), or 5cos(-3x^2) or
5cos(-3x)cosx?

Just plugging the three into the formula as it stands doesn't get me the
right answer, from what I see.

Also, I couldn't think of any trig ID's to simplify the issue.

James Stein, Oct 1, 2004

2. ### David MoranGuest

You're close, but check your differentiation of 5sin(-3x). I don't think
there's any way of simplifying this.

Dave

David Moran, Oct 1, 2004

3. ### Rich CarreiroGuest

Quotient rule is:

d(uv)/dx = (v du/dx - u dv/dx)/v^2

u(x) = 5sin(-3x) = -5sin(3x) [I don't like having negatives in
the argument of trig functions]

v(x) = 1 + cos(x)

du/dx = (-5)(3)cos(3x) = -15cos(3x) [you forgot about the 3]

dv/dx = -sin(x)

so

df/dx = -15[1 + cos(x)]cos(3x) - 5sin(3x)sin(x)
---------------------------------------
[1 + cos(x)]^2
Why do you want to FOIL it? Just plug in x=3.
No, it doesn't. Why do you think it might (I'm not being
sarcastic -- I'm trying to see why you think that's even
a possibility)?
See above.
That's it. Aside from using the
cos(a)cos(b) = [cos(a+b) + cos(a-b)]/2
formula to write it as a sum of individual
cosines, that's about all you can do with it.

Rich Carreiro, Oct 1, 2004
4. ### James SteinGuest

Forget about it? I don't even know where it came from. Is that a chain rule
multiplying by d/dx 3x?

I have a very shaky grasp of basic trig. I manage to pull off high 90's on
my calc tests anyway, but mostly through rote memorization of
problem-solving strategies and certain recurrent 'tricks' - not through
understanding of the fundamentals. Which is unfortunate, and something I'm
working to correct.

James Stein, Oct 1, 2004
5. ### Rich CarreiroGuest

Yes, it's because of the chain rule.

Remember, d[f(u(x))]/dx = (df/du)(du/dx)

For sin(3x), f(u) = sin(u), u(x) = 3x, so
d(sin(3x))/dx = d[sin(u)]/du * d(3x)/dx
= cos(u) * 3
= 3cos(u)
= 3cos(3x)

Rich Carreiro, Oct 1, 2004
6. ### Stan BrownGuest

(ahem) I venture to offer my Trig Without Tears, at

--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
A: Maybe because some people are too annoyed by top-posting.
Q: Why do I not get an answer to my question(s)?
A: Because it messes up the order in which people normally read
text.
Q: Why is top-posting such a bad thing?

Stan Brown, Oct 1, 2004
7. ### VirgilGuest

One point question on notation. May one presume that your function is
equivalent to
f(x) = 5 sin(-3 x) / (1 + cos(x))= 5*sin(-3*x) / (1 + cos(x))
?

It is customary in ASCII representations to either leave a space or use
something like "*" to indicate multiplications, to enclose the arguments
of all functions, including trig functions, in parentheses, and to
enclose divisors or denominators of more than one term or factor in
parentheses, as I have done.

Virgil, Oct 1, 2004
8. ### William ElliotGuest

= 5sin(-3x) + cos x
f'(x) = -15cos(-3x) - sin x

William Elliot, Oct 1, 2004