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.
You're close, but check your differentiation of 5sin(-3x). I don't think there's any way of simplifying this. Dave
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.
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.
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)
(ahem) I venture to offer my Trig Without Tears, at http://oakroadsystems.com/twt/ -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com 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?
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.
