Anyone know how to differentiate a trigonometric fuction?

im doing the same thing in calc too man, so maybe i can help.

you use the derivatives of the sin functions exactly as you would a normal function, so for ex, dy/dx of sin^5(x) is 5sin(x)cos(x), just like f(x)^5 is 5f(x)f'(x)

secondly, here are the derivatives of the other functions you asked for:

sin'=cos
cos'=-sin
tan'=sec^2
cot=-csc^2
csc'=-cot*csc
sec'=tan*sec

hope that helped man!

Those are all pretty standard derivatives.

d(sin(x))/dx = cos(x)dx, do you understand that notation? It basically means that the derivative of the sin(x) is the cosx multiplied by the derivative of x, which is the chain rule. So if you had sin(5x) and you need to take the derivative of it, you take the derivative of sin() function first, which is the cos() and then you use the chain rule, which means you have to take the derivative of the inner function. Thus:

d(sin(5x))/dx = cos(5x)*5 = 5cos(5x), the 5 comes from taking the derivative of the inner function 5x.

As for the other functions:

d(cos(x))/dx = -sin(x)dx
d(tan(x))/dx = sec^2(x)dx
d(cot(x))/dx = -csc^2(x)dx
d(sec(x))/dx = sec(x)tan(x)dx
d(csc(x))/dx = -csc(x)cot(x)dx

Some of those get pretty complicated to derive, and I don't really know the best way of getting into those, do you actually need to know how the derivation works or just the formulas?

my head hurts just looking at that

It's alot simpler than it looks.

Thanks, I think I can differentiate the functions using the formulas they gave us. I just have to derive the three formulas. I think I should be able to get it.

One thing that might make this a little bit less confusing, is that if you have something like, sin^5(x), you can visualize it differently than that.

sin^5(x) = (sin(x))^5, so basically like he said, you are really just taking the derivative of some function, f(x) to the 5th power, or f(x)^5, which when you derive it, is 5f(x)^4 and then you have to multiply that by the derivative of the inner function, f'(x) because of the chain rule.

f(x)^5 = 5f(x)^4 * f'(x)

or:

sin(x)^5 = 5sin(x)^4 * cos(x)

Hope that helps.

I"m on the lesson right before I learn the chain rule so I'ms sure this info will come in handy before the night's end.

yeah i would recommened getting that shit down pat mad kids in my class had trouble with the chain rule. 

I just got done with my first lesson, and now I'm starting on the chain rule. It makes no sense to me so far.