Okay, we’ve had a big argument about how do find a derivative of a calculus problem. Maybe you more mathmatically gifted dopers can help.
Here’s the problem:
Find the derivative of 3cos (y-e)+ln 3y
Thanks for helping. It’s getting ugly.
I could tell you, but then I’d have to kill you. 
Seriously, if you could say what part of it you don’t understand, or what the controversy is over, it might raise this above the level of a “Help me with my homework” question.
(P.S. Is that the derivative with respect to y, or is there some other independent variable?)
3sin(y-e)+1/y
Yes, I know this could fall under “Help me with my homework,” but we’ve actually worked this problem out and now having an argument over it, so I think it’s legit. We thought it was -3sin y+3ey RayMan, how did you come to that? It very well could be right, I’m just curious as to how you worked it.
Yeah, what’s the argument? The derivative of 3cos(y-e) is straightforward as is the derivative of ln(3y)
Wouldn’t that be **-**3sin(y-e)+(1/y)?
Actually, the main argument is about the last part, the derivative of ln 3y. There’s an argument about whether that changes to e or if it becomes something else. Unfortunately, the guy arguing the something else option isn’t real specific about what this something else would be.
Generally, the derivative of ln|x| is 1/x. Using the chain rule, when taking the derivative of a more complicated function of x, you multiply the general form by the derivative of the function replacing x in the general case.
In this case, d/dy(ln|3y|)=1/(3y)*d/dy(3y)=3/3y=1/y
Make sense?
Fuji is correct (assuming we’re not misinterpreting the OP’s attempt at typing in mathematical notation).
Key rules: the derivative of cos(something) is -sin(something) times the derivative of the something, and
the derivative of ln(something) is 1/(something) times the derivative of the something.
Alternatively, ln(3y) = ln(3) + ln(y), whose derivative is 0 + 1/y
(Since ln 3 is a constant, its derivative is 0.)
The derivative of f(g(y)) = g’(y) * f’(g(y))
So, the derivative of 3 * cos(y - e) + ln(3y) = d(y-e)/dy * 3 * -sin(y - e) + d(3y)/dy * 1/3y
= -3 sin(y - e) + 1/y
Like Fuji said.
In the second part set u = 3y the derivative of ln(u) is du/dy1/u which I get to be 3/3*y.
Okay, our guy arguing it’s something else is agreeing it’s 1/y. This sounds familiar to me, too, now. Now I guess the question is, why is this?
The derivatives of sin and cos functions are cyclic, such that, in the general cases:
d/dx(sin(x))=cos(x)
d/dx(cos(x))=-sin(x)
d/dx(-sin(x))=-cos(x)
d/dx(-cos(x))=sin(x)
et cetera.
Okay, we’ve finally figured it out, thanks to all your input. Thank you all very much.
Because ln(y) is defined to be integral(1/t, 1 < t < y) for y > 1, and integral(1/t, y < t < 1) for 0 < y < 1. Take that and the fundamental theorem of calculus, and you get that d/dy(ln(y)) = 1/y.
At the risk of injuring the dead horse further:
ln(3y) = ln(3) + ln(y) = K + ln(y)
So it’s reasonable that the derivative of ln(3y) is the same as the derivative of ln(y)