# Mathmatical Argument. **URL:** \ **Category:** Factual Questions\ **Created:** [October 18, 2004, 12:45am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741 "2004-10-18T00:45:24Z")\ **Posts on this page:** 16\ **Page:** 1
**Author:** ![JimSox5](https://avatars.discourse-cdn.com/v4/letter/j/b5a626/32.png) [@JimSox5](https://boards.straightdope.com/u/JimSox5)\ **Post date:** [October 18, 2004, 12:45am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/1 "2004-10-18T00:45:24Z")
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. ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [October 18, 2004, 12:52am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/2 "2004-10-18T00:52:43Z")
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?) ---
**Author:** ![RayMan](https://avatars.discourse-cdn.com/v4/letter/r/e495f1/32.png) [@RayMan](https://boards.straightdope.com/u/RayMan)\ **Post date:** [October 18, 2004, 12:54am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/3 "2004-10-18T00:54:40Z")
3sin(y-e)+1/y ---
**Author:** ![JimSox5](https://avatars.discourse-cdn.com/v4/letter/j/b5a626/32.png) [@JimSox5](https://boards.straightdope.com/u/JimSox5)\ **Post date:** [October 18, 2004, 1:00am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/4 "2004-10-18T01:00:20Z")
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. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [October 18, 2004, 1:02am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/5 "2004-10-18T01:02:19Z")
Yeah, what’s the argument? The derivative of 3cos(y-e) is straightforward as is the derivative of ln(3y) ---
**Author:** ![Fuji\_Kitakyusho](https://avatars.discourse-cdn.com/v4/letter/f/f0a364/32.png) [@Fuji\_Kitakyusho](https://boards.straightdope.com/u/Fuji_Kitakyusho)\ **Post date:** [October 18, 2004, 1:06am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/6 "2004-10-18T01:06:15Z")
> [@RayMan](#): > > 3sin(y-e)+1/y Wouldn’t that be \*\*-\*\*3sin(y-e)+(1/y)? ---
**Author:** ![JimSox5](https://avatars.discourse-cdn.com/v4/letter/j/b5a626/32.png) [@JimSox5](https://boards.straightdope.com/u/JimSox5)\ **Post date:** [October 18, 2004, 1:09am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/7 "2004-10-18T01:09:12Z")
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. ---
**Author:** ![Fuji\_Kitakyusho](https://avatars.discourse-cdn.com/v4/letter/f/f0a364/32.png) [@Fuji\_Kitakyusho](https://boards.straightdope.com/u/Fuji_Kitakyusho)\ **Post date:** [October 18, 2004, 1:13am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/8 "2004-10-18T01:13:29Z")
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? ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [October 18, 2004, 1:13am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/9 "2004-10-18T01:13:30Z")
**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.) ---
**Author:** ![Punoqllads](https://avatars.discourse-cdn.com/v4/letter/p/d2c977/32.png) [@Punoqllads](https://boards.straightdope.com/u/Punoqllads)\ **Post date:** [October 18, 2004, 1:13am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/10 "2004-10-18T01:13:39Z")
> [@JimSox5](#): > > Find the derivative of 3cos (y-e)+ln 3y 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. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [October 18, 2004, 1:14am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/11 "2004-10-18T01:14:18Z")
In the second part set u = 3_y the derivative of ln(u) is du/dy_1/u which I get to be 3/3\*y. ---
**Author:** ![JimSox5](https://avatars.discourse-cdn.com/v4/letter/j/b5a626/32.png) [@JimSox5](https://boards.straightdope.com/u/JimSox5)\ **Post date:** [October 18, 2004, 1:14am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/12 "2004-10-18T01:14:33Z")
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? ---
**Author:** ![Fuji\_Kitakyusho](https://avatars.discourse-cdn.com/v4/letter/f/f0a364/32.png) [@Fuji\_Kitakyusho](https://boards.straightdope.com/u/Fuji_Kitakyusho)\ **Post date:** [October 18, 2004, 1:19am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/13 "2004-10-18T01:19:55Z")
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. ---
**Author:** ![JimSox5](https://avatars.discourse-cdn.com/v4/letter/j/b5a626/32.png) [@JimSox5](https://boards.straightdope.com/u/JimSox5)\ **Post date:** [October 18, 2004, 1:34am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/14 "2004-10-18T01:34:02Z")
Okay, we’ve finally figured it out, thanks to all your input. Thank you all very much. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [October 18, 2004, 2:03am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/15 "2004-10-18T02:03:51Z")
> [@JimSox5](#): > > 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? 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. ---
**Author:** ![Punoqllads](https://avatars.discourse-cdn.com/v4/letter/p/d2c977/32.png) [@Punoqllads](https://boards.straightdope.com/u/Punoqllads)\ **Post date:** [October 18, 2004, 2:05am UTC](https://boards.straightdope.com/t/mathmatical-argument/269741/16 "2004-10-18T02:05:45Z")
> [@JimSox5](#): > > 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? 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)