Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[Peewee]

Can anyone help me find the second derivative of

f(x) = cosx/(2+sinx)

And also its NSPI points? (or when f''(x)=0)

Not exactly overly challenging, but I've been having a bit of trouble on it.

Try u = cos(x), v = (2+sin(x)) for the first derivative.

Spoiler: if you get stuck on that (click to show/hide)

f(x)' = -((1 + 2 Sin

• )/(2 + Sin
• )^2)

Oh, and someone should probably link to the math help thread.

[Virex]

Can anyone help me find the second derivative of

f(x) = cosx/(2+sinx)

And also its NSPI points? (or when f''(x)=0)

Not exactly overly challenging, but I've been having a bit of trouble on it.

d(f(x)*g(x))/dx = f(x)*dg(x)/dx + g(x)*df(x)/dx, applied twice. That should get you on your way.

[Darvi]

This is a division, which can be simplified thusly:

(f(x)/g(x))'=(f'(x)g(x)-f(x)g'(x)/g(x)²

[Virex]

That is actually the same formula as the multiplication rule I gave, but with the division worked into it. Both will give the same answer, it'll just look different at a first glance (which is also the reason I always forget the division rule)

[Darvi]

Yeah, it's because it's jus' (f(x)h(x)^-1)'
But this is kinda easier to remember imo.

[Karlito]

So here's a trickier question:

What is the divergence of Vector?

[noodle0117]

I know the formulas and stuff, but I'm just not very good at simplifying my results.

[Cecilff2]

|x^xdx

Do it.

[Blargityblarg]

Is that pipe meant to mean an integral?

[Cecilff2]

Indeed it is.

[Virex]

|x^xdx

Do it.

Wolfram chokes on that and only gives a series expansion. I am disappointed...

[Peewee]

|x^xdx

Do it.

Let there be a function F, such that dF/dx = x^x.

Your answer is F(x). 

[Nadaka]

I have one for sufficiently vague definitions of calculus.

What is a/the algorithm capable of generating any/all valid sudoku board(s) that does not suffer from the halting problem.

I suspect that there is not one.

[Virex]

It should be possible, but the exact way to do it eludes me at the moment. The trick is to recurse from the top left corner to the bottom right corner and at each square generate a random, valid value. If no valid value is possible, you back up to the parent node in the recursion tree and try a different value at that level (gotta remember what values have been tried at each level). Since you iteratively exclude impossible solutions that way and there is a finite amount of impossible solutions (proof is left as an exercise to the reader) you should eventually hit a valid solution and thus you can't get stuck in infinite recursion.

Actually for any problem with a finite solution set and an objective way to determine the validity of the solution, if it is possible to generate all solutions without having a chance to generate a single solution an infinite amount of times, the problem can be solved by generating all possible solutions and checking their validity. This does not mean said approach is going to be viable to implement in a real-world context, but in theory it is possible.

[Nadaka]

Now I feel embarrassed. I should have thought of that.