Note that you can also find solutions where u + v + w ≠ 6, so the premise is wrong. (Example: just replace ⁶√6 with ¹²√36.) The question should be something like: “Find u, v, w with u + v + w =6 such that ...”.
Hi Prime Newtons, I gave this question a try. I did show that u+v+w=6. But my working had a slight variation. Whereas you took the negative sign into the csc(6x), from "-csc(6x)" to csc(-6x)" ( which is also ok because csc is an odd function). I took the negative sign in the square root. Just like 4^(-1/2) = (-2)th root of 4, so is 6^(-1/6) = (-6)th root of 6. Both methods prove u+v+w=6, only values of u and v change in each method. Just wanted to point this out. Appreciate your work!👍
Very nice solution. It looked slightly baffling at first...but following through the differentation steps it become clearer where it was leading. Nice trick to spot that the minus sign could go inside the csc function.
I got to the top level of high school maths in Scotland, and did 1st year maths as an under grad at uni. I never encountered cosecent, cotangent, etc, until I went on youtube. It feels like a whole part of maths was just left out.
NOTE that we can make sure that u = -6,however ,we could just find that w^(1/v) = 6^(1/6).for example there is a solution that w=36 and v = 12. in this condition u+v+w = 42
Basically, you can’t declare, that u=-6, v=w=6. You just have that f’(x,u,v,w) = f’(x). One equation for 3 variables. For example, the equation A*sin(x)= B*sin(y) doesn’t guarantee, that A=B and x=y is the only solution.
a) u + v + w = - 6 + 6m + 6ᵐ // u = -6, v = 6m, w = 6ᵐ b) u + v + w = 6 + 6m + (1/6)ᵐ // u = 6, v = 6m, w = (1/6)ᵐ 𝒇'(x) = csc(-6x)/𝒍𝒏(⁶√6) = csc(-6x)/𝒍𝒏(⁶ᵐ√(6ᵐ) = = csc(+6x)/𝒍𝒏(⁶√(1/6)) = csc(6x)/𝒍𝒏(⁶ᵐ√((1/6)ᵐ), m∈ℕ
Love how excited you get when you solve a problem. Awesome.
Note that you can also find solutions where u + v + w ≠ 6, so the premise is wrong. (Example: just replace ⁶√6 with ¹²√36.)
The question should be something like: “Find u, v, w with u + v + w =6 such that ...”.
I agree.
One of my favorite videos of yours; the excitement building as you approach the solution and the flash when it becomes clear 10/10 would watch again
Very nice problem and nice presentation. This is one of your best videos. Thank you!
Hi Prime Newtons, I gave this question a try. I did show that u+v+w=6. But my working had a slight variation. Whereas you took the negative sign into the csc(6x), from "-csc(6x)" to csc(-6x)" ( which is also ok because csc is an odd function). I took the negative sign in the square root. Just like 4^(-1/2) = (-2)th root of 4, so is 6^(-1/6) = (-6)th root of 6. Both methods prove u+v+w=6, only values of u and v change in each method. Just wanted to point this out. Appreciate your work!👍
lol i did that too
I also did the same hehe
You are a great mathematician .Keep it up !!!
Thanks for your great content! Keep it up. Like your very pedagogic explanations @ perfect speed!
AWEEEESOMEEEE 🎉🎉🎉🎉🎉🎉
Very nice solution. It looked slightly baffling at first...but following through the differentation steps it become clearer where it was leading. Nice trick to spot that the minus sign could go inside the csc function.
Excellent explanation Sir. Thanks 👍
I got to the top level of high school maths in Scotland, and did 1st year maths as an under grad at uni. I never encountered cosecent, cotangent, etc, until I went on youtube. It feels like a whole part of maths was just left out.
Definitely required the biblical pasy after getting 666 at the end of the solution!
Jokes aside, love the vids and your enthusiasm!
NOTE that we can make sure that u = -6,however ,we could just find that w^(1/v) = 6^(1/6).for example there is a solution that w=36 and v = 12. in this condition u+v+w = 42
Thanks man. Love your vids, and that you use a blackboard like when I was in school!
Amazing stuff
It was beautiful the derivative can be simplified very easily
What a amazing exercise!!!!
Basically, you can’t declare, that u=-6, v=w=6.
You just have that f’(x,u,v,w) = f’(x). One equation for 3 variables.
For example, the equation A*sin(x)= B*sin(y) doesn’t guarantee, that A=B and x=y is the only solution.
Easy problem. As always, you did a great job in solving it!
Looked pretty weird to me at first, but I got it.
f(x) = ln cot(3x)/ln6
f'(x) = 1/cot(3x) * -csc^2(3x) * 3 / ln6
= -1/sin^2(3x) * sin(3x)/cos(3x) * 3/ln6
= -3/ln6 * 1/sin(3x)cos(3x)
= -3/ln6 * 1/(1/2)sin(6x)
= csc(6x)/ (-1/6*ln6)
= csc(6x) / (ln6^(-1/6))
= csc(6x)/ -6th root(ln6)
6 - 6 + 6 = 6
Not sure how you properly rewrite negative fractional powers, but that seems about right
So satsfine❤❤
9:19 Ohhhhh!!!!! You saw what I just saw!! 😎
You win the race 😮
Csc(x)/(ln(1/6)^(1/6)) if put negative on bottom instead of top sums to 4/3
Added wrong 43/6
@@robertcotton8481 But it's 6x instead of x inside csc.
My point was that with the negative being able to go both up and down there is not just one sum
@@robertcotton8481 I know, when I did the sum I got 73/6 because I put the -ve sign inside the ln instead of csc.
Hi sir ❤❤
First
Do you have any online courses?
The problem can have more than 1 solution it is not necessarily the case that u and v and w have to be just 6 or -6.
The problem just asks to show the sum of these three numbers to be 6. It doesn’t matter the value of the numbers just the sum
@@ppbuttocks2015 understood my point is the total sum changes, it is not the case that it’s always 6 that’s my point.
Check check check!!!! Love that.
beautiful
Can i get the source forr this question?
I got 73/6. u=6, v=6, w=1/6. Instead of putting the negative sign inside cosec(6x), I put it inside the ln function.
why is w omega
What will be the inverse function of f(x) = xe^x?? Someone slove please🙏🏻🙏🏻🙏🏻.
Easy !!!
AMEN TO THAT
noice
666 and so this problem has summoned the demon here
god forbids... this equation is fomented by the devil.
Love the math, could do without the evangelism.
Thanks. I wish I could.
a) u + v + w = - 6 + 6m + 6ᵐ // u = -6, v = 6m, w = 6ᵐ
b) u + v + w = 6 + 6m + (1/6)ᵐ // u = 6, v = 6m, w = (1/6)ᵐ
𝒇'(x) = csc(-6x)/𝒍𝒏(⁶√6) = csc(-6x)/𝒍𝒏(⁶ᵐ√(6ᵐ) =
= csc(+6x)/𝒍𝒏(⁶√(1/6)) = csc(6x)/𝒍𝒏(⁶ᵐ√((1/6)ᵐ), m∈ℕ