This is really nice. For anyone who was wondering about using trig identities to find the same result: Let A = arctan(x) Let H = tan((1/2) A) Note x = tan(A) From tan half-angle identity H= (sqrt(1+x^2) - 1)/ x From tan subtraction tan((1/2) A - pi/4) = (H - tan(pi/4)) / (1 + H * tan(pi/4)) = (H - 1) / ( H + 1) Note H*x = sqrt(1+x^2) - 1, so (H-1)/(H+1) = (Hx-x) / (Hx+x) = (sqrt(1+x^2) - 1 - x)/(sqrt(1+x^2) - 1 + x) Multiply top and bottom by sqrt(1+x^2) + 1 - x = (1+x^2 + (1+x)^2 - 2sqrt(1+x^2)(1+x) ) / (1+x^2 - (1-x)^2) = (1 + x^2 - 2x (sqrt(1+x^2) - (1-x)(1+x)) / (2x) = (2x^2 - 2x(sqrt(1+x^2))) / 2x = x - sqrt(1+x^2) tan((1/2) arctan(x) - pi/4) = x - sqrt(1+x^2) So arctan(x-sqrt(1+x^2)) = (1/2) arctan(x) - pi/4.
Brilliant reasoning! Makes me remember the good old days of studying calculus at my university for the mere fun of it. In reality they were mostly bad old days of course, but studying math was a bright exception, making them more enjoyable. I could definitely enjoy calculus again, real calculus, complex calculus, matrix calculus (multivariate calculus), functional calculus (differentiating functions taking functions themselves of a function space as arguments, not merely parameters being functions of some variables), p-adic calculus with a different non-archimedean metric, and characteristic p calculus in a formal variable with its own metric (which almost necessitates hyperderivatives in order to avoid losing higher order derivatives to the zero term introduced by d/dx(x^(p*n)) = 0 because p*1 = 0 in char p).
Eres el único que conozco por estos lados que re busca estos ejercicios (y sirve para ENTENDER los principios...)! Bien ahí! Jaja! BlackPenRedPen yeaaah💪💪💪
Put the false chain rule example on a shirt! Or anything like that, false methods leading to correct answers are great ;) the simplest being fraction reductions 64/16 -> cancel out the 6's -> 4/1 -> 4 which is the correct answer
Wow, What's really interesting about this result is that let's say we replace x by some other f(x), then if we are required to work the LHS then instead we can work the RHS because its comparatively easier.
First function is ½arctan(x)-π/4. In arctan(x-√(x²+1)) as the range of the tangent function is R, we can substitute x=tan(u). Then we get -arctan(sec(u)-tan(u)) which is -arctan( {1-sin(u)}/cos(u)). 1-sin(u) = {cos(u/2)-sin(u/2)}² and cos(u)= cos²(u/2)-sin²(u/2). So one cos-sin term cancels and if we then divide the numerator and denominator by cos(u/2) we get -arctan({1-tan(u/2)}/{1+tan(u/2)}). 1= tan(π/4) so we have {tan(π/4)-tan(u/2)}/{1+tan(π/4)tan(u/2)}= tan(π/4-u/2) whose negative arctanget we want which is simply u/2-π/4 or arctan(x)/2-π/4.
There is also an algebraic way to prove this. First, let arctan(x-√(x²-1))=y, x-√(x²-1)=tan(y), [solve for x ], x=(tan²y -1)/(2tan y) = [usual trig. ]=-cot (2y) =[inverse reduction]=tg(2y+π/2); arctan x = 2y + π/2, y=1/2*arctan x - π/4 .
So I went for it using good ol' calc 3. For me, the most natural thing upon seeing tan⁻¹ was applying tan to both sides to cancel it out: tan(y)=x-sqrt(x²+1) F(x,y)=tan(y)-(x-sqrt(x²+1))=0 Applying the total derivative operator, it follows dF(x,y)=(1+tan²(y))dy-(1-x/sqrt(x²+1))dx=0 (1) Let's focus know on the functions: by the power invested by HS algebra, we can prove tan(y)=x-sqrt(x²+1) -> 1+tan²(y)=1+[x-sqrt(x²+1)]²=2x²+2x*sqrt(x²+1)+2 1-x/sqrt(x²+1)=[2x²+2x*sqrt(x²+1)+2]/(x²+1) Using this information on expression (1) we get [2x²+2x*sqrt(x²+1)+2]dy=[2x²+2x*sqrt(x²+1)+2]/(x²+1)dx It follows that dy/dx=1/2(x²+1) but this is the function we were looking for, hence concluding the answer. PS: I watched the video to double-check not gonna lie
If two real functions have the same derivative on an interval, they differ by a constant on that interval (not everywhere). Let F(x) = Arctan(x), for all real x, and G(x) = - Arctan(1/x), for all non-zero, real x. Then, F'(x) = G'(x) = 1/(x^2 + 1), for all non-zero, real x. However, F(x) = G(x) + pi/2, for x > 0, and F(x) = G(x) - pi/2, for x < 0.
Well, off by a constant, but yeah. d(f(x)+a)/dx = g(x) = d(f(x)+b)/dx Integral of g(x)=f(x) + c Derivative of f(x)+h(x)+a, for some h(x) not a constant, will be g(x)+h'(x), and h'(x) is by premise nonzero, therefore identical derivatives imply a difference of only a constant term which is zero for identical functions
When you find C, how do you know that it is -pi/4 ? Given that tan^-1(-1) = -pi/4 + 2pi*n , shouldn't we pick another nice number to find n ? (Or another technique, given that there will always be a lurking 2*pi*n). Or, when we write the inverse trigonometric functions, so we assume the "first" result (n = 0) ?
Sorry for late answer 😝 If you plot both funtions, the two funtions would be excactly the same except the graph one is moved upwards compared to the other
Why -pi/4? Inverse tangent is defined in all quadrants. The C could of been 3pi/4 or 7pi/4. It isn’t common to use negative reference angles (even though it is still correct). So my question is why can’t the C be the other 2 values?
Gaussian Entity oh I see, and this can also be because the angle doesn’t just oscillate like it would in a trig function (such as 7pi/4 is the same as -pi/4 inside any normal trig function) but instead this angle is the actual “y” dependent value, therefore we can’t use the unit circle logic of clock wise/counter clockwise. Would you also say this is correct?
Give us a (tricky) problem with multiple questions. Ask us to email you the answers. Short out top x solvers who would get the t-shirts. You can also keep the contest to have multi stages, not all involving problem questions necessarily. Some stages could include some casual fun tasks related to maths and your channel specifically. Idea 2- Give us the task of making a video explaining some maths and then organize a peer review stage in which each contestant would rate each other(not himself/herself). And finally you decide the winner based or not based on that rating metric.
this seems like u sub-ing should help but i'm not sure how. trying to figure out how to show that they're off by pi/4 with just algebra. x = sinh(u) tan^-1 ( sinh(u) - sqrt(sinh^2(u) + 1)) = tan^-1 (sinh(u) - sqrt(cosh^2(u))) = tan^-1 (sinh(u) - cosh(u)) not sure if you can simplify it more from here (sinhx - coshx = -e^-x but can't think of anything to do here) cos(x) = sin(x + pi/2) which seem like it should be relevant somehow, where you can get a pi/4 instead of pi/2 with the hyperbolic versions maybe? or possibly a double/ half angle formula situation? alternatively: x = tan(u) tan^-1 ( tan(u) - sqrt(tan^2(u) + 1)) = tan^-1 (tan(u) - sqrt(sec^2(u))) = tan^-1 (tan(u) - sec(u)) which seems even closer.
(outline only, but this works) Write z = sqrt(1 + x^2) - x. The target expression is arctan(-z) = arctan(x)/2 - pi/4; because tan is odd it's the same as arctan(z) = pi/4 - arctan(x)/2. So that's what we're aiming for here. It can be proven using trig functions & geometry. Note that sqrt(1 + x^2) is the hypotenuse of a right triangle with legs x and 1. If the angle opposite x is called t, we have tan(t) = x. Let A be the vertex associated with t, B be the vertex on the other end of the hypotenuse, and C be the right angle. Then build a triangle on BC, with a third vertex D on the hypotenuse at a distance x from B. Then BCD is isoceles (two sides equal to x), and the length of AD is equal to z. Some angle math gives us that the angle ACD = pi/4 - t/2. So we are looking for tan(ACD). Use law of sines and law of cosines on triangles ACD and BCD to get expressions for sin(ACD) and cos(ACD) in terms of sin(t) and x. We can write sint(t) in terms of x also. Do that and simplify to get tan(ACD) = z, and take arctan of both sides to reach the goal. There is probably a more direct construction to do this though (I started by constructing a line of length z but it turned out I needed to expand z to be able to simplify, so it kind of just made it take longer)
Same Derivatives Implies Same Functions ? Yes, there is the problem of the constant but there is also the problem of the domain of definition. It must be defined on the same domain: ln (x)! = ln (| x |) or x * x / x! = identity or any restricted functions it is nevertheless the case for the functions that you give on C \ {i; -i} but it was necessary to say it.
well, you could give a mathematical problem, and whoever get's it first is the lucky one. but that might be not so fair, since some people simply watch your vids earlier than others
I thought this was a video with a super hard derivative challenge, and the first person to solve the problem wins a shirt. You should do something like that soon, it would be very interesting!
For the giveaway, you should have everyone who wants to participate leave a comment with a joke or math pun and whatever joke makes you laugh the most, that person wins :) (crumby idea but seems fun) #YAY
This is really nice. For anyone who was wondering about using trig identities to find the same result:
Let A = arctan(x)
Let H = tan((1/2) A)
Note x = tan(A)
From tan half-angle identity
H= (sqrt(1+x^2) - 1)/ x
From tan subtraction
tan((1/2) A - pi/4) = (H - tan(pi/4)) / (1 + H * tan(pi/4)) = (H - 1) / ( H + 1)
Note H*x = sqrt(1+x^2) - 1, so
(H-1)/(H+1) = (Hx-x) / (Hx+x)
= (sqrt(1+x^2) - 1 - x)/(sqrt(1+x^2) - 1 + x)
Multiply top and bottom by sqrt(1+x^2) + 1 - x
= (1+x^2 + (1+x)^2 - 2sqrt(1+x^2)(1+x) ) / (1+x^2 - (1-x)^2)
= (1 + x^2 - 2x (sqrt(1+x^2) - (1-x)(1+x)) / (2x)
= (2x^2 - 2x(sqrt(1+x^2))) / 2x
= x - sqrt(1+x^2)
tan((1/2) arctan(x) - pi/4) = x - sqrt(1+x^2)
So arctan(x-sqrt(1+x^2)) = (1/2) arctan(x) - pi/4.
This deserves more apricciacion (idk how to spell that)
@@איתןגרינזייד "appreciation"
I can hear my calc teacher shouting "DONT FORGET THE PLUS SEA!!!" Never did find out why she was so obsessed with the ocean tho.
LOL!
This also works for d/dx( ln(ax)) =1/x for any a>0. This can be easily proven by properties of logarithms or by just taking the derivative
Brilliant reasoning! Makes me remember the good old days of studying calculus at my university for the mere fun of it. In reality they were mostly bad old days of course, but studying math was a bright exception, making them more enjoyable.
I could definitely enjoy calculus again, real calculus, complex calculus, matrix calculus (multivariate calculus), functional calculus (differentiating functions taking functions themselves of a function space as arguments, not merely parameters being functions of some variables), p-adic calculus with a different non-archimedean metric, and characteristic p calculus in a formal variable with its own metric (which almost necessitates hyperderivatives in order to avoid losing higher order derivatives to the zero term introduced by d/dx(x^(p*n)) = 0 because p*1 = 0 in char p).
Eres el único que conozco por estos lados que re busca estos ejercicios (y sirve para ENTENDER los principios...)! Bien ahí! Jaja! BlackPenRedPen yeaaah💪💪💪
As an idea
Just make the t shirt the working out for the integral of the cube root of tan x
#YAY : )
Yay
instaBlaster.
Put the false chain rule example on a shirt! Or anything like that, false methods leading to correct answers are great ;)
the simplest being fraction reductions
64/16 -> cancel out the 6's -> 4/1 -> 4 which is the correct answer
Wow, What's really interesting about this result is that let's say we replace x by some other f(x), then if we are required to work the LHS then instead we can work the RHS because its comparatively easier.
First function is ½arctan(x)-π/4. In arctan(x-√(x²+1)) as the range of the tangent function is R, we can substitute x=tan(u). Then we get -arctan(sec(u)-tan(u)) which is -arctan( {1-sin(u)}/cos(u)). 1-sin(u) = {cos(u/2)-sin(u/2)}² and cos(u)= cos²(u/2)-sin²(u/2). So one cos-sin term cancels and if we then divide the numerator and denominator by cos(u/2) we get -arctan({1-tan(u/2)}/{1+tan(u/2)}). 1= tan(π/4) so we have {tan(π/4)-tan(u/2)}/{1+tan(π/4)tan(u/2)}= tan(π/4-u/2) whose negative arctanget we want which is simply u/2-π/4 or arctan(x)/2-π/4.
My new favorite math video!
This is absolutely wonderful
There is also an algebraic way to prove this. First, let arctan(x-√(x²-1))=y, x-√(x²-1)=tan(y), [solve for x ],
x=(tan²y -1)/(2tan y) = [usual trig. ]=-cot (2y) =[inverse reduction]=tg(2y+π/2); arctan x = 2y + π/2,
y=1/2*arctan x - π/4 .
Inside the arc tan is RHS of fiirst Euler substitution which rationalizes hyperbola
I love his t-shirt.
This is my favorite video so far! I love seeing trig proofs using calculus! #YAY
So I went for it using good ol' calc 3.
For me, the most natural thing upon seeing tan⁻¹ was applying tan to both sides to cancel it out:
tan(y)=x-sqrt(x²+1) F(x,y)=tan(y)-(x-sqrt(x²+1))=0
Applying the total derivative operator, it follows
dF(x,y)=(1+tan²(y))dy-(1-x/sqrt(x²+1))dx=0 (1)
Let's focus know on the functions: by the power invested by HS algebra, we can prove
tan(y)=x-sqrt(x²+1) -> 1+tan²(y)=1+[x-sqrt(x²+1)]²=2x²+2x*sqrt(x²+1)+2
1-x/sqrt(x²+1)=[2x²+2x*sqrt(x²+1)+2]/(x²+1)
Using this information on expression (1) we get
[2x²+2x*sqrt(x²+1)+2]dy=[2x²+2x*sqrt(x²+1)+2]/(x²+1)dx
It follows that
dy/dx=1/2(x²+1)
but this is the function we were looking for, hence concluding the answer.
PS: I watched the video to double-check not gonna lie
The same is very similar if you take the positive square root of x^2+1. You just add pi/4 instead of subtracting it
i LOVE THIS SHIRT SO MUCH!!!!!!!!
I am Mohammed from Algeria and I am not a terrorist. I love your classes very much
If two real functions have the same derivative on an interval, they differ by a constant on that interval (not everywhere). Let F(x) = Arctan(x), for all real x, and G(x) = - Arctan(1/x), for all non-zero, real x. Then, F'(x) = G'(x) = 1/(x^2 + 1), for all non-zero, real x. However, F(x) = G(x) + pi/2, for x > 0, and F(x) = G(x) - pi/2, for x < 0.
Well, off by a constant, but yeah.
d(f(x)+a)/dx = g(x) = d(f(x)+b)/dx
Integral of g(x)=f(x) + c
Derivative of f(x)+h(x)+a, for some h(x) not a constant, will be g(x)+h'(x), and h'(x) is by premise nonzero, therefore identical derivatives imply a difference of only a constant term which is zero for identical functions
when you said you forgot the derivative of inverse tan, and started to turn around, i thought it was gonna be a plug for the shirt! #YAY
You are my favorite youtuber! #YAY
Giving away t-shirts is a derivative idea, but integral to growing your channel.
When you find C, how do you know that it is -pi/4 ? Given that tan^-1(-1) = -pi/4 + 2pi*n , shouldn't we pick another nice number to find n ? (Or another technique, given that there will always be a lurking 2*pi*n). Or, when we write the inverse trigonometric functions, so we assume the "first" result (n = 0) ?
inverse tangent is not periodic, it only has one value everywhere
nathanisbored Oops, sorry ^^'
The case if function a) on cartesian and b) in polar coordinates. Even "same derivative", functions different in nature.
Principal from Baldi
"No skipping maths class in the hall. Derivative for you"
#YAY I was here before this channel had 1 million subs
What would be a geometric (i.e. visual) interpretation of the difference of those functions being pi/4?
Sorry for late answer 😝
If you plot both funtions, the two funtions would be excactly the same except the graph one is moved upwards compared to the other
@@caetanogarelli6657 Not sure why I didn't realize that. Wondering if I was trying to ask something else, but signs are pointing to no...
1:03 WERE YOU trying to show your shirt?
I love this T-shirt
Integrals are more fun than derivatives though why don't you make an integral shirt :(
maybe one with a beatiful resolution, such as integral from 0 to inf of cos(x)/(1+x^2)
Felipe Lorenzzon don't u mean that integral from -inf to inf?
Eliot Arguello
Idk, but I remember flammable maths made a video on that. "A beautiful result in calculus" the answer is pi/e
Me too it was from -inf to inf and I love that answer too! ❤️
but is so painful to work with cuz u need to choose value for u
Cool but i recognize the inside expression its the same as cosh^-1(x) if ww turned it to the ln version is there any cool connection we can deduce
Another cool derivatives
Why -pi/4? Inverse tangent is defined in all quadrants. The C could of been 3pi/4 or 7pi/4. It isn’t common to use negative reference angles (even though it is still correct). So my question is why can’t the C be the other 2 values?
The domain is all reals but the range is between -pi/2 to pi/2. Inverse tangent will never allow you to obtain 3pi/4 or 7pi/4.
Gaussian Entity oh I see, and this can also be because the angle doesn’t just oscillate like it would in a trig function (such as 7pi/4 is the same as -pi/4 inside any normal trig function) but instead this angle is the actual “y” dependent value, therefore we can’t use the unit circle logic of clock wise/counter clockwise. Would you also say this is correct?
@@tabatuby Yes
Fabulous!!
There are much simpler examples that answer the title question. The simplest version is the functions f(x) = x and g(x) = x + 1.
I believe the integral of sin(x)cos(x) has 3 valid solutions as well all of which are off by a constant.
Actually there are infinitely many solutions to any (valid) integral that are all off by a constant :P
So do they belong to a special family of functions ?
Like family of lines.
why must we find the +c part?
We can also write x as tan a and we get the same result.
THANK YOU FORM THE SUDAN
is there any non-calculus way to prove the identity at the end?
you should start a merch store
#yay
y: you
a: get
y: your derivatives right
i want that tshirt!!! from italy tho
They can't be the same function because ½arctan(x)>0 for x>0, while arctan(x-√(1+x²))0. But they could be off by a constant!
What is d2y/dx2 when x=t-Sin(t) and y=t+Cos(t) ?
You need use a graphic of these to curves side by side.
I don't get why tan to the -1 power is arctan, but not ctg
d/dy(x²)=2x
d/dy(x²+1)=2x
x²=/=x²+1
done
Give us a (tricky) problem with multiple questions. Ask us to email you the answers. Short out top x solvers who would get the t-shirts.
You can also keep the contest to have multi stages, not all involving problem questions necessarily. Some stages could include some casual fun tasks related to maths and your channel specifically.
Idea 2-
Give us the task of making a video explaining some maths and then organize a peer review stage in which each contestant would rate each other(not himself/herself). And finally you decide the winner based or not based on that rating metric.
Thanks you for your videos ! You are amazing ! #YAY
3:46 You could have just multiplied and divided (sqrt(x^2+1)). Would give the answer in the next step. noice video anyway.
Will you please make this shirt available on ebay or any online shopping website? It will be bought by a lot of people in a short period of time
I wanna that the t-shirt say #yay
Is my idea
this seems like u sub-ing should help but i'm not sure how. trying to figure out how to show that they're off by pi/4 with just algebra.
x = sinh(u)
tan^-1 ( sinh(u) - sqrt(sinh^2(u) + 1)) = tan^-1 (sinh(u) - sqrt(cosh^2(u))) = tan^-1 (sinh(u) - cosh(u))
not sure if you can simplify it more from here (sinhx - coshx = -e^-x but can't think of anything to do here)
cos(x) = sin(x + pi/2) which seem like it should be relevant somehow, where you can get a pi/4 instead of pi/2 with the hyperbolic versions maybe? or possibly a double/ half angle formula situation?
alternatively:
x = tan(u)
tan^-1 ( tan(u) - sqrt(tan^2(u) + 1)) = tan^-1 (tan(u) - sqrt(sec^2(u))) = tan^-1 (tan(u) - sec(u))
which seems even closer.
They look super cool!!! I want one.....#YAY
6:27 when i sit for half hour in jee advanced
Keep posting man! lOVE THE VIDEOS
This is so coool
I like how he uses minus as a verb :D
What is the derivative of -2tan^-1(sqr((1-x)/(1+x))?
A derivative? For ME? Oh, you shouldn't have!
LOL
I know I'm late but
f(x)= x+1
g(x)= x
They have the same derivative, but ofc they are different.
How can I send you a nice integral + the solution?
Would it be considered cheating to wear that to a first year calc exam? :)
You said that's just one way to prove it.
Can you prove it without differentiation?
(outline only, but this works)
Write z = sqrt(1 + x^2) - x. The target expression is arctan(-z) = arctan(x)/2 - pi/4; because tan is odd it's the same as arctan(z) = pi/4 - arctan(x)/2. So that's what we're aiming for here.
It can be proven using trig functions & geometry. Note that sqrt(1 + x^2) is the hypotenuse of a right triangle with legs x and 1. If the angle opposite x is called t, we have tan(t) = x. Let A be the vertex associated with t, B be the vertex on the other end of the hypotenuse, and C be the right angle.
Then build a triangle on BC, with a third vertex D on the hypotenuse at a distance x from B. Then BCD is isoceles (two sides equal to x), and the length of AD is equal to z.
Some angle math gives us that the angle ACD = pi/4 - t/2. So we are looking for tan(ACD).
Use law of sines and law of cosines on triangles ACD and BCD to get expressions for sin(ACD) and cos(ACD) in terms of sin(t) and x. We can write sint(t) in terms of x also. Do that and simplify to get tan(ACD) = z, and take arctan of both sides to reach the goal.
There is probably a more direct construction to do this though (I started by constructing a line of length z but it turned out I needed to expand z to be able to simplify, so it kind of just made it take longer)
if arctan is tan^-1, how how u write cotan?? isnt cotan 1/tan = tan^-1?????
+blackpenredpen EACH OF US CAN MAKE VIDEO ABOUT MATH AND YOU SELECT SOME GOOD ONES! #YAY #YAY #YAY #YAY
@ 3:55 u cud have saved urself some time by multiplying and dividing by just √(1+x^2), instead of the conjugate. Anyways, great vid!
Same Derivatives Implies Same Functions ?
Yes, there is the problem of the constant but there is also the problem of the domain of definition. It must be defined on the same domain:
ln (x)! = ln (| x |) or x * x / x! = identity or any restricted functions
it is nevertheless the case for the functions that you give on C \ {i; -i} but it was necessary to say it.
2:07
3:00
*stops joking, back to serious business* 😂
Good job 😊 #YAY
If you give any of these shirts to one of your students, does that make them Student's-t shirts?
Excellente démonstration! P.-S. J'aime beaucoup ton t-shirt aussi!
Thank you : )
well, you could give a mathematical problem, and whoever get's it first is the lucky one. but that might be not so fair, since some people simply watch your vids earlier than others
AndDiracisHisProphet exactly...
I think I will do a "bprp parody", suggested by a viewer
can you parody yourself?
AndDiracisHisProphet maybe lol
if I was mean....
It's okay. I just (shamelessly) gave myself a heart.
在哪可以买到这个T恤啊
F dd 目前我還沒有要賣 我只想要做個giveaway先
还以为是在外面买的
All over My head
I thought this was a video with a super hard derivative challenge, and the first person to solve the problem wins a shirt. You should do something like that soon, it would be very interesting!
Nature loves symmetry , which it expels in the for of maths
For the giveaway, you should have everyone who wants to participate leave a comment with a joke or math pun and whatever joke makes you laugh the most, that person wins :) (crumby idea but seems fun) #YAY
+higher mathematics
Lol, we both have a small maths channel and we both watch bprp, hahaha
Derive and subscribe! #YAY If I don't win a shirt, would I be able to buy one? It's so awesome!
Do a math competition and lets say the 5 people who will email you the correct answer first win a derivative t-shirt
Awesome identity!!
#Yay do some PDEs!!!!
What math problem would you like ME to solve? #YAY #Yay #yAy #yaY #YAy #yAY #YaY #yay
Moral of the story: the hardest part of calculus is algebra :)
#yay !!!!
f(x) = 1/(2x^2 + 2) and g(x) = -x^2/(2x^2 + 2) brought me here.
i need that shirt 👌 #YAY
: )
deriv
ativ
You should do a streaming and give it randomly :p #YIAY :v (#YAY. 😻)
#YAY for my favorite YTber!
if we find the Derivative of x and x+1, we can draw the same conclusion
I don't care if I win the shirt, but I'd at least want to buy that shirt you have on where can I get one?
I am only planning a giveaway for now tho, sorry...
Then can the shirt you have on be the one for the give away it's so cool!!!
So beautiful hahaha
#YAY!!!
f(x) = ln(x) and f(x) = ln(2x) have the same derivative.
8:07
- What's this? ÓwÓ
That t-shirt!
: )
won’t be able to wear that shirt during any math exam that’s for sure haha