@@gtc4189 I'm 14 in an Italian school and they have just started doing equations with algebraic fractions. I like to always study things before others so I have a better time understanding them when they're eventually taught. I've just finished my trigonometry book and before this comment, I had seen a video explaining what derivatives are, so I had wondered what the derivative of different functions were. I had wondered if the derivative of sin(x) was cos(x) ever since I first thought about it
I am kind of sad that you chose to present this fact this way, because the derivation of that fact can be seen even nicer in a visual form, by looking at a point P on the unit circle with angle theta from the x-axis and moving it a tiny step Δtheta to obtain P'. Draw a vertical line through P and a horizontal line through P' which intersect at Q to obtain a right triangle PQP' such that angle QPP' ≈ theta. As |PP'| ≈ Δtheta, we obtain that sin'(theta) ≈ |PQ|/|PP'| ≈ cos(theta). This proof is not rigorous but can easily be made into a rigorous argument.
Yes. I like that one. I’ll try that one in long format. This one is part of a series about graphing derivatives as you walk along the curve. This is more for my students as we think about increasing/decreasing behavior and how the derivative captures that. The shorts format is still tough for me to say something very interesting.
@@MathVisualProofs Yes, I understand! It's hard to convey nice insights in less than a minute. But having a teacher like you putting so much effort into these visualizations must be awesome :D
i found this out 2 days ago when i was messing around in desmos trying to find the tangent line of a sine function. i knew the graph would be a sine wave of some sort, and i found out that the gradient of the tangent line of sin x = sin x-π/2, which is equal to cos x
Thnx for this! And have you considered make a "Essence of conic sections geometrically"? There are many old books on internet for free access like geometry of conic sections please have a look at them
Anyway, to anyone who wants to try this, you can use y=f'(t)(x-t)+f(t), and initialise f(x) as whatever differentiable function you want, and t as a parameter. This function will generate a tangent line on the curve at point (t,f(t)). This works on desmos btw. If you can't write f'(t), try d/dt(f(t)) instead.
theorems you could solve in a minute: for any function that is not e to the x, f'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''(x) is equal to 0. thank you.
Here’s the idea/concept : ua-cam.com/users/shortseE4IGCAzmqA?si=mljokGicigflQKgE . I’m working on an animation of the sine derivative slope another way.
theres a cool proof or just this using a unit circle, if you have a vector from the center to any point on the circumference and then you take itself multiplied by i and put it on its head you can visually see it
@@MathVisualProofs yeah but when the base ( which is 1) is in second quadrant if we plot it it will on negative side of x-axis like in Perpendicular of slope which is positive in second quadrant as we plot it is in the upper y axis( which is positive)
Thanks for the demonstration sir, but can u explain in terms physics, if first derivate is velocity, second derivate acceleration, third derivate is jerk, i dont know understand if wave is unable to calculate, by providing tangent with rate of change it gives another wave, how it is simplifying our task , please help me to rectify my thought , thank you
What he's done here, is a visual inspection of the key points. The slope of sin(x) at x=0 is +1, and at sin(pi/2) it's zero, and then -1 at sin(pi). Mapping the derivative, and we see it resembles cos(x). To prove this formally, you construct the definition of the derivative from first principles. d/dx sin(x) = (sin(x + h) - sin(x))/h Then use the angle sum identity to unpack sin(x + h). This gives us: sin(x)*cos(h) + sin(h)*cos(x) Reconstruct: [sin(x)*(cos(h) - 1) + sin(h)*cos(x) ]/h In the limit as h goes to zero, cos(h) approaches 1, so the first term goes away. We're left with: cos(x)*sin(h)/h Now we just need to prove sin(h)/h = 1. This can be proven by the squeeze theorem (yes, that really is its name). By bounding h between sin(h) and tan(h), you can show that sin(h)/h is trapped between two expressions that both approach 1 as h approaches zero. I'll leave it to you to look up the method.
Why are the progressions on one axis more than the ones on the other? In an activity which I have to write down in which we have to make the graph of inverse of sin(x) with the graph of sin(x), we are instructed to make the progressions on x-axis 1.25 times that on y-axis, and i don't understand why. It appears that the graph here is the one we're supposed to draw, and i just can't because I need to understand why first.
Seems like a completely arbitrary choice to make it that way. I don't see any advantage to that particular ratio. For trig functions, it usually helps to make your graph on one axis in terms of pi/4, which is approx 0.785. Or even pi/2, which is approx 1.57. This is because the points of interest on the sine and cosine functions happen at x-values that are multiples of pi/4. But I don't see the advantage of a ratio of 1.25 between the two axis scales.
Anyway you could do a section on the man the myth the traveling suitcase PAUL ERDOS sorry I can't find the thing to put over the o in ERDOS to make it sound erdish or erdesh
@@MathVisualProofs ty it would be epic just to see what you put forth when u put ur mind to it definitely you will have my sponsorship with $ every month that's for dam uncle Sam n big red sure ty if u give it a go
No, for that you need to use the chain rule. That said, your answer was mostly correct, you just need to multiply it by the derivative of the internal function.
just draw a line that is tangent to the curve at that point, this is the tangent line, and the values he used were the values of the slopes of those tangent lines along the curve
@@Z7youtube I want to intuitively understand the concept behind the use of that line in graphs, what is the meaning like you want to explain to a little child and understand
@@solaokusanya955 heres how it was explained to me. Imagine a function on a graph that isnt linear, like sin(x) or x^2. If we choose 2 points on the function, we can calculate the rate of change. For example, if we plot a line that goes through (0,0) and (10,100) over x^2 we can see that the rate of change for that interval is 10. Now, what if we move x=10 to 4? The line goes through (0,0) and (4,16) and the rate of change is now 4. A derivative is what happens when you move the second point to a point which is basically the first plus an infinitesimal amount. So, the line goes through (0,0) and (0+h, (0+h)^2) where h = 1/infinity. Now, the rate of change is 0.
@@solaokusanya955 pick a point on the curve. A tangent line is a straight line that only touches that one point you picked. The tangent line will the same slope as the instantaneous slope of the curve at that point. In other words, the “rise over run” (slope) of the tangent line is the same as the “rise over run” (slope) of the curve at that point. The derivative of the function is just making a new function that keeps the same x-value but makes the slope of the original graph the new y-value. Derivative = function that shows the rate of change of the original function Tangent line = A line that shows the slope of a curve at a single point, which also only intersects the curve at that one point locally.
Same reasoning, just shifted pi/2 radians to the left. Derivatives of trig functions follow a 4-part cycle: sin(x) -> cos(x) -> -sin(x) -> -cos(x) and back where we started
@@franbh94 the derivative is the slope of that line. There are many ways to compute the slope but I am selecting points so they are 1 unit apart on x-axis so the slope is the y-difference
You take a function of x, and you call it y Take any x-naught, that you care to try You make a little change and call it delta x The corresponding change in y is what you find next And now, take the quotient, and now carefully Send delta x to zero, and I think you'll see That what the limit gives us, if our work all checks Is what we call...dy/dx! It is dy/dx.
@@crazychicken8290 Quotient of ∆y/∆x. That was Tom Lehrer's lyrics. He wrote the definition of the derivative in a song, set to a tune coincidentally called, "There'll be some changes made."
Derivative of cos(x) is -sin(x). Trig derivatives of the two main trig functions follow a 4-part cycle: sin(x) -> cos(x) -> -sin(x) -> -cos(x) -> repeat Taking derivatives, shifts a trig function to the left by a quarter cycle each time. Taking integrals does the opposite, and shifts a quarter cycle to the right.
Thanks for the feedback. I use these visualizations in my class as students think about how to see the derivative of a function from its graph. Thought it could be helpful. I have another in the queue that shows how the derivative is defined. But it’s hard to get into shorts format so far so I just put these up :)
If you are asking whether sine and cosine are each others' derivatives, the short answer is almost. For their "cousin" functions of cosh(x) and sinh(x), they are each others' derivatives, but that's a topic for another day. The derivatives of sine and cosine follow a 4-part cycle: f(x) = sin(x) f'(x) = cos(x) f"(x) = -sin(x) f'''(x) = -cos(x) The 4th derivative begins the next repeat of this cycle.
Tangent is a quotient of sine and cosine. Using the quotient rule, you can show that sec^2(x) is tangent's derivative. It is only the simple sine wave shapes, where a derivative is as simple as a quarter cycle phase shift to the left.
False. If you compute the derivative at that point then you can draw the tangent line with one point (of course the def of the deriv requires many points ;) ).
@@MathVisualProofsIs _"only one point is required if you have the slope"_ a visual-proof type of thing? I ask because I have something in two points...but proofs are a foreign land to me.
@@oddlyspecificmath I mean if you know calculus and the techniques involved, the idea is you can compute the derivative of a function and that function outputs the slope of the tangent line. So all you need is the derivative function and you can draw the tangent line with one point. However, this is a bit circular because to define the derivative function you need a limiting process finding slopes of secant lines. This particular visualization is not a proof (despite my channel name sometimes I just show visualizations that I like or utilize in my teaching). This one is to give the suggestion that the derivative of sine is cosine, but this is not a proof.
@@MathVisualProofs If I had a way to generate tangents without taking the derivative, i.e., no limits, just an algorithm, does this become a proofs area? _(Hoping I'm representing correctly, I'm not trying to squeeze you for info; rather checking to see how / if I have something of use to creators here)_
@@oddlyspecificmath I mean, the derivative is, by definition, a limit. However, many modern differential calculus classes focus on how to find derivatives without using limits. So I am not sure what you mean.
This way of ecplaining is really arrogant and its like playing hard to get. Just ecplain it with eg y=15 and a few other easy formulas. In middle school it is also not teached this way!!!!!
Ok EVERYONE needs to be briefed here on the notation of a function. 2 cases if you have an f and an x in your sentence 1) you are writing f(x). x needs to be introduced before. So you are writing a NUMBER just like 7 or 1/π. 2) you are writing f. You are therefore writing a FUNCTION. In linear algebra, it's a vector and no one would ever allow you to say that f(x) is a vector. You don't have any idea of what does it look like, without EVALUATING in a point (within its set of definition) Examples: i) "Because f' is positive on D, then f is an increasing function on D" In other words " Because for all x in D, f'(x)≥0, then f in an increasing function on D." ii) " Let f defined by for all x in R, f(x) =sin(x). Then, as f is derivable on R, for all x in R, f'(x) = cos(x). Or f= sin therefore f'= cosx. I hope that I had been easy to understand. Let me know if there's any questions.
I've been wondering about this ever since learning about derivatives. Thanks for confirming my suspicion
You weren't taught about trig derivatives?
@@gtc4189 I'm 14 in an Italian school and they have just started doing equations with algebraic fractions. I like to always study things before others so I have a better time understanding them when they're eventually taught. I've just finished my trigonometry book and before this comment, I had seen a video explaining what derivatives are, so I had wondered what the derivative of different functions were. I had wondered if the derivative of sin(x) was cos(x) ever since I first thought about it
@@kirahen0437 Ah I see this makes much more sense. Good luck on your studies my dude
@@gtc4189 thanks
@@kirahen0437 don't take study as pressure rather love it here in India many people are studying just to earn money
I am kind of sad that you chose to present this fact this way, because the derivation of that fact can be seen even nicer in a visual form, by looking at a point P on the unit circle with angle theta from the x-axis and moving it a tiny step Δtheta to obtain P'. Draw a vertical line through P and a horizontal line through P' which intersect at Q to obtain a right triangle PQP' such that angle QPP' ≈ theta. As |PP'| ≈ Δtheta, we obtain that
sin'(theta) ≈ |PQ|/|PP'| ≈ cos(theta). This proof is not rigorous but can easily be made into a rigorous argument.
Yes. I like that one. I’ll try that one in long format. This one is part of a series about graphing derivatives as you walk along the curve. This is more for my students as we think about increasing/decreasing behavior and how the derivative captures that. The shorts format is still tough for me to say something very interesting.
@@MathVisualProofs Yes, I understand! It's hard to convey nice insights in less than a minute. But having a teacher like you putting so much effort into these visualizations must be awesome :D
@@pengin6035 :)
This is the proof I found as well!
i found this out 2 days ago when i was messing around in desmos trying to find the tangent line of a sine function. i knew the graph would be a sine wave of some sort, and i found out that the gradient of the tangent line of sin x = sin x-π/2, which is equal to cos x
Love desmos for this!
@@MathVisualProofs same! i re-discovered a lot of equations thanks to it.
@@MathVisualProofsI don’t really Understand
What was your set up in Desmos and what things did you plot?
exact same here for me
Thnx for this! And have you considered make a "Essence of conic sections geometrically"? There are many old books on internet for free access like geometry of conic sections please have a look at them
Clear.for.teaching.. Good.job
Excelllent video 😊😊
Thanks!
Amazing explanation
Love it! Visualization helps!!
Awesome
Amazing
“We can see that…” is doing a lot of heavy lifting here
Yes. For shorts format we are just using the visual. Of course it needs more work to actually convince ourselves it is true.
Absolutely 😅
Nice! Love your visual 😊
Thanks!
My jaw dropped and I don't even know why. I knew what would be the outcome but it still surprised me.
this is literally the video i have been searching for the past 6 hours
Thanks! I am just learning about derivatives and this video helped me understand it a little bit better
Anyway, to anyone who wants to try this, you can use y=f'(t)(x-t)+f(t), and initialise f(x) as whatever differentiable function you want, and t as a parameter. This function will generate a tangent line on the curve at point (t,f(t)).
This works on desmos btw. If you can't write f'(t), try d/dt(f(t)) instead.
It was beautiful ❤
dsin(x)/dx =(sin(x+dx) - sin(x)) /dx
=( sin(x).cos(dx)+sin(dx)cos(x)-sin(x))
/dx
=sin(x).(cos(dx)-1)/dx+cos(x).sin(dx)/dx
=sin(x)×0 + cos(x) × 1
=cos(x)
[if dx tends to 0 sin(dx) /dx tends to 1 and (cos(dx) - 1)/dx to 0]
I'm curious on how you create theese visualisations what software u use ?
Love it!
Absolutely beautiful
Omg this is the most beautiful explanation of derivatives i have seen
Amazing!
Thanks!
amazing
Beautiful
Nice! Can you show us the derivative for e^x?
Yes, soon!
ua-cam.com/users/shortsMr5cpZyYHX0?feature=share
This is what teaching should be , without letting children memorize everything
Yoo this is crazy af, l have always been trying to visualize all these but now I found one... Cool
theorems you could solve in a minute:
for any function that is not e to the x,
f'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''(x) is equal to 0.
thank you.
Nice , could you tell me please how to compute the slope
Here’s the idea/concept : ua-cam.com/users/shortseE4IGCAzmqA?si=mljokGicigflQKgE . I’m working on an animation of the sine derivative slope another way.
👍👍👍
theres a cool proof or just this using a unit circle, if you have a vector from the center to any point on the circumference and then you take itself multiplied by i and put it on its head you can visually see it
Wow I forget that it's been almost 20 yrs since I used a derivative in a physical problem
Amazing... Which program do you use for animation
This is done with manimgl
@@MathVisualProofs Thank you 🏵️🌹
To find the derivative, plot in a graph following here:
f(x) = (your common function here)
y = d/dx f(x)
So you get the derivative!
Why does that 1 is always positive hoping for answer 😊😊
We are measuring a slope. So I fixed the base of the triangle to be 1 in the positive direction.
@@MathVisualProofs yeah but when the base ( which is 1) is in second quadrant if we plot it it will on negative side of x-axis like in Perpendicular of slope which is positive in second quadrant as we plot it is in the upper y axis( which is positive)
@@MathVisualProofs pls bro solve mine problem
@@MathVisualProofs how x can be always positive it's direction is changing
Can demonstrate the integral of trigulometic functions visually?
That would be a miracle, I believe.
Thanks for the demonstration sir, but can u explain in terms physics, if first derivate is velocity, second derivate acceleration, third derivate is jerk, i dont know understand if wave is unable to calculate, by providing tangent with rate of change it gives another wave, how it is simplifying our task , please help me to rectify my thought , thank you
What he's done here, is a visual inspection of the key points. The slope of sin(x) at x=0 is +1, and at sin(pi/2) it's zero, and then -1 at sin(pi). Mapping the derivative, and we see it resembles cos(x).
To prove this formally, you construct the definition of the derivative from first principles.
d/dx sin(x) = (sin(x + h) - sin(x))/h
Then use the angle sum identity to unpack sin(x + h). This gives us:
sin(x)*cos(h) + sin(h)*cos(x)
Reconstruct:
[sin(x)*(cos(h) - 1) + sin(h)*cos(x)
]/h
In the limit as h goes to zero, cos(h) approaches 1, so the first term goes away. We're left with:
cos(x)*sin(h)/h
Now we just need to prove sin(h)/h = 1. This can be proven by the squeeze theorem (yes, that really is its name). By bounding h between sin(h) and tan(h), you can show that sin(h)/h is trapped between two expressions that both approach 1 as h approaches zero. I'll leave it to you to look up the method.
So a wave waves as u derive
Why are the progressions on one axis more than the ones on the other? In an activity which I have to write down in which we have to make the graph of inverse of sin(x) with the graph of sin(x), we are instructed to make the progressions on x-axis 1.25 times that on y-axis, and i don't understand why.
It appears that the graph here is the one we're supposed to draw, and i just can't because I need to understand why first.
Seems like a completely arbitrary choice to make it that way. I don't see any advantage to that particular ratio.
For trig functions, it usually helps to make your graph on one axis in terms of pi/4, which is approx 0.785. Or even pi/2, which is approx 1.57. This is because the points of interest on the sine and cosine functions happen at x-values that are multiples of pi/4. But I don't see the advantage of a ratio of 1.25 between the two axis scales.
can you please make a course on manim?
I am not sure I am the best person to do that....
But the question is, how are you going to create a tangent line at a point if the derivative of the function is needed for that
You prove it using Sin addition formulae right?
Yes. And then investigating a couple of special limits.
Anyway you could do a section on the man the myth the traveling suitcase PAUL ERDOS sorry I can't find the thing to put over the o in ERDOS to make it sound erdish or erdesh
Not sure if I can find good visualizations, but I'll keep an eye out.
@@MathVisualProofs ty it would be epic just to see what you put forth when u put ur mind to it definitely you will have my sponsorship with $ every month that's for dam uncle Sam n big red sure ty if u give it a go
@@MathVisualProofs lol an eye out for visualisation, I'm confident it will come to you
How did you make these kinds of animation?
I use manimgl. That’s the python package created by 3blue1brown
So what does it mean to derive in this sense?
does this apply to functions? the derivative of a sine of a function equal to the cosine of a function?
No, for that you need to use the chain rule. That said, your answer was mostly correct, you just need to multiply it by the derivative of the internal function.
Got an explanation for why cos = -sin but sin= cos?
The slopes of cosine starting at 0 are negative.
NO WAYYYYY HAJAHAJAJAJA NOO WAYYYYYYYYYYYYYYY IT BLEWW MY FK MINDDD
What I don't understand is the "tangent line" and the value you have it...
I don't understand that tangent line
just draw a line that is tangent to the curve at that point, this is the tangent line, and the values he used were the values of the slopes of those tangent lines along the curve
@@Z7youtube I want to intuitively understand the concept behind the use of that line in graphs, what is the meaning like you want to explain to a little child and understand
@@solaokusanya955
do you know what a slope is? or how to calculate the slope of a line?
@@solaokusanya955 heres how it was explained to me. Imagine a function on a graph that isnt linear, like sin(x) or x^2. If we choose 2 points on the function, we can calculate the rate of change. For example, if we plot a line that goes through (0,0) and (10,100) over x^2 we can see that the rate of change for that interval is 10. Now, what if we move x=10 to 4? The line goes through (0,0) and (4,16) and the rate of change is now 4. A derivative is what happens when you move the second point to a point which is basically the first plus an infinitesimal amount. So, the line goes through (0,0) and (0+h, (0+h)^2) where h = 1/infinity. Now, the rate of change is 0.
@@solaokusanya955 pick a point on the curve. A tangent line is a straight line that only touches that one point you picked. The tangent line will the same slope as the instantaneous slope of the curve at that point. In other words, the “rise over run” (slope) of the tangent line is the same as the “rise over run” (slope) of the curve at that point. The derivative of the function is just making a new function that keeps the same x-value but makes the slope of the original graph the new y-value.
Derivative = function that shows the rate of change of the original function
Tangent line = A line that shows the slope of a curve at a single point, which also only intersects the curve at that one point locally.
So with cos I can measure the slope of sin?
Can I apply to stocks graph? 😅
Could u explain
cos x derivative
Same reasoning, just shifted pi/2 radians to the left.
Derivatives of trig functions follow a 4-part cycle:
sin(x) -> cos(x) -> -sin(x) -> -cos(x) and back where we started
I was so proud of myself the day I figured this myself ❤
Isnt this how everyone is taught derivatives? Or do some teachers really just give you a bunch of symbols and rules with no context?
I dont get what the numbers "1" and the other (from -1.00 to +1.00 represent)
I’ve drawn a triangle. The base is always 1 and so the height is the slope (as slope is rise over run)
@@MathVisualProofs Oh, I see. And the derivative is taken from the points where the base and the slope meet, right?
@@franbh94 the derivative is the slope of that line. There are many ways to compute the slope but I am selecting points so they are 1 unit apart on x-axis so the slope is the y-difference
WHATS A DERIVATIVE
You take a function of x, and you call it y
Take any x-naught, that you care to try
You make a little change and call it delta x
The corresponding change in y is what you find next
And now, take the quotient, and now carefully
Send delta x to zero, and I think you'll see
That what the limit gives us, if our work all checks
Is what we call...dy/dx! It is dy/dx.
@@carultch quotient of what
@@crazychicken8290 Quotient of ∆y/∆x.
That was Tom Lehrer's lyrics. He wrote the definition of the derivative in a song, set to a tune coincidentally called, "There'll be some changes made."
@@carultch thanks
Then what’s the derivative of cos?
Derivative of cos(x) is -sin(x).
Trig derivatives of the two main trig functions follow a 4-part cycle:
sin(x) -> cos(x) -> -sin(x) -> -cos(x) -> repeat
Taking derivatives, shifts a trig function to the left by a quarter cycle each time. Taking integrals does the opposite, and shifts a quarter cycle to the right.
Im in the 8th grade doing algebra and functions, can u upload a video explaining what this means and how to solve it?
So what’s the derivative of tangent?
y = tan(x)
dy/dx = sec^2(x)
@@professionalcatgirl8592Thanks; I went off to Desmos and got it in terms of cot², which didn't look so pretty. Made me go check my identities 😊
@professional catgirl thanks!
After scrolling through hundreds of shitposts, I thought it was one of them
Love your videos, but this derivative series is just not doing it for me. You’re barely visualizing anything here.
Thanks for the feedback. I use these visualizations in my class as students think about how to see the derivative of a function from its graph. Thought it could be helpful. I have another in the queue that shows how the derivative is defined. But it’s hard to get into shorts format so far so I just put these up :)
Hey say no to DRUGS
vise verza?
Your comment turned into "did you see the cabbage?" when I clicked translate to English.
If you are asking whether sine and cosine are each others' derivatives, the short answer is almost. For their "cousin" functions of cosh(x) and sinh(x), they are each others' derivatives, but that's a topic for another day.
The derivatives of sine and cosine follow a 4-part cycle:
f(x) = sin(x)
f'(x) = cos(x)
f"(x) = -sin(x)
f'''(x) = -cos(x)
The 4th derivative begins the next repeat of this cycle.
And then there's tan which is sec^2 and not cot😂
Tangent is a quotient of sine and cosine. Using the quotient rule, you can show that sec^2(x) is tangent's derivative.
It is only the simple sine wave shapes, where a derivative is as simple as a quarter cycle phase shift to the left.
nah bro that graph is 5 sin(x)
tick marks are at multiples of 0.2 :)
@@MathVisualProofs ah
cannot draw a tangent line with one point
False. If you compute the derivative at that point then you can draw the tangent line with one point (of course the def of the deriv requires many points ;) ).
@@MathVisualProofsIs _"only one point is required if you have the slope"_ a visual-proof type of thing? I ask because I have something in two points...but proofs are a foreign land to me.
@@oddlyspecificmath I mean if you know calculus and the techniques involved, the idea is you can compute the derivative of a function and that function outputs the slope of the tangent line. So all you need is the derivative function and you can draw the tangent line with one point. However, this is a bit circular because to define the derivative function you need a limiting process finding slopes of secant lines. This particular visualization is not a proof (despite my channel name sometimes I just show visualizations that I like or utilize in my teaching). This one is to give the suggestion that the derivative of sine is cosine, but this is not a proof.
@@MathVisualProofs If I had a way to generate tangents without taking the derivative, i.e., no limits, just an algorithm, does this become a proofs area? _(Hoping I'm representing correctly, I'm not trying to squeeze you for info; rather checking to see how / if I have something of use to creators here)_
@@oddlyspecificmath I mean, the derivative is, by definition, a limit. However, many modern differential calculus classes focus on how to find derivatives without using limits. So I am not sure what you mean.
This way of ecplaining is really arrogant and its like playing hard to get. Just ecplain it with eg y=15 and a few other easy formulas. In middle school it is also not teached this way!!!!!
Best use f'(x)= lim h→0 [f(x+h)-f(x)/h]
Ok EVERYONE needs to be briefed here on the notation of a function.
2 cases if you have an f and an x in your sentence
1) you are writing f(x). x needs to be introduced before. So you are writing a NUMBER just like 7 or 1/π.
2) you are writing f. You are therefore writing a FUNCTION. In linear algebra, it's a vector and no one would ever allow you to say that f(x) is a vector. You don't have any idea of what does it look like, without EVALUATING in a point (within its set of definition)
Examples:
i) "Because f' is positive on D, then f is an increasing function on D"
In other words
" Because for all x in D, f'(x)≥0, then f in an increasing function on D."
ii) " Let f defined by for all x in R, f(x) =sin(x). Then, as f is derivable on R, for all x in R, f'(x) = cos(x). Or f= sin therefore f'= cosx.
I hope that I had been easy to understand.
Let me know if there's any questions.
I'm curious on how you create theese visualisations what software u use ?
I'm curious on how you create theese visualisations what software u use ?
I'm curious on how you create theese visualisations what software u use ?
I'm curious on how you create theese visualisations what software u use ?
I use manim.