Visualizing the derivative of sin(x) 

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

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

Yes. For shorts format we are just using the visual. Of course it needs more work to actually convince ourselves it is true.

Absolutely 😅

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.

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.

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

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

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!

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.

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.

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.

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 :)

tick marks are at multiples of 0.2 :)

@@MathVisualProofs ah

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 ;) ).

​@@MathVisualProofs​Is _"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.

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!

Yes. And then investigating a couple of special limits.

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.

Yes, soon!

ru-vid.comMr5cpZyYHX0?feature=share

I use manimgl. That’s the python package created by 3blue1brown

Thanks!

I use manim.

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.

Here’s the idea/concept : ru-vid.comeE4IGCAzmqA?si=mljokGicigflQKgE . I’m working on an animation of the sine derivative slope another way.

The slopes of cosine starting at 0 are negative.

This is done with manimgl

@@MathVisualProofs Thank you 🏵️🌹

Thanks!

Thanks!

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

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

That would be a miracle, I believe.

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 am not sure I am the best person to do that....