I have been struggling in ib math and this helped me clear up what my teacher taught for the past week. I watched your videos for chemistry, physics, and now math. You have saved my life sir. 🙏
I like your channel so much. You explain everything in the best and the easiest as well as in an interesting way. Thanks Professor Dave. Make more videos.
This is exactly why people like me are actually asking for when we ask what this has to do with anything or what will I actually use this for. Our minds work backwards. If you tell me the rule first, then the application, I forgot the rule because it had no relationship to anything I understand. By showing the application and what the variables relate to, it's so much easier to remember the rules or learn them while applying them. I wish all math videos would explain according to this concept like this video. There are already too many that do things the other way.
Dave! Your videos are incredible, I was quite stuck on Optimization/Related-Rates using derivatives, but you've cleared it up almost effortlessly! Great professor, truly, lol. Thanks again!
Wait! I think I understand now. Derivative is actually the SLOPE OF TANGENT LINE of a function. So when it's 0. That means the tangent line is going horizontally. This means that the function's y value can't be any greater as it is contained by the tangent line. Thus the f(x) is largest when dy/dx =0. Am I babbling??
no you pretty much get it, if you look at a f(x) graph, the points where f(x) change from increasing to decreasing and vice versa have a derivative of 0 and are the points of maximum or minimum of the graph, it could also be seen as when dy/dx goes from positive to negative and vice versa
For the fence problem, I did it differently. I said dA/dx=y and dA/dy=x. Upon reaching this calculation, I intuitively thought, "no matter which variable we choose to differentiate A(meaning no matter how we try to find the rate of change of A with respect to either variable), they must be equal to one another ". That being the case, I said x=y. From there, I said 2x+x=2400(since y=x).
Maximizing the area doesn't mean making a square like you have. For this problem we have 2x+y = 2400. Solving for x you will get (2400 - y)/(2) and solving for y you will get 2400 - 2x. Their derivatives will be different, and solving for x will get you 600 and solving for y will get you 1200. What you're doing doesn't really make sense.
@professor dave sir....for the same question how can I find both maximum and minimum values...suppose I have 2400 m fence...how can I find the minimum area...pls explain it sir??
In this problem, the minimum area occurs in the degenerate case that you make the plot of land either infinitesimally thin or infinitesimally narrow. The corresponding area would be zero.
Thank you for this serier.. But i have a doubt. if rather than covering the max area we have to cover the minimum area with same fence. Then how would we solve this ??? Thanks in advance.
8 months late, but essentially what he's doing here is calculating a Maxima, to calculate the smallest area you would need to calculate the Minima, which is done practically the same way. Both Maxima and Minima have a derivative equal to 0, you just need to check which one your equation gave you.
In the first example using 2400m of fencing the constraint was that the enclosure was to be rectangular with one side being the river, but what if instead of a rectangle the fence was arranged as half a circle with the river being equal to the diameter. In this case the area enclose would be over 916,000m --nearly 200K more than the rectangle. One of the problems with questions like this is knowing what the constraints are and in this case the constraint was clearly stated that the enclosure was to be rectangular, but what if that constraint was not specified and the question was: what would be the greatest area enclosed with 2400m of fencing if some portion of the enclosure was defined by the river. How do you determine the ultimate geometry to use?
@@ProfessorDaveExplains -- Well it makes sense that a semi-circle would be pretty good as a circle is known to have the greatest area for a given circumference or perimeter. I guess my question or point though is that in some max/min cases the geometry or constraint is not well known so narrowing down or pinching the constraint is a bigger problem than the Algebra or Calculus. BTW, thanks for doing these videos -- its been a long time since I studied these topics (mid 80's) so its nice to refresh my somewhat atrophied brain cells.
@@Raptorman0909 A circle has the optimal area-to-perimeter optimization, but that is only if we are interested in the entire perimeter. In this problem, one given side is the linear shore of the river, and the shore of the river is "free fencing" for this problem. We may be able to do better than a semicircle for how to optimize shape of the fence. I've attempted this using Ramanujan's approximation for perimeter of an ellipse, and surprise/surprise, a perfect semicircle is the optimal semi-ellipse for this problem. There could be a shape other than an ellipse, that optimizes the area of this fence.
Because 605+605+1195=2405, even though we only have 2400m of fence to use. You got a much higher area simply because you have 2405m of fence, not 2400 (like the problem does).
Mr Dave Rave. you helped me get into Uni, and now helping me go through it (biochem). Ive always been known as the "not the sharpest tool in the shed" girl, cause i really am and there is no shame in that so thats a testament to your amazing italian teaching skills that im finally learning. I got into one of the best unis here with your help. Maybe late started (24). but still. I cant even speak my native language well cause i have trouble with language cause i hit my head really hard when i was young. (no joke, but you are allowed to laugh), but hey this is an achievment for me, so let me be happy. thank you very much Mr David, and thanks to Italy for bringing great things. like you, the godfather and pisa (the tower not the food)...no but really. Thanks sir, and cheers from Chile! youve helped me a lot!
did he knew ahead of time that x = 600 is a maximum point, what if it's a minimum or not an extremum point at all, what I know is we must check at first using the first derivative test or use second derivative to check for concavity instead.
By intuition, you probably can tell that it would be a maximum point, rather than a minimum point. You know that for a full rectangle, the optimal solution for area to perimeter ratio occurs when the rectangle is a square. So a very long and narrow rectangle, or very wide and short rectangle, are likely not the answer. You know that a solution with maximum area would be somewhere in between. And since this is a continuous and differentiable function in the entire domain, you know that the only critical point you find, will correspond to the maximum. You can do the second derivative test to confirm it is a local maximum, and after ruling out all other critical points, and end points, you can confirm that it is a maximum.
Then you investigate the end-points of the domain. One of those will be the maximum. As an example. Two positive integers have a product of 100. What is the maximum possible sum? You'll find a critical point at x=10 and y=10, where the sum is 20. This is a local minimum to solving this problem with calculus. We were interested in a maximum, rather than a minimum. But 5 and 20 are another pair of integers whose product is 100, and they add up to 25, which is greater than 20. To get the maximum sum, we have to look at the end of the domain. The smallest positive integer is of course 1, and the corresponding other number in our constraint is 100 itself. 1*100 = 100, and 1+100 = 101. This is our maximum.
Dave - you have created disharmony in my head. For 50 years now I have been under the impression that the common tuna fish can was optimized for materials. I can remember sitting in my calculus class in junior college and being amazed by the proof of this from the instructor. I watched this video of yours and realized that these optimal dimensions (h=2r) don't match with a tuna can in my cupboard. I measured it, and h=r approx. What the heck ? Did I misunderstand my teacher all those years ago, or is a tuna can not optimized ?
There is actually a lot more that goes into designing a tuna can. I saw a tweet about it a long time ago, but someone was trying to tell a cat food manufacturer that they could save money by optimizing the surface area and whatever. One of their engineers replied and basically said there are like 10 other factors that go into the cost of manufacturing that all outweigh this optimization.
Calculus used to be a generic term for calculations in general (obsolete meaning of the term), with infinitesimal calculus being what we call calculus today. The way the term calculus is used today in mathematics, it is all infinitesimal calculus. There are two separate branches of calculus, which are integral calculus and differential calculus, both of which are related through the fundamental theorem of calculus. It means something completely different in dentistry, but that is the only meaning of the word that is unrelated to the way the word is used in mathematics.
In this problem, he's defining x to be the vertical side. Since there are two vertical sides and one horizontal side, we need to use 2*x + y = 2400 m, rather than the other way around. He could've just as easily assigned x to be the horizontal side length and y to be the vertical side length. It was an arbitrary choice.
The second derivative test. Or more generally, the even order derivative test. If the second derivative is negative, it is a local maximum, with concave-down (negative) curvature. If the second derivative is positive, it is a local minimum, with concave-up (positive) curvature. A second derivative of zero, is inconclusive. You have to continue taking derivatives until you get to an even-ordered derivative that is non-zero. A forth derivative can conclude the same thing, given a second derivative of zero. If all even-ordered derivatives are zero, then you have an inflection point that coincides with a flat point on the graph, like you see for y=x^3. An inflection point is a location where the curvature switches from being concave-up to concave-down, or vice-versa.
The area is being differentiated relative to the independent dimension that is to be determined. He's looking at the land area as a function of the x-dimension perpendicular to the river.
Whether x is positive or negative, has nothing to do with whether the point is a local maximum or minimum. The value of x just indicates what the horizontal position of the point is, representing the input of a function. It is the second derivative d^2y/dx^2 that determines whether the point is local maximum or local minimum. And the reason is that the second derivative indicates the curvature of the function, and the sign of it indicates whether it is concave-up or concave-down. A positive second derivative indicates concave-up, and a negative second derivative indicates concave-down. A local maximum should slope down away from the maximum point, thus being concave-down, and vice-versa for a local minimum.
@@ProfessorDaveExplains It's not true that "when the derivative of a function is zero, there is a maximum or minimum at that point." I think you meant to say the converse, which is true provided the derivative exists.
I don't understand what you're saying. You cited x^3 at x = 0, where the derivative of the function is not zero. That's why there is not a local maximum or minimum there. I am not aware of any example where a derivative of zero does not indicate a local maximum or minimum other than a constant function.
He doesn't Explain why area would be maximum when the derivative of the function is 0. Does he understand why? Coz I don't know either. AND I DIDN'T SAY ONE PEOPLE ASK WHY DERIVATIVE OF A FUCNTION IS 0 WHEN IT'S MAXIUM VALUE
It’s 0 because it means the functions slope has stopped increasing, the tangent line at that point would be a flat line meaning the function must decrease afterwards.
Dear sir, how can i concrete Sketching derivatives function on the graph, i feel that i still need some more, will you do some other videos about this?
My maths teacher taught me a great rule about this that can help you instantly figure out how much this was when for example have x+y=100 the maximum product of this two numbers will be 100 divided by 2 and this works every time.
![How to Solve ANY Optimization Problem [Calc 1]](http://i.ytimg.com/vi/cUMvwG7wvzM/mqdefault.jpg)
Optimisation problems is arguably the best real-life application of calculus
related rates
@@chris2217I feel like diff eq is more useful for that though…
I have been struggling in ib math and this helped me clear up what my teacher taught for the past week. I watched your videos for chemistry, physics, and now math. You have saved my life sir. 🙏
I’m doing IB too and Prof. Dave is 🙏
I like your channel so much. You explain everything in the best and the easiest as well as in an interesting way. Thanks Professor Dave. Make more videos.
This is exactly why people like me are actually asking for when we ask what this has to do with anything or what will I actually use this for. Our minds work backwards. If you tell me the rule first, then the application, I forgot the rule because it had no relationship to anything I understand. By showing the application and what the variables relate to, it's so much easier to remember the rules or learn them while applying them. I wish all math videos would explain according to this concept like this video. There are already too many that do things the other way.
Look em in the eye and say scheduling major league baseball. OK. not calculus, but more advanced stuff.
I felt like this too, but it gets better over time. I guess I developed some kind of „trust“ in maths, to be useful in the real world. :D
I like how you put it, "our minds work backwards". true.
Instead of saying "our minds work backwards," you can say you prefer a "top-down approach."
I FINISHED DIFFERENTIATION LAST YEAR BUT UR VIDEOS HELPED ME MASTER IT DAVE THANKS!
You are seriously the best. Thank you so much for these videos.
Instead of watching the useless online lecture videos my professor posts, I come straight here to get edumacated
Dave! Your videos are incredible, I was quite stuck on Optimization/Related-Rates using derivatives, but you've cleared it up almost effortlessly!
Great professor, truly, lol. Thanks again!
Thank you very much mathematics is always simple with professor Dave.
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Wait! I think I understand now. Derivative is actually the SLOPE OF TANGENT LINE of a function. So when it's 0. That means the tangent line is going horizontally. This means that the function's y value can't be any greater as it is contained by the tangent line. Thus the f(x) is largest when dy/dx =0. Am I babbling??
no you pretty much get it, if you look at a f(x) graph, the points where f(x) change from increasing to decreasing and vice versa have a derivative of 0 and are the points of maximum or minimum of the graph, it could also be seen as when dy/dx goes from positive to negative and vice versa
man your videos are helping me pass finals in computer science ! thank you mate keep going
Every time I hear your intro song, my thoughts turn happy and I know that this difficult subject will soon be understandable. Thanks Professor Dave :)
Math is like my bff for life because I'll use it to create a new cure for cancer when I'm a chemical engineer
@@brettandy9718 it wants money
It's been 4 years, how is that going?
Another very helpful video! Thanks Professor Dave :)
At 7:22 how do you get the common denominator r^2?
really interesting. I thought the area should be maximized when simply having the shape of a square. The farmer example shows DO THE MATHs
Thanks for changing my views on calculus!
Thank you sir for your dedication and for making this free! 🙏
if only the farmer knew his farm would be bigger if it were a half circle
Dude... Your a true god. Love your vids :D
preparing for my calc test today
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For the fence problem, I did it differently. I said dA/dx=y and dA/dy=x. Upon reaching this calculation, I intuitively thought, "no matter which variable we choose to differentiate A(meaning no matter how we try to find the rate of change of A with respect to either variable), they must be equal to one another ". That being the case, I said x=y. From there, I said 2x+x=2400(since y=x).
ua-cam.com/video/XPCgGT9BlrQ/v-deo.html 👍💐💐💐
Maximizing the area doesn't mean making a square like you have. For this problem we have 2x+y = 2400. Solving for x you will get (2400 - y)/(2) and solving for y you will get 2400 - 2x. Their derivatives will be different, and solving for x will get you 600 and solving for y will get you 1200. What you're doing doesn't really make sense.
10:17 where did the A=xy came from?
A represents the product, so the multiplications of two integers (x and y) equals the product (or A)
@@astronm64 Ohh, i see. Thanks
the area of a rectangle
He uses A bc it’s ol reliable but you can also use P for product seems stupid but helps me
@professor dave sir....for the same question how can I find both maximum and minimum values...suppose I have 2400 m fence...how can I find the minimum area...pls explain it sir??
In this problem, the minimum area occurs in the degenerate case that you make the plot of land either infinitesimally thin or infinitesimally narrow. The corresponding area would be zero.
Thanks for the video Professor Dave. Brilliant explanation.
optimization is by far the hardest topic in calculus 1, it was going well until then
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me too
Even harder than calculus 2 i think
you're so good at teaching calculus may God bless you Sir❤
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Thanks, Professor Dave..
Thank you very much I understood this lesson so fast it helped me a lot
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sir your tutorials are very helpful
I LOVVVE PROFESSOR DAVEE
Thank you for this serier.. But i have a doubt. if rather than covering the max area we have to cover the minimum area with same fence. Then how would we solve this ??? Thanks in advance.
8 months late, but essentially what he's doing here is calculating a Maxima, to calculate the smallest area you would need to calculate the Minima, which is done practically the same way. Both Maxima and Minima have a derivative equal to 0, you just need to check which one your equation gave you.
In the first example using 2400m of fencing the constraint was that the enclosure was to be rectangular with one side being the river, but what if instead of a rectangle the fence was arranged as half a circle with the river being equal to the diameter. In this case the area enclose would be over 916,000m --nearly 200K more than the rectangle. One of the problems with questions like this is knowing what the constraints are and in this case the constraint was clearly stated that the enclosure was to be rectangular, but what if that constraint was not specified and the question was: what would be the greatest area enclosed with 2400m of fencing if some portion of the enclosure was defined by the river. How do you determine the ultimate geometry to use?
Hmm, I guess you just try different shapes and see what happens! I'm pretty sure a semi-circle will maximize the area though.
@@ProfessorDaveExplains -- Well it makes sense that a semi-circle would be pretty good as a circle is known to have the greatest area for a given circumference or perimeter. I guess my question or point though is that in some max/min cases the geometry or constraint is not well known so narrowing down or pinching the constraint is a bigger problem than the Algebra or Calculus. BTW, thanks for doing these videos -- its been a long time since I studied these topics (mid 80's) so its nice to refresh my somewhat atrophied brain cells.
@@Raptorman0909 A circle has the optimal area-to-perimeter optimization, but that is only if we are interested in the entire perimeter. In this problem, one given side is the linear shore of the river, and the shore of the river is "free fencing" for this problem. We may be able to do better than a semicircle for how to optimize shape of the fence. I've attempted this using Ramanujan's approximation for perimeter of an ellipse, and surprise/surprise, a perfect semicircle is the optimal semi-ellipse for this problem.
There could be a shape other than an ellipse, that optimizes the area of this fence.
Dancing to "checking comprehension" music when my answer is right
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@@beoptimistic5853 ok
✨Thank you, Sir, 🙏🌺✨for all of yours beautiful videos🙏🏼🌺✨Please keeping teaching us🙏🏼🌺✨
I'd always wondered how the solver function on Excel worked, didn't know it was something I could actually do by hand lol.
The part of Calculus that actually might be useful to my life.
if extrema is not the highest or lowest point of a function. Then why they are the most optimized?
What this guy taught me in 10 mins is what my teacher failed to teach me this week
Thanks professor dave😊😊😊
gracias por salvar mi vida 😭
in the first problem, if the sides are for example 605m x 1195m = 722,975. I get a much higher area, why?
Because 605+605+1195=2405, even though we only have 2400m of fence to use. You got a much higher area simply because you have 2405m of fence, not 2400 (like the problem does).
Thank you!
thank you for so awesome videos
thank u for this !!
Thanks, Prof!
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Mr Dave Rave. you helped me get into Uni, and now helping me go through it (biochem). Ive always been known as the "not the sharpest tool in the shed" girl, cause i really am and there is no shame in that so thats a testament to your amazing italian teaching skills that im finally learning. I got into one of the best unis here with your help. Maybe late started (24). but still.
I cant even speak my native language well cause i have trouble with language cause i hit my head really hard when i was young. (no joke, but you are allowed to laugh), but hey this is an achievment for me, so let me be happy. thank you very much Mr David, and thanks to Italy for bringing great things. like you, the godfather and pisa (the tower not the food)...no but really. Thanks sir, and cheers from Chile! youve helped me a lot!
Hey prof, can the second derivative be 0, if so would that mean the second derivative test failed?
The second derivative can be zero, and that means we can't use the second derivative test to find the local min/max for that point.
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'tis was v informative
Or maybe it was dx/dt informative?
Make more videos to co-realte with theory.
AMAZIN VIDEO
did he knew ahead of time that x = 600 is a maximum point, what if it's a minimum or not an extremum point at all, what I know is we must check at first using the first derivative test or use second derivative to check for concavity instead.
By intuition, you probably can tell that it would be a maximum point, rather than a minimum point. You know that for a full rectangle, the optimal solution for area to perimeter ratio occurs when the rectangle is a square. So a very long and narrow rectangle, or very wide and short rectangle, are likely not the answer. You know that a solution with maximum area would be somewhere in between. And since this is a continuous and differentiable function in the entire domain, you know that the only critical point you find, will correspond to the maximum.
You can do the second derivative test to confirm it is a local maximum, and after ruling out all other critical points, and end points, you can confirm that it is a maximum.
What should we do if we want to find the maximum,but then do the second derivative test and find out that we are actually finding the minimum?
Then you investigate the end-points of the domain. One of those will be the maximum.
As an example. Two positive integers have a product of 100. What is the maximum possible sum?
You'll find a critical point at x=10 and y=10, where the sum is 20. This is a local minimum to solving this problem with calculus. We were interested in a maximum, rather than a minimum. But 5 and 20 are another pair of integers whose product is 100, and they add up to 25, which is greater than 20.
To get the maximum sum, we have to look at the end of the domain. The smallest positive integer is of course 1, and the corresponding other number in our constraint is 100 itself. 1*100 = 100, and 1+100 = 101. This is our maximum.
Dave - you have created disharmony in my head. For 50 years now I have been under the impression that the common tuna fish can was optimized for materials. I can remember sitting in my calculus class in junior college and being amazed by the proof of this from the instructor.
I watched this video of yours and realized that these optimal dimensions (h=2r) don't match with a tuna can in my cupboard. I measured it, and h=r approx.
What the heck ? Did I misunderstand my teacher all those years ago, or is a tuna can not optimized ?
There is actually a lot more that goes into designing a tuna can. I saw a tweet about it a long time ago, but someone was trying to tell a cat food manufacturer that they could save money by optimizing the surface area and whatever. One of their engineers replied and basically said there are like 10 other factors that go into the cost of manufacturing that all outweigh this optimization.
Dear professor,may l ask you a question, is optimisation calculus a part or an exception of infinitésimal calculus ?
Calculus used to be a generic term for calculations in general (obsolete meaning of the term), with infinitesimal calculus being what we call calculus today. The way the term calculus is used today in mathematics, it is all infinitesimal calculus. There are two separate branches of calculus, which are integral calculus and differential calculus, both of which are related through the fundamental theorem of calculus.
It means something completely different in dentistry, but that is the only meaning of the word that is unrelated to the way the word is used in mathematics.
Wow what a nice video 👌
How would you know to do 2x+y=2400m instead of x+2y=2400m?
In this problem, he's defining x to be the vertical side. Since there are two vertical sides and one horizontal side, we need to use 2*x + y = 2400 m, rather than the other way around. He could've just as easily assigned x to be the horizontal side length and y to be the vertical side length. It was an arbitrary choice.
thanks
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What if you only get one fixed solution to the derivative.. then how would you determine whether it's maximum or minimum
The second derivative test. Or more generally, the even order derivative test.
If the second derivative is negative, it is a local maximum, with concave-down (negative) curvature.
If the second derivative is positive, it is a local minimum, with concave-up (positive) curvature.
A second derivative of zero, is inconclusive. You have to continue taking derivatives until you get to an even-ordered derivative that is non-zero. A forth derivative can conclude the same thing, given a second derivative of zero. If all even-ordered derivatives are zero, then you have an inflection point that coincides with a flat point on the graph, like you see for y=x^3. An inflection point is a location where the curvature switches from being concave-up to concave-down, or vice-versa.
Just wow
If for both Max and min you set the derivative to zero, how do you know if you’re getting the max or min?
2nd derivative test! Or sometimes it'll just be obvious.
If you have a quadratic, just look at the sign. + means you have a min and - means you have a max.
Your work is terrific but still need more in detail, if it’s possible please
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While differentiating the area, with respect to what we are differentiating? Help me with this.
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The area is being differentiated relative to the independent dimension that is to be determined. He's looking at the land area as a function of the x-dimension perpendicular to the river.
why is it that when x is positive, it is the local minima and when x is negative it the local maxima?
Whether x is positive or negative, has nothing to do with whether the point is a local maximum or minimum. The value of x just indicates what the horizontal position of the point is, representing the input of a function.
It is the second derivative d^2y/dx^2 that determines whether the point is local maximum or local minimum. And the reason is that the second derivative indicates the curvature of the function, and the sign of it indicates whether it is concave-up or concave-down. A positive second derivative indicates concave-up, and a negative second derivative indicates concave-down. A local maximum should slope down away from the maximum point, thus being concave-down, and vice-versa for a local minimum.
8:35 Not true. f(x) = x^3 at x = 0
Huh?
@@ProfessorDaveExplains It's not true that "when the derivative of a function is zero, there is a maximum or minimum at that point." I think you meant to say the converse, which is true provided the derivative exists.
I don't understand what you're saying. You cited x^3 at x = 0, where the derivative of the function is not zero. That's why there is not a local maximum or minimum there. I am not aware of any example where a derivative of zero does not indicate a local maximum or minimum other than a constant function.
@@ProfessorDaveExplains The derivative of f(x) = x^3 is f'(x) = 3x^2, and f'(0) = 0. It's a stationary point, but not a local max/min.
Shoot, I guess that's true. But is it an inflection point? When the derivative is zero there has to be some special behavior, no?
thankyou lord calculus
Done.
After learning calculus i am slowly becoming a MAN😊
He doesn't Explain why area would be maximum when the derivative of the function is 0. Does he understand why? Coz I don't know either. AND I DIDN'T SAY ONE PEOPLE ASK WHY DERIVATIVE OF A FUCNTION IS 0 WHEN IT'S MAXIUM VALUE
It’s 0 because it means the functions slope has stopped increasing, the tangent line at that point would be a flat line meaning the function must decrease afterwards.
wallah you re the best habibii
I laughed so hard at 8:49 for some reason 😂
thank god
I wrote it.
shen gennacvale dave
I'm too dumb for this
Thank you Science Jesus !!
Dear sir, how can i concrete Sketching derivatives function on the graph, i feel that i still need some more, will you do some other videos about this?
i have a tutorial on graphing functions and their derivatives! check it out.
My maths teacher taught me a great rule about this that can help you instantly figure out how much this was when for example have x+y=100 the maximum product of this two numbers will be 100 divided by 2 and this works every time.
As it did in x+y=400
i love you
The funny thing is i solved comp. without calculus so lol
Thank you Jesus
Will you do intergration (Calculus II)
Yep! Starts next week!
cool!
neat
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200,200
This is the first time I can't understand
👍
math jesus
:)
👍
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