Amazing Way to Graph the Gradient Function (Derivative)
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- Опубліковано 27 тра 2015
- It is likely that you have never seen this in all your schooling. Certainly, in our schools it seems to be an unknown skill.
Usually, when we are asked to draw a gradient function (graph of the derivative of a function), we are not asked for great accuracy.
We are expected to find the stationary points (locations of horizontal tangents). We are also expected to identify whether the gradient is positive or negative between each of those points of zero gradient. If the original graph has no identified points and no scale on the axes, any more detail than that is not required.
What you are about to see is a graphical/geometric method for drawing quite accurate gradient functions (graphs of derivatives), using a very simple geometric technique (parallel lines) and triangulation concept (gradient = rise/run).
I read about it many years ago in a publication of the NSW Department of Education called "The Mathematics Teacher." The contributer was G I Miller from Corowa (near the NSW/Victorian border). I tried to find out more about this person but could not. I think he deserves some cudos for having shared this method.
You may download a PDF copy of his article from my website via this link: www.crystalclearmaths.com/wp-c....
If you also wish to have a useful collection of functions to practise your differentiation skills, I have created a FREE PDF FILE containing a wide variety of exercises (and their solutions). You might like to work through them either on your own or with friends. You may download the file from here: www.crystalclearmaths.com/wp-c....
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For more information about mathematics or how to study, visit my website, Crystal Clear Mathematics at www.crystalclearmaths.com/
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Best wishes for your study and your mathematics!
Thank you.
7 years later and this is still helping people, What an amazing way to deal with the slope, a brilliant idea and a perfect explanation, Thank you very much sir.
Thank you very much for your kind words, Amer.
Kind regards from Australia!
Thank you very much. You explained this concept very clearly to me.
+Gabriel Lobo I am glad this helped, Gabriel. It is a very useful idea/concept.
Your feedback is very encouraging.
Best wishes for your studies.
I teach calculus and will include this in my class demonstrations, if not at least in my personal work. Thanks for hanging on to this concept and keeping it alive to pass it on. I'll keeping passing it on too now.
Thank you very much, BingtheLizard.
It is encouraging to find other teachers/instructors, like you, who wish to keep demonstrations of concepts like this alive for their students.
I greatly appreciate your feedback and encouragement and wish you well with your mission and burden to teach.
You may also find my work on asymptotes and hyperbolae (and their link to polynomials) interesting ~ ua-cam.com/video/zWBQNBypoOk/v-deo.html.
Warm regards from Australia and best wishes to you,
Graeme
I'm brand new to Differential Calculus and often find graphing helps me understand new concepts. This is far easier to understand than how my lecturers are explaining it. Nice one for the toolbox and thanks!
You are welcome, Al, and thank you very much for providing me with such encouraging feedback. It always helps to know that my videos are of use to people.
Like you, I am a very visual learner. Graphing is a wonderful skill to develop and it is especially useful in calculus. Every so often, when you have solved a particularly interesting or unusual or difficult problem, it is a good idea to take the time to graph the function. Not only is the exercise valuable, but the insights that you gain will help you enormously in your mathematical development.
Kind regards to you from Australia!
Absolutely brilliant and clear. Many thanks.
You are very welcome, SKY JUMP.
I am glad that the video was useful to you (and clear!).
Kind regards from Australia!
Crystal Clear, indeed! Thank you!!
You are welcome, DdaHienster.
I am glad that it was so clear to you.
Excellent stuff Sir. Very much appreciated and an unusual shortcut I've not come across before.
Thank you, Paul. I'm glad that you liked it.
What I like is that it uses your understanding of gradients to map the gradient function in a very direct way.
Kind regards to you and thank you for leaving your comment/feedback!
Pretty cool and modest way!
thank you so much for your impressive concern!
You are welcome, דסה Dessie‘l.
I am glad that you found the video helpful.
This is splendid - thank you.
You are very welcome, Doug.
I'm glad that you found it helpful.
Very thorough and intuitive. Thank you.
Thank you, Sri Goli. I am glad that you enjoyed the video.
Kind regards to you.
Awesome! keep up with the good work! Thank you!
You are welcome, Merry Hailu. Thank you for your encouragment.
Hey Graeme
just what I was looking for! Super clear - love it.
You are an amazing teacher!!
+queenforever Thank you very much, indeed, QF.
You are such an amazing encourager (and your enthusiasm is greatly appreciated)!
I only just found your message because there was a period of 2-3 week where UA-cam did not notify me of any comments that were made on any videos ... so I am sorry for the delay.
Warm regards to you and your family.
Graeme
+Crystal Clear Maths Keep posting more videos Graeme - they help so many people :-) A detour via linear algebra for me but will revise calculus and work my way thru the Coroneos integrals in a few weeks. Warm regards to you and family too! G
+queenforever Hello, again, G :-).
I wish to produce an entire series of videos about linear algebra, and want to finish Jim Coroneos' integrals soon. Of course, there are many other ideas 'in the pipeline' too!
I anticipate resuming within the next 24 hours.
My family will be glad to hear of your greetings and look forward to visiting some time.
Warm regards,
Graeme
+Crystal Clear Maths YAY! for linear algebra - I really look forward to your new videos. Go Graeme!!
+queenforever You are full of surprises, QF! :-) Thank you.
Best explanation. It is really crystal clear. !!! Thank you soo much. I’m from South Australia and I have test this week.
I'm glad you liked it, Captain.
All the best for your test this week.
best explanation on youtube... thank's for helping me for my test tomorrow.
Thank you, Shyzah.
Best wishes for your mathematical future (and your test tomorrow)!
Please keep in touch and let me know how you go.
I'm so grateful for this wonderful method. My students loved it and felt confident in using it to find any derivative.
I am delighted to learn that this insight is helping students (and teachers) elsewhere in the world, Izabela.
Thank you so much for taking the time to let me know that your students are finding it useful.
Although I have not added anything for about five years (due to some life crises), your students may find useful material at my website (crystalclearmaths.com/) and my FaceBook page (facebook.com/crystalclearmathematics/). Everything is free on both platforms. There are no advertisements and no data kept or used. I hope to resume adding material around August this year.
Very kind regards to you from Australia.
Graeme
thank you so much, so easy to follow.
You are very welcome, Isabel :-)
Brilliant explanation, thank you very much.
You are welcome, Reyadh.
Thank you for your feedback.
Clear knowledge... 👍 Thanks for this wonderful lecture .❤️
Thank you very much, Ramiz.
I am glad that you enjoyed the video.
Thank you very much for taking the time to send me this encouraging message. It is appreciated.
Kind regards to you from Australia.
Thank you for the clear explanation.
You are very welcome, friend.
Thank you for leaving your message.
Kind regards from Australia.
Very nice. Thank you for the video!!
You are welcome, Ace. I'm glad that you liked it :-).
You are blessed with Talent in explaining and working with mathematics.
Thank you, Agam. I am glad that you enjoyed the video :-)
thank you so much!!! this is something that has confused me all term and now that i have an exam tomorrow - so thankful i found this before i took it. thanks again!
All the best for your exam, XGG. I am glad that my video was of use to you :-).
Excellent,superb,thanx sir,thank u very much.
You are welcome, Subodh ... and thank you for your very kind comment.
Wonderful. Sir, I really appreciate your way of teaching. U r an excellent teacher.
Thank you, Mahendra.
I am glad that my videos are making a difference for you.
Very best wishes for your studies.
That's very useful, thank you.
You are very welcome, Chen Yanyu.
Very best wishes to you.
Graeme
I have never seen this before and I will be sharing it with my teaching colleagues.
I am glad, Mrs Ng, to learn that this will be used among your colleagues. It is a useful way of incorporating our understanding of gradient in graphing the gradient function.
Thank you very much for letting me know. Your feedback is most encouraging.
Thanks a lot sir!
I was searching for this kind of explanation since 3 days and finally i found. Thank you so much! Stay blessed sir!
You are welcome, Muhammad.
I am glad that my video was able to answer your questions.
Grace and peace to you!
@@CrystalClearMaths thanks for your kind cooperation sir!
@@muhammadirfan8035 You are very welcome, Muhammad.
This is an even a better explanation than the one I just saw b4 presented by a lass at MIT lectures...
Thank you, Stephen.
thanks man appreciate this a lot :)
You are welcome, Raheem!
I appreciate your leaving feedback for me (and others). Thank you.
Thanks heaps - really helpful
You are welcome, Jedh.
Clear in the brain clear in speach bravo
Thank you, friend. It is good to 'hear' from you again. I hope all is going well.
BIG THANKS
You are very welcome, fe4rlesssimo.
This is so good, it's evil. Thanks so much for sharing. And I've discovered something thorugh my studies. There's a property of functions called concave-ity (I would guess) which can be extrapolated from the second derivative, and I think it's amazing. Any thought on that topic? Also, great to see you making new videos, and just constantly pumping them out. Also, I want to discuss a certain topic with you, but I'll only have time for that after I came back. Any suggestions where to talk about that? It's not a particularly beefy topic, and it has to do with suggestions about carreer and my thoughts on that, and wether you can help me out with that. I don't know, talking about that under a video like this would be a bit out of place.
~Steve
+The Skooma Cat Hi Steve. I'm glad that you liked this video, too.
The second derviative of a function does, as you say, provide a measure of its 'concavity' (or 'curvature'). I will be producing videos, eventually, that will help in understanding this. I am not sure when I will start, but I would like to complete the Coroneos Integrals first.
If you wish to send me a private message, just click on the [About] button above and then on the [Messages] button to the right of the screen. Messages sent this way are not public ... so we can chat about your career matters without others seeing any of the conversation.
Best wishes, friend ... and I look forward to hearing from you.
Graeme
Aussie ! Aussie Aussie! So relevant for my syllabus. As the name indicates, he is crystal clear in his approach / explanation. Thank you very much.
Oi, oi, oi!
I am glad that this video helped you, SC.
Thank you very much for leaving your comment. It really helps and is appreciated.
Kind regards and best wishes for 2021!
@@CrystalClearMaths Thank you Sir ! Best wishes to you too.
@@staycalm9979 Thank you very much, SC.
Thank you sir
You are welcome, Roshay.
I appreciate you taking the time to send me this message.
Kind regards to you.
this is veeeeeeeeery amazing. it is the integration of easy, accurate and assumable which i have never been able to combine in any math topic... thank you sir for this piece of information... i will sure pass this along
Hi dope geek. I am glad that this video helped you on your mathematical journey.
Thank you for letting me know, and best wishes to you!
Thanks! From a fellow Aussie!
Excellent! You are welcome, Azfar.
Best wishes to you.
So much easier than memorizing a bunch of rules or trying to intuitively map it, much appreciation from a old dog just getting back into maths :)
Thank you very much from another 'old dog' :-).
It is always better to understand the principles. Memorising rules with no (or little) understanding is of little value (in my opinion).
I greatly appreciate your comment, BCT, and your taking the time to leave it here. Thank you.
Thanks I enjoyed that
You are welcome, Dwight.
Do you know of any videos on drawing the parent function from a graph of the second derivative without y values?
At this stage, I have not produced such videos, Tim. They are "in the pipeline" but I have had some major family events intervene and my "studio" is no longer available. I am hoping that I will be producing videos again before the middle of next year.
The main thing to remember is that, when graphing a parent function, all zero values for your current function (i.e. where it encounters the horizontal axis) become stationary points for the parent function. In order to determine whether those stationary points are maxima, minima or inflections, you will need to determine the gradient between each stationary point. I won't go into more detail here ... otherwise my reponse will become a long essay.
Hopefully, what I said makes sense to you ... and I hope that you have friends/mentors/teachers who can help you master this very useful skill.
Kind regards to you,
Graeme
very good, coherent presentation of drawing a detailed derivative function. Could we see it via a graphics software? Like desmos or geogebra? Congratulations, one again
Thank you, adminassos.
I had not thought to demonstrate this using Desmos (for example). That is an excellent idea!
Unfortunately, due to personal circumstances, I will not be posting videos for some time. I hope to resume around the end of the year or early next year.
I will add your suggestion to my 'to do' list.
I appreciate your input and your encouraging support. Thank you.
Kind regards,
Graeme
Very clear thanks I have a test tomorrow
You are welcome, Fly Guy.
I am glad that the video helped you and wish you well for your test!
Great
Thank you, mathematicaATD.
That's so amazing taking -1 on x axis for getting the gradient 👍👍
Thank you, Biswaranjan. It is simply "rise over run," isn't it?
I remember, when I first saw this, how simple and amazing and obvious it seemed.
Thank you for commenting. I am glad that you like the video.
yep thanks ..but how do you make an equation for a curved line done at random ..is it possible?
I don't believe it is possible, johnpro2. Does a formula exist for a close approximation to any finite random curved line? Probably. Do we have techniques or skills to find such an equation? No.
We do have some skills. I will be producing another video explaining how to generate a polynomial to pass through a finite number of points that are equally spaced horizontally (along the x-axis). Generating a polynomial for a finite number of randomly scattered points would be another matter entirely. Generating other function (or relation) types can be exceedingly complicated.
Unfortunately, it will be another 6-12 months before I have access to my studio and the time to resume producing videos. If you can hold off until then, you should see this polynomial video appear as one of the first that I will produce.
Kind regards and thank you for leaving your comment and question.
Brilliant !!!!!
Thank you, Felix. I am glad that this was useful to you.
Kind regards,
Graeme
Crystal Clear Maths Hi My Grahams, I am glad you saw my comment. I was wondering if you could help me with a project idea to do with this course. I will happily contribute in donations to your website. Do you mind me sending you an email? Could you provide me with an address please?
I would like to help but, unfortunately, my real-life commitments at the moment are great. It is unlikely that I will be producing any more material or be able to help via the Internet for another year. I have quite a few other matters that I must deal with first. I am sorry, but I am unable to help you at this time.
At the moment, all that I manage to do on UA-cam (or FaceBook) is reply to comments.
Crystal Clear Maths thank you still and I appreciate you taking the time to reply
Very useful!
You are welcome, X00000370.
Does your 'name' (370) relate to cubic numbers?
@@CrystalClearMaths Nothing that significant, it refers to an old college ID number. I guess "X-numbers" kept everyone's personal information a bit more concealed...
@@X00000370 Codes are interesting :-)
I had wondered whether you knew of the 'game' of summing the cubes of digits within a number.
E.g., if you choose a random number, say 83, then the next iteration would be 8³ + 3³ = 64 + 27 = 91. Then 91 gives rise to 9³ + 1³ = 729 + 1 = 730.
Then 730 gives rise to 7³ + 3³ + 0³ = 343 + 27 + 0 = 370. and, finally, 370 gives rise to 3³ + 7³ + 0³ = 27 + 343 + 0 = 370, and you see that we will simply cycle away on the number 370.
It turns out that there are three other such numbers, all of them three digits.
Just something interesting about your college ID number :-) ...
@@CrystalClearMaths X number 370 and a connection to the "'game' of summing the cubes of digits within a number"...I had no idea. I guess they should have reserved this number for one of the Mathematics Professors.
@@X00000370 Hahaha :-)
love your vids and i love your crystal clear hair. is that how you got the youtube name "crystal clear maths" because of your hair? because i love it xoxo
Hahahahahaha ... I never thought of that 'angle,' Jack.
I think my hair is the result of age and genetics.
"Crystal Clear Mathematics" was named this way because I hoped that students would understand and practise their mathematics until it was "crystal clear" to them. I hope my explanations are a "crystal clear" as possible.
I am glad that you love the videos (and appreciate your letting me know). Thank you.
jeeeeez like thank you- this was impossible before I found this video
Hi, fwb.
I am glad that this video cleared up a few things for you! May you continue to soar with your calculus.
Thank you very much for your encouraging comment/feedback.
Best.
Thank you.
I have a question sir, if gradient is increasing and approaches zero then why we are making positive curve for gradient i mean in curve 1st gradient will increase till max. value and then gradually drops to zero. while when we look at the graph of function then gradient is on increasing for that curve. I hope you will get my point and will tell the answer as early as possible.
Hello, Muhammad.
As the (continuous) curve increases towards a maximum, the gradients will be positive and will gradually DECREASE towards zero. The function values will increase, but the gradients will be decreasing. It is very important that you understand the distinction between the function value and the gradient at each point on the curve.
I hope I have understood your point and that my answer is not too brief or too difficult to understand.
Best wishes to you,
Graeme
Thank you once again sir..!
@@muhammadirfan8035 It is a pleasure to help you, Muhammed.
Kind regards,
Graeme
Nice trick!
Thank you, Djimm.
Kind regards from Australia :-)
@@CrystalClearMaths You are welcome, from western new york usa.
I am starting an electrician apprenticeship soon and want to pursue the study of electromagnetism, which has led me to the gradient, and hence your video!
@@djimms5644 Excellent! I wish you well with your venture, Djimm S. Including electromagnetism specifically in an electrician's apprenticeship course sounds as though you already have a direction in which you plan to head. I hope it gives you a bright future.
During the strange days in which we live, you are probably a lot safer in western New York rather than NYC. Stay safe!
Kind regards to you.
@@CrystalClearMaths Thank you. Yes, there are less than 50 cases of the covid in my county and we havent been affected really.
I look forward to viewing some more videos on your channel. Looks like you specialize in math. My wife and I homeschool our children. Perhaps I can find some material on your channel!
@@djimms5644 My wife and I home schooled our daughter for over six years ... wonderful days. In Australia, the profession most represented (proportionally) among home schooling parents are teachers! We know what the system is like.
I have about 240 videos on my channel and they range in mathematical difficulty from early primary to university level. You will find some material that will interest your children.
I have not added anything for some years due to a cascade of family and private matters. I hope to resume in the next year or so, but a lot of other things are requiring my attention at the moment. It will happen!
Best wishes to you and your family.
just out of curiosity which mathematics magazine are you referring to
Hi fwb. Sorry it took so long for me to respond.
It's good to chat with you again (I am sure that I recognise your name).
The magazine that I referred to was (I think) produced by the NSW Department of Education here in Australia (during the 1950s to 1980s). It was called the "Mathematics Bulletin." The article in question was in Volume 11, pp 23-26.
There is little reference to this magazine online, so I may attempt to digitise and post it on my website if I can get permission to do so ... i.e. if it is out of copyright or if the Department is happy for me to do this.
Thank you for asking.
PS ... if you wish to contact me via my website and leave your e-mail address, I will send you a copy of the article.
@@CrystalClearMaths definitely will connect through email!
@@funwithbrittanyofficial Wonderful!
awesome this reminds me my graphics lecture during engg !!!!
+Abhi Mazire I'm glad that you liked it, Abhi, and appreciate your letting me know!
Best wishes to you,
Graeme
Hi good video, but i have a question: how would you use that technique if the original f(x) graph also had negative x values. The one in the video does not and i was wondering if it would work with negative x values.
Hello Varuos. That is an excellent question! I should have addressed that matter in the video.
You will find that the same procedure applies. From the point (0,-1) on the derivative graph, simply draw a line parallel to the tangent on the original graph and the y-intercept will represent the gradient of the tangent. This will become the y-coordinate on the derivative graph in the same way that it would for the positive x-axis.
Cheers bud you're a great teacher :)
You are welcome, Varuos ... and thank you!
would you please explain Varuos' question with one more video.
Also how about going from the derivative graph to a function graph.
There is not (2) I.P. points on d^2y/dx^2 of M though. There is one point and it is of undulation only at a vertex on f". I think I have a book error here.
I did not get as far as drawing a graph of the second derivative, Joseph. I am impressed with your further application of this idea.
Unfortunately, I am not certain that I understand your observation. I think this graphing method is sufficiently imprecise (it is an approximation) that it may not pick up the subtleties of small variations in gradient and, therefore, may miss some inflection points.
Kind regards from Australia.
this is amazing way of teaching you are a great teacher sir but why x=-1 is taken can we take -3 also please explain sir i have doubt
Thank you, Sahel.
We use x = -1 so that the base of the triangle is one unit long. Thus means that, when we show the gradient (rise/run), the run will always be one unit.
This makes all our calculating and our geometry so much easier.
I hope I have understood and answered your question.
Kind regards from Australia.
Sir how we consider m=2/3
A gradient of 2/3 would be reproduced by locating one point on the x-axis at (-1,0) and the other on the y-axis at (0,2/3).
This means that, if the original function has a tangent gradient of 2/3 for some value of x ... then the gradient function will have a value of 2/3 for that same value of x (as shown in the video).
I like that he doesn't plot f and f' on the same coordinate plane as it is often done. It's a bad habit. Just imagine that x is time and y=f(x) is the location, then f' is the velocity. The units for the y-axis don't match!
Thank you, CI.
Your observation about the units is quite correct. I appreciate your taking the time to leave a (very pertinent) comment.
It's a Drafting Skill projecting points I get it! Mmmmmmmmmmmmmmmmmm lol lmao
It is a good drafting skill, Joseph, isn't it? It helps students understand and apply the concept of gradient being a rise/run to creating a gradient curve ... a very direct application, and a good teaching/learning tool.
I wish I could press 1000 like .
Thank you, Fatema 🙂.
Kind regards and thank you for taking the time to leave such a lovely comment.
Absurdly tedious
Perhaps, but insightful.
f is negative quartic graph, power n=4, so f' is cubic graph. power n=3. but lecturer draw a strange graph and power n over 4. that's misleading students.
The second graph certainly LOOKS like a negative quartic graph, but it is not necessarily one. I randomly sketched it and no formula was provided.
The object of the video was to explain and teach a principle or method for sketching gradient functions when the formula for the original function is unknown.
It is quite possible for a graph of that shape (not a negative quartic) to have a gradient function as I depicted.
You are, however, quite correct in this ... IF it was a negative quartic graph, the derivative function would most certainly be a cubic function.