Differentiation (Maxima and Minima)
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- Опубликовано: 25 авг 2024
- Exam Questions: www.1stclassma...
In this video I explain how to find stationary points and determine their nature. This video is intended for those studying AQA's Level 2 Further Maths GCSE.
Good luck to everyone who has the exam tomorrow!!! Youre going to do great!
Help I didnt know anything at the beginning of today I never attended any of my further maths lessons 😢
@@sumayyahkhan8897good luck o7
thanks mate
THANK YOU SO MUCH!
Thank you!
it’s the night before the exam and you’re saving my life like jesus THANK YOU SO MUCH.
Good luck!!
@@1stClassMathsI GOT AN 8 THANK YOU SO MUCHHH
@@mutilatettenicee❤
@@mutilatettewas that further maths
Thank u btw Jai shree ram
I just want to say I really appreciate your hard work in helping us with the exams. Your content is informative,very well explained and captivating . On behalf of all my classmates thank you
It's my pleasure! Good luck tomorrow.
THE CARRY of 1stclass maths getting me that 9
GL
Stationary points came up today as a 5 marker. We didn't go over stationary points in any detail in school, but thanks to your videos on differentiation, i was able to do the question incredibly easily. Thank you again for these great videos!
That is awesome!
I jus wanna thank you brother u have jus explained me the entire concept in 8 mins where my teacher takes 40 mins
U are very efficient 😢❤❤
You are most welcome
Finally all my doubts got cleared with the help of this video. thankyou so much sir
My anxiety after understanding this whole concept 📉,it was crystal clear,thankyouu!
your expalanaion is too good
dy/dx of negative power gives negative power since you subtract 1 to the power and multiply base by the original power e.g x^-1: drop -1 to have -1x and subtract 1 from -1 to get power -2 combining the two the full derivative is -1x^-2 giving you -1 multiplying x^-2 or -1/x^2. I used power rule and laws of indices to write negative power x^-2 to positive power x^2
Thank you so muchhh .... It really helped me to understand.... I thought that this topic was hard to understand but you just made it a lott simple and easy to understand!!!!! ☺☺☺☺
You're welcome!
Thank you so much 😊it made my topic crystal clear✨
Glad it helped!
literal legend
This video was so helpful. Thank you so much
my factoring skills are trash, I'm mixing up the signs a lot
Thanks!
A very helpful video . Thank you. I've just subscribed .
Awesome, thank you!
Thank you. It clearly explains the whole concept.
You are welcome!
Great content, keep it up
Appreciate it!
Thank you sir
Great explanation but i have a question,
What if there are more than 2 stationary points, suppose i have to find maxima and i get three values of x for which first derivative is equal to zero,
and for two values of x, second derivative is negative.
Then how do i find on which value of x would there be maxima?
Do i have to put both of these value in original function to check which one gives greater value, or is there any other way to know?
Like the example you gave in starting of video at 0:34
Hi. You can have multiple maxima and minima. When we say "maxima" we are talking about a localised maximum point rather than the maximum for the entire function/graph.
So the maxima is just where the graph reachs a peak then comes back down again but this could happen multiple times. E.g. the sin/cos graphs have infinitely many maxima and minima.
What if the second derivative is equal 0? is it a maxima or minima?
Great question. In this situation more investigation is needed. It could be either or also a point of inflection. This is not needed for this course though so I have not covered it in this video.
Neither maxima nor minima
@@premanandapatel4042 incorrect it could still be either.
6:38 can you tell me how you got -1/x^2 when differentiating x^-1 plz
Multiply by -1 so we have -x but reduce the power from -1 to -2 so it is -X^-2 but due to negative indices x^-2 = 1/x^2
Yes power rule and laws of indices combined.
thank uuuu
Thank you!
You're welcome!
@@1stClassMaths The exam went really good thanks to you :)
Excellent!!
what to do when 2nd derivative is equals to Zero?
This is not needed for this course but is at A-Level. If it is 0 you need to check the gradient either side of the point as it max be either or a point of inflection.
Sir , i kindly request please explain the step 3xsquare -12x +9=0 to xsquare-4x+3=0 to( x-1)(x-3)=0 . Please explain these three steps sir 🙏🏻
You want to look up factorising quadratic expressions. I haven't got a video on this yet but I will one day :)
@@1stClassMaths thanks sir 🙏🏻
You just split (-12) into such a way that when added it gives (-12) and when multiplied it gives the product of 3 and 9.
This is called factorisation.
We split 12 into -9 and -3. Now adding these both will give the number that we split i.e., -12 and when multiplied we get 27.
The new equation would be 3x² - 9x -3x + 9 = 0. Let's simplify it by dividing LHS and RHS by 3. So we got x² - 3x - x + 3 = 0.
Let's take x² -3x and make it x(x-3) by taking common x out. Now let's so the same with -x + 3 , making it -1 ( x-3). Now remember our aim to do this was to make the contents of both the brackets same, which we did.
So the final equation would be,
(X-1) (X-3) = 0.
Thus putting each bracket equal to 0 we got x=1 and x=3.
derivative=0 is inflection point
This is not correct I am afraid. I assume you're referring to the second derivative but this is also not correct.
For example if y = x^4 then the second derivative is equal to 0 when x = 0 but this is a minimum point.
If the second derivative is 0 then it *could* be an inflection point but it not guaranteed. Furthermore inflection points do not have to be stationary points.
i am so confused in skl we got taught that you only find the first derivative and find out the stationary point then calculate the gradient before and after the stationary point to determine the nature but here it says to find the second derivative . do both ways work or no ? pls help me🥲
Hi. The method you have been told is a valid way of checking for the nature of points but perhaps lacks rigour compared to using the second derivative. It makes the assumption that between the values you select either side of the stationary point nothing crazy happens e.g. a discontinuity or another stationary point.
The teaching guidance says for specfication point 4.7 states:
"prove whether a point at which the gradient is zero is a maximum or minimum point using either increasing/decreasing functions or d^2y/dx^2"
I think it would be worth learning this method as you will need to know what the second derivative is come A level maths, if you chose to do it. The other method will be absolutely fine for this exam though :D
@@1stClassMaths oh okay thankyouu :D