thanks!! this was really helpful. i hope you can do for the whole differentiation chapter bcs i really need help. chain rule, product rule, quotient rule, basically everything in the form 4 syllabus. really really would appreciate.
I know this is late. But the concept revolving around differentiation is about finding the gradient on a specific point on a graph. But before finding the gradient on a specific point, we first need to find a gradient of a line that is between two points of the graph. We'll say the horizontal distance between the two points is δx and the vertical distance is δy. By doing this, you should know that the gradient on this line is the vertical distance divided by the horizontal distance. i.e. δy/δx. But, the δx and δy is still unknown to us, we don't know the actual value of these two distances. And you can't just plug these into the function itself, as the function only works for one specific point (an object) only, not the whole distance. As such, we'll need to try to find our δx and δy by doing something on the Cartesian Plane. Let's say, there's a point on a graph of the function with the coordinate (x, y). Then, in order to find the gradient for a line in the graph, as I said before, you'll need two points, and now we just need to find where that second point is. First, let's put our second point δx units away (horizontally) from our first point, this causes the x-coordinate of the second point to be (x + δx). Now, in order to find the y-coordinate of this point, we'll need to plug in the x-coordinate of our second point into our function. And that way, we'll get our y-coordinate. And we'll say that it's δy units away (vertically) from our first point. This causes the same effect as before, the y-coordinate of our second point is (y + δy). So in conclusion, (y + δy) is just the image of the object (x + δx). Just like you're saying y is the image of the object x. I hope this clears it up for you :) If you still have any questions, just ask. I'll try to answer them asap.
Its so clear for me n much more understand than my teacher..thank you..
Thank you so much. I definitely love your 4 methods, which makes first principles of differentiation becomes much easier.
thanks!! this was really helpful. i hope you can do for the whole differentiation chapter bcs i really need help. chain rule, product rule, quotient rule, basically everything in the form 4 syllabus. really really would appreciate.
thank you so much.I'm truly understand...Glad i watch it before my spm this wednesday😁
Thank u so much..I am using all ur video to revise my form 4 chapter as I am in form 5...Tq so much...it helps me a lot...👏👏
sir is it okay to use the method that you explained if the formula f(x+h)-f(x) is given?
thank you !!! i really understandddd !!
Is it important to do the simulteneous?
Great!!! Really helpful !
thanks sirr
thank you so much boss i love you
thanks so much !!! it is reaaaally helpful!!!!!!!
this was dope bruh, really enjoyed the way you show why dy/dx
Thank you!
老师你好，请问一下，有没有中文版的。
plz solve x^2-6/3x by 1st principle
Tchr how about the question for the y is kuasa tiga
Sie why must be adding dy and dx after y?
I know this is late. But the concept revolving around differentiation is about finding the gradient on a specific point on a graph.
But before finding the gradient on a specific point, we first need to find a gradient of a line that is between two points of the graph. We'll say the horizontal distance between the two points is δx and the vertical distance is δy.
By doing this, you should know that the gradient on this line is the vertical distance divided by the horizontal distance. i.e. δy/δx.
But, the δx and δy is still unknown to us, we don't know the actual value of these two distances. And you can't just plug these into the function itself, as the function only works for one specific point (an object) only, not the whole distance.
As such, we'll need to try to find our δx and δy by doing something on the Cartesian Plane.
Let's say, there's a point on a graph of the function with the coordinate (x, y). Then, in order to find the gradient for a line in the graph, as I said before, you'll need two points, and now we just need to find where that second point is.
First, let's put our second point δx units away (horizontally) from our first point, this causes the x-coordinate of the second point to be (x + δx). Now, in order to find the y-coordinate of this point, we'll need to plug in the x-coordinate of our second point into our function. And that way, we'll get our y-coordinate. And we'll say that it's δy units away (vertically) from our first point. This causes the same effect as before, the y-coordinate of our second point is (y + δy).
So in conclusion, (y + δy) is just the image of the object (x + δx). Just like you're saying y is the image of the object x.
I hope this clears it up for you :)
If you still have any questions, just ask. I'll try to answer them asap.
@@kevinlim3395 understood thanks
Thx
dude ur stuff z on point buh tht flue z disgustin
But his knowledge can easily overpower you
i cant hear clearly.. anyway thank you :D