do u know what, I love you so much man. this is the type of teaching that I was looking for, explaning one by one, low key, man I love you so much thanks a lot I wish u had more of these classes becaise I can see on the list the rest of those are just 4 mins classes. but anyways perfect way of teaching very clear. thanks a lot
I always remember it in a similar order to the quotient rule's numerator, because then I won't screw up. For product, the order of addition doesn't matter. So my way is: f'(x)g(x) f(x)g'(x).
Some people may be in 8th and 9th grade struggling with this but I am in 5th and now that I have done a couple of these and rewatched the video I am really good at it and I have no problem with it obvious I learned this by myself and there isn’t calculus homework in 5th grade but I just wanna learn it now so then I have no problem with it later
In the last problem, towards the last step; why is it that you added only x's with similar exponents? That's where I got it wrong, you know, I simplified it altogether because I assumed the x as a base. Isn't that the way it should be?
13:42 Can i just type (3x+1)`(2x²+5x)+(3x+1)(2x²+5x)` This would be the same thing i think? but with more simple and easier to memorize (like it's just take all of it and add it to it self but the first time the first () is prime and second time the second () is prime) ??
I wish I understood step 4 of the last problem. -x^-2 + -2x^-2 = -x^-2. I get -2x^-2. I don't follow where the -2 went in your solution. Been trying figure it out.
If you look at the the entire thing, you just have to simplify (x^-2+x^-3) +(-2^-2+x^-3) You can rewrite it to match the powers for example (x^-2 - 2x^-2) + (x^-3 + x^-3) For (x^-2 - 2x^-2) you will get -x^-2 since x-2x = -1x then for (x^-3 + x^-3) you will get 2x^-3 since x+x = 2x and then you just rewrite as -x^-2+2x^-3 which is what he got If that helps
George C Arvanitis Sure there is. If your initial function is a position function, then it’s derivative is the velocity. The derivative of THAT is called the acceleration. So it definitely has physical meaning.
You can of course believe what you like, but it doesn’t mean you are correct. We have sent space probes to every planet in this solar system. I think we have a good handle on how calculus works for bodies in motion.