Thank you! I've been trying to get a more intuitive understanding of this rule because I don't like the blind 'plug and chug' formula thing. I'm still not 100% sure on it, but I feel a lot closer to understanding.
Nicely explained! Something worth mentioning in the video though is why you multiply the derivatives, instead of, say, add or subtract them; in 2:06 you say it's made of 2 smaller processes and then you just write them next to each other but it's not very clear why multiplication is the right action. If one of the derivatives is twice as big, then the whole derivative should, too, which makes sense but is not a very trivial argument. l think it'd be better if you added some note about it, either in the description or as an annotation. Thanks!
This makes way more sense. I'm semi-studying for Calc next year and I'm in Honors Algebra II right now. I've seen like 3 videos on the chain rule and they never explain it. They just say it as it is. Thank youuuuu
Excellent. Thank you. I've watched a number of chain rule videos, and this is the first time it clicked really for me. I sort of understood it, but didn't feel comfortable. Now it makes sense to me. I had a bit of trouble following the triple, but I'll watch it again and maybe write it out with the f, g, k from the previous part.
I keep referring to this video as it makes the chain rule much easier to understand. I now understand the "mechanics" of what is going on. The only confusing aspect is (for example on the first equation) the number 4 is still added to the inner equation which still goes to 3 but we also have the derivative of 2 from the first input so we are using 4 and 2 rather than just one of the numbers
this makes a while lot of sense now. I was trying to find the d/dx of ln(sinx). Which, going by what this says, spit me out cotx, which is correct. Thanks!
With regards to my earlier question about where do you get 2x from - I see that you multiplied the exponent by the inside function of "X" or "U" depending upon notation
Mr Eric, is there a way we can donate as a token of our gratitude? This may also help you continue with your good work. Please start a patreon account :)
I have a question that may come up as somehow dumb, but, Isn't (2x - 5)² the same as (2x - 5) × (2x - 5) .. and just using the product rule is supposed to work it out?
i was going to give you a thumbs up anyway, but i thought it was cool when it went from 199 to 200. i kinda wasn't paying attention to the 199 but i sure noticed that it was now a nice round 200. i know, i get too excited over simple numbers, but hey, this calculus stuff is too fancy for me.
The multiplication has primarily to do with how the functions are linked to one another. When two functions are combined using composition the output of one becomes the input of the next one. input -> g -> f -> output So any change that happens in the first function potentially get amplified when it arrives at the second function. A good "real world" example of this is two gears that are connected. If we double the turning speed of the first gear, the result will multiply the rotation of the second gear accordingly. Derivatives are about keeping track of this rate of change, so with composition we see the derivatives are multiplied. Hopefully that helps out a bit. :^D
It was power rule, which states that the derivative of a power function is nx^(n-1) if n is the initial power, so x^2 becomes 2x^(2-1). That simplifies to 2x^1 or just 2x.
Harvey Williams try this use dy/dx and use the intermediate variable z that cancels when thinking of these as fractions (dz/dx)(dy/dz) so what's happening here is ur taking the derivative of the intermediate variable in terms of x multiplied by the derivative of y in terms of z
Harvey Williams so if you had lets say f(g(x)) and you wanted to find the derivative of that you would make z=g(x) and find it like such f'(z)=(f'(z))(z') then replace z with g(x) which we can do because we said z=g(x) so f'(g(x))=(f(g(x)))(g'(x)) apply this to your function and bam u got it
Thanks :) I realised I was actually just looking for a really consise way of explaining why normal differentiation works (rather than taking it as a given in this video)
