I am sorry to hear you are getting confused its quite normal with this topic. I will try to revise the video to make things more clear but its the nature of the beast.
you can see it as a tree. Z in function of x and y en both are in function of s and t. so if you have d/ds(df/dx) i like to imagine you have the path from z to x and you want it in function of s. So you have d2z/d2x. dx/ds obviously but Z is also in function of y and y from s so you have a second path. thats the reason why you get the second term (d2z/(dx.dy).dx/ds). Sorry if i am wrong
Its referred to as Clairaut's theorem or Young's theorem also known as Schwarz's theorem. It looks a the symmetry of the 2nd order mixed partial derivatives of a continuous function
Don't skip the product rule step, it's the only problem in the example.
How does the product rule work at 1:31?
How do you get the part that is in red?? ate 3:12???
can you please explain the chaim rule at 2:37? i really dont understand...by the way good video bro..
For guys who don't understand the product rule, you use the method of A'B + B'A
Thank you, my professor's example left my brian dusty
Thank you :)
how do you get d2f/dydx for the chain rule part? how do you even get y?? very confused T.T
I am sorry to hear you are getting confused its quite normal with this topic. I will try to revise the video to make things more clear but its the nature of the beast.
you can see it as a tree. Z in function of x and y en both are in function of s and t. so if you have d/ds(df/dx) i like to imagine you have the path from z to x and you want it in function of s. So you have d2z/d2x. dx/ds obviously but Z is also in function of y and y from s so you have a second path. thats the reason why you get the second term (d2z/(dx.dy).dx/ds). Sorry if i am wrong
Treat dF/dx as a function, say G. dG/ds = (dG/dx dx/dt) + (dG/dy dy/dt). Replace G with dF/dx and you'll get what's in the first set of pink brackets.
Great help. Cheers
really helpful thank you sir
Guys like u save my career.
thank u very much.
waw very nice explanation sir.. thank u sooo much
Thanks💗💗💗
Thinking of the first chain rule df/ds as d/ds (f) and substituting df/dx and then df/dy helps
this SOOO much, really clears it up
Nice video sir
Can someone *PLEASE!!* explain to me the chain rule happening at 3:05. What is that????!!
i got no clue
That's fucken difficult equations 💔😭
I was about to drop my class. I'm glad I didn't cause this isn't too bad if well explained.
Thanks for great explanation with great detail!
Your hand Writing is so hard to read
Thaaaank youuu!!
bruh, why does the df/dx equal to f
it took me a while to get this thing my myself. I couldn't really find a video explaining this, so you could do one making it clearer
Thanks for this...
thank you glad it was helpful
you really are the man
8:49 who's theorem?
Its referred to as Clairaut's theorem or Young's theorem also known as Schwarz's theorem. It looks a the symmetry of the 2nd order mixed partial derivatives of a continuous function
Such a bad tutorial u are just doing urself on this problem without any explanation, no details around
unfortunately s very confusing
wow!
Your handwriting is barely legible
Yurrrrr good vid!
Thanks
Wtf