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.
For guys who don't understand the product rule, you use the method of A'B + B'A
How does the product rule work at 1:31?
How do you get the part that is in red?? ate 3:12???
Thank you, my professor's example left my brian dusty
can you please explain the chaim rule at 2:37? i really dont understand...by the way good video bro..
waw very nice explanation sir.. thank u sooo much
Guys like u save my career.
Great help. Cheers
really helpful thank you sir
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.
Thanks for great explanation with great detail!
thank u very much.
Thaaaank youuu!!
Nice video sir
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
Thanks💗💗💗
That's fucken difficult equations 💔😭
Thank you :)
I was about to drop my class. I'm glad I didn't cause this isn't too bad if well explained.
Can someone *PLEASE!!* explain to me the chain rule happening at 3:05. What is that????!!
i got no clue
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
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
you really are the man
Thanks for this...
thank you glad it was helpful
Yurrrrr good vid!
wow!
Your hand Writing is so hard to read
unfortunately s very confusing
Thanks
Wtf
Such a bad tutorial u are just doing urself on this problem without any explanation, no details around
Your handwriting is barely legible