Multivariable Calculus 13 | Schwarz's Theorem [dark version]

[@hearteyedgirl]

i don't understand 5:08
why (f(h_1, h_2) - f(h-1, 0))/h_1 then let h_1 approach to 0 will equal to d/dx_1 f(x_1, x_2) at (0, 0)
isn't d/dx_1 = (f(h_1, h_2) - f(0, h_2))/h_1 as h_1 approach 0?

[@brightsideofmaths]

Do you mean the application of the mean value theorem here?

[@lazarusunkwon6]

Nailed it! Outstanding work!

[@brightsideofmaths]

Thank you! Cheers!

[@vitalivorbenko670]

Nice trick!

[@brightsideofmaths]

Thanks! :D

[@melodii8377]

nice german video

[@brightsideofmaths]

Thanks, I guess.

[@Just_kundan]

Use two variables as x and y to make it look easy

[@brightsideofmaths]

We have n variables :)