Multivariable Calculus | Implicit derivatives with the chain rule.

[@drewmacha]

At minute 2:03... y is also dependent on x! Why don't we draw little branches down so that y is a function of x and y? Then the derivative of F with respect to x has an additional term, namely, partial F / partial y * dy/dx. (I know your solution is correct, I just can't seem to justify in my mind why we ignore the y dependence on x.)

[@drewmacha]

Nevermind... I realized we are looking for the *partial* derivative of z with respect to x, not the ordinary derivative of z with respect to x.

[@lionelronaldo1574]

Great teacher can you include videos on special functions?

[@eugenesiu9226]

This is really helpful. Many thx!

[@junxionglu4079]

Algebraic mistake at 3:36. Should be -Fx/Fz

[@u12bdragon]

Yes this mistake cost me over 45 minutes of stress. I have an exam 7am tomorrow, and not I have studied 45mins less things.

[@Mario_Altare]

Excellent!

[@kash8519]

Common Michael Penn W