Implicit Differentiation | Differentiation when you only have an equation, not an explicit function

Description:
An equation like x^2 + y^2 =1, which plots a circle of radius 1, isn't nearly written as y=f(x) for a single explicit function f(x). How then can we compute derivatives, and thus slopes of tangent lines at all points? We develop the trick of implicit differentiation, thinking of y as some (often unknown) function of x called y(x) and then taking the derivative of both sides anyways. This lets us rearrange for y'(x).
Learning Objectives:
1) Compute the derivative of a function given implicitly.
Now it's your turn:
1) Summarize the big idea of this video in your own words
2) Write down anything you are unsure about to think about later
3) What questions for the future do you have? Where are we going with this content?
4) Can you come up with your own sample test problem on this material? Solve it!
Learning mathematics is best done by actually DOING mathematics. A video like this can only ever be a starting point. I might show you the basic ideas, definitions, formulas, and examples, but to truly master calculus means that you have to spend time - a lot of time! - sitting down and trying problems yourself, asking questions, and thinking about mathematics. So before you go on to the next video, pause and go THINK.
This video is part of a Calculus course taught by Dr. Trefor Bazett at the University of Cincinnati.
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[@FRC_CR]

I have to say up until this point the course was great but you seemingly assumed that from 4:03 adding some dydx randomly made sense. I have no clue how you got to that conclusion and you didn't really explain it. Can you help me please?

[@gordongoodwin6279]

Agree this part was poorly explained compared to everything else. The dy/dx is the derivative of the inside function (y), which flows from the chain rule. In other words, y(x)^2 is a composite function. The outside is the squaring act, so derivative of outside with respect to inside is 2y(x). Then we multiply it by the derivative of the inside. Well, the inside is just (y), so the derivative is dy/dx. Putting it all together we get 2y(x) multiplied by dy/dx

[@Celtics-x4w]

@@gordongoodwin6279h

[@gordongoodwin6279]

I love your videos but it would be great if you could re-do this one when you have time. As others have noted, it wasn't very clear how you arrived at the 2y(x) dy/dx as the derivative of the composite function. I understand it now but it wasn't clear in the video

[@NINJA-tf6bf]

Thank you very much ❤
Your explanation is mind blowing 🤯

[@SandeepKumar-qu9dh]

Happy teacher's day sir

[@studentsofmathematics8995]

dear which software are you used for presenting lecture? please guid me. thanks

[@mnada72]

Great 👍, very informative, thanks.

[@gustavopalma9451]

Nice explanation.

[@mohammadfallahzade2110]

You are a great Teacher thank you so much

[@andybrice2711]

Thinking of y as the function y(x) was very helpful here. I've never quite understood before why this can be treated as an application of the chain rule.

[@captainphysix]

great video!

[@eyesack3606]

Old Sheldon.

[@qualquan]

bad

[@florentinosanchez3969]

Very nice video

[@twisted4872]

That was a horrible explanation . . .you murdered it.

[@nebilzeleke3703]

are u edmonds cc student ????

[@NINJA-tf6bf]

You truly are twisted