Basic differentiation: second derivative

I'm currently teaching myself some differential equations by watching the MIT OCW series on the topic. In This video, at 21:50mins, the lecturer calculates the following derivatives:

1st $y'=x^2-y^2$

2nd $y''=2x-2yy'$

My simple question is, how he came to the second one. Is this a "total derivative" and why is it required? If I try to calculate the total derivative of y', I get:

$y''=(2x-1)dx+(1-2y)dy=2x dx - 2y dy$

I'm pretty sure that I made a silly mistake. Thanks for your help!

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This is implicit differentiation. –  Daniel R Jan 22 at 14:30

What's going on in the video, and in your posted problem, is what we refer to as implicit differentiation.

We view $y$ as a function of $x$, and thus, need to use the chain rule: $$y'(x) = x^2 - [y(x)]^2 \implies y''(x) = 2x - 2y(x)y'(x)$$ The author simply omits the parenthetical argument $(x)$: $$y'' = 2x - 2yy''$$

You can also see that your suggested answer just stops short one step: divide your $y''$ through by $dx$ and simplify! So you are not wrong; you "simply" haven't simplified!

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What's going on in the second part is the chain rule. $x^2$ has derivative $2x$, but $y^2$ is a function of $y$, which itself is a function of $x$. If you rewrite it as $$f(z) = z^2, \ \ z = g(x),$$ you get $(f \circ g)^{\prime}(x) = 2g(x)g^{\prime}(x)$ by the chain rule, and then replacing $g(x)$ with $y$ gives you the answer he got. You could also see this by writing $y(x)$ instead of $y$. As others have pointed out, the general technique is called "implicit differentiation", and the chain rule is applied in essentially every instance of implicit differentiation.

Also note that the work you did is, essentially, the same as the work he did. Divide through by $dx$ in your answer to see how the two answers correspond.

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