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!

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.
