We started with x2 + y 2 = 25 Then we take the derivative of both sides: d d 2 d d 2 (x + y 2 ) = (25) =⇒ (x ) + (y 2 ) = 0 dx dx dx dx d(y 2 ) dy =⇒ 2x + =0 dy dx | {z } chain rule 0 =⇒ 2x + 2yy = 0 If it’s not clear how the chain rule applies, it might help to think of y as a function of x. So like, you can think of another, easier function. For example, y = 2x + 5 as y(x) = 2x + 5 because you are plugging in an x. This is like writing f (x) = 2x + 5. So then when we take the derivative of y 2 , you can think of it as the derivative of (y(x))2 (think (f (x))2 ), so like d (y(x))2 dx So taking the derivative and applying the chain rule gives us d (y(x))2 = 2(y(x))y 0 (x) which you can write as 2yy 0 dx