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implicit differentiation

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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
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