I have seen how people implicitly differentiate the equation x^2+y^2=c. d/dx(x^2)+d/dx(y^2)=d/dx(c) treating "y" as "f(x)" and using the chainrule we get 2x+2y(y′)=0 and solving for y′ y′=−2x/2y The problem is that I just don´t understand implicit differentiation, I do know the rules but they don´t make any sense to me. The fact that it is valid to differentiate both "x" and "y" on the same side of the equation is what´s bothering me and even if I see "y" as a function of "x" I just end up imagining x^2+(−x^2+c)=c which doesn´t help me. I also don´t know very much about partial derivatives but I´m willing to learn about them if that helps me understand implicit differentiation.

Amira Serrano 2022-10-13 Answered
I have seen how people implicitly differentiate the equation x 2 + y 2 = c.
d / d x ( x 2 ) + d / d x ( y 2 ) = d / d x ( c )
treating " y" as " f ( x )" and using the chainrule we get
2 x + 2 y ( y ) = 0
and solving for y
y = 2 x / 2 y
The problem is that I just don´t understand implicit differentiation, I do know the rules but they don´t make any sense to me. The fact that it is valid to differentiate both " x" and " y" on the same side of the equation is what´s bothering me and even if I see " y" as a function of " x" I just end up imagining
x 2 + ( x 2 + c ) = c
which doesn´t help me. I also don´t know very much about partial derivatives but I´m willing to learn about them if that helps me understand implicit differentiation.
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Answers (2)

Tiberlaue
Answered 2022-10-14 Author has 7 answers
Maybe this helps. Take the function f ( x ) = x 2 and some other functions g ( x ) and h ( x ).
Let's differentiate the expression
f ( x ) + ( g ( x ) ) 2 = h ( x )
with respect to x (not: "differentiate x").

We use the fact that you can differentiate each summand individually, and the chain rule for ( g ( x ) ) 2 , to get
f ( x ) + 2 g ( x ) g ( x ) = h ( x ) .
In your case, f ( x ) = x 2 , g ( x ) = y ( x ), and h ( x ) = c, i.e.:
2 x + 2 y ( x ) y ( x ) = 0 .
Sometimes, the argument ( x ) is omitted -- that's what you have there; but that's just notation (no deeper meaning, afaik).
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Aarav Atkins
Answered 2022-10-15 Author has 2 answers
We are just using the chain rule
If I said:
f ( x ) = ( 2 x + 1 ) d f d x = 2 g ( x ) = x 2 d g d x = 2 x d d x ( g f ) ( x ) = d g d f d f d x = 2 ( 2 x + 1 ) ( 2 )
Do have a problem with that?
So, y is a function of x. y 2 is a function of y
d d x y 2 = d d y ( y 2 ) d y d x = 2 y d y d x
And now for the last little bit:
d d x ( y 2 = c x 2 ) 2 y d y d x = 2 x
Is that so different from
d d x ( x 2 + y 2 = c ) d d x ( x 2 ) + d d x ( y 2 ) = d d x c 2 x + 2 y d y d x = 0 d y d x = x y
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