# Why can we define y as a function of x in implicit differentiation?

I understand the method used in implicit differentiation, it's just an application of the chain rule. But why can you define $y$ as a function of $x$?

In this equation for example:
$x^2 + y^2 = 1$

$2x + 2yy' = 0$

Why isn't it just this?:
$2x + 2y = 0$

• short answer: you differentiate with respect to $x$ not to a mixture of $x$ and $y$. – user251257 Feb 8 '17 at 23:25
• another short answer: implicit differentiation involves you assuming that $y$ is a function of $x$ – Timothy Cho Feb 8 '17 at 23:26
• @user251257 But why not assume y is constant when differentiating with respect to x? – SadSeven Feb 8 '17 at 23:35
• @SadSeven: Why should it? If it is constant, then $\frac{d}{dx} x^2 + y^2 = 2x$. – user251257 Feb 8 '17 at 23:38
• @SadSeven We treat $y$ as a constant in partial differentiation but not in ordinary differentiation. – projectilemotion Feb 8 '17 at 23:41

(1). $\{(x,y): x^2+y^2 =1\}$ is NOT the graph of a function.
(2). There IS a function $f(x)$ for $|x|<1$ such that $f(x)$ is differentiable and $x^2+f(x)^2=1.$ In fact there are 2 such functions. And we have $$0= \frac {d}{ dx} (1) =\frac {d}{ dx} (x^2+f(x)^2)=2x + \frac {d}{ dx }(f(x)^2)=2x+2f(x)f'(x).$$
Note: In differentiation, it matters what you differentiate BY. Just as in division, where it matters whether you divide by x or by y. The derivative of $y^2$ with respect to $y$ is $2y.$ The derivative of $y^2$ with respect to $x$ is $2y \frac {dy}{dx}.$
Consider $$F(x,y)=0$$ and compute $F'_x(x,y)$ considering $y$ as a constant and then $F'_y(x,y)$ considering $x$ as a constant. Now, the implicit function theorem $$\frac{dy}{dx}=-\frac{F'_x(x,y) }{F'_y(x,y) }$$