Implicit differentiation I want to differentiate 
$x^2 + y^2=1$ with respect to $x$.
The answer is $2x +2yy' = 0$.
Can some explain what is implicit differentiation and from where did $y'$ appear ? I can understand that $2x +2yy' = 0$ is a partial derivative but then it becomes multi calc not single.
This is in a chapter about chain rule so I assume there is some use of the chain rule here but I can't see any composite functions here. We can express y in terms of x but y is not composite.
P.S: I am NOT looking for HOW to solve the problem, I am looking for the WHY as stated above.

The function $x^2 + y^2 = 1$ defines $y$ implicitly as a function of $x$. In this case, we have $y^2 = 1 - x^2$. Thus, instead or writing $y$ in the equation we can write $f(x)$ where $f(x)^2 = 1 - x^2$. This leaves the problem of differentiating $x^2 + f(x)^2 = 1$. In this form, we can see how to apply the chain rule
$$ \begin{align*}
        & \frac{d}{dx} \left( x^2 + f(x)^2 \right) && \\
 ~ = ~  & \frac{d}{dx} x^2 + \frac{d}{dx} f(x)^2   && [\text{linearity}] \\
 ~ = ~ & 2x + \frac{d}{dx} f(x)^2                  && [\text{power rule}] \\
 ~ = ~ & 2x + 2f(x)\frac{d}{dx}f(x)                && [\text{chain rule}]
\end{align*}$$
In the fourth line, we can apply the chain rule because $f(x)^2$ is the composition of two functions, namely the function $g(x) = x^2$ and $h(x) = f(x)$. The chain rule says that to differentiate $(g \circ h)(x)$, we first differentiate $g(x)$ with respect to $x$, then differentiate $h(x)$ with respect to $x$, then form the product $(g' \circ h)(x)h'(x)$ as the result. In this case, differentiate $g(x)$ with respect to $x$ gives $2x$. Differentiating $h(x)$ with respect to $x$ gives $h'(x)$. Thus, $(g \circ h)(x) = 2f(x)f'(x)$.
