Implicit Differentiation I was just wondering where the y'/(dy/dx) in implicit differentiation comes from.
$$
x^2 + y^2 = 25
$$
$$
(d/dx) x^2 + (d/dy) y^2 **(dy/dx)** = 25 (d/dx)
$$
$$
2x + 2y (dy/dx) = 0
$$
$$
(dy/dx) = -x/y
$$
Where does the bold part come from? Wikipedia says it's a byproduct of the chain rule, but it's just not clicking for me.
 A: Isaac and Ryan have already answered your question in words.  Now, in symbols, the chain rule gives:
$$\frac{d(y^2)}{dx} = \frac{d(y^2)}{dy}\frac{dy}{dx} = 2y\frac{dy}{dx}$$
A: When you implicitly differentiate $x^2+y^2=25$, you are differentiating with respect to a particular variable—in this case, $x$, so:
$$\begin{align}
\frac{d}{dx}(x^2+y^2)&=\frac{d}{dx}25
\\
\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)&=0
\\
2x+2y\frac{dy}{dx}&=0
\\
2y\frac{dy}{dx}&=-2x
\\
\frac{dy}{dx}&=-\frac{x}{y}
\end{align}$$
From the 3rd line to the 4th line, $\frac{d}{dx}(y^2)$ is the derivative with respect to $x$ of $y^2$, in which (as in Ryan Budney's comment) we assume that $y$ is some function of $x$, so we apply the chain rule, differentiating $y^2$ with respect to $y$ and multiplying by the derivative of $y$ with respect to $x$ to get $2y\frac{dy}{dx}$.

edit: Based on the comments below, I think it might be useful if I introduced a slightly different notation: Let $D_x$ be the differential operator with respect to $x$, which you have previously written as $\frac{d}{dx}$ (and, similarly, $D_y$ is the differential operator with respect to $y$).  When we apply the differential operator to something, we read and write it like a function: $D_x(x^2)=2x$ is "the derivative with respect to $x$ of $x^2$ is $2x$."
Now, rewriting the work above in this notation:
$$\begin{align}
D_x(x^2+y^2)&=D_x(25)
\\
D_x(x^2)+D_x(y^2)&=0
\\
2x+D_y(y^2)D_x(y)&=0
\\
2x+2yD_x(y)&=0
\\
2yD_x(y)&=-2x
\\
D_x(y)=\frac{dy}{dx}&=-\frac{x}{y}
\end{align}$$
And, to your question of finding $\frac{dx}{dy}$:
$$\begin{align}
D_y(x^2+y^2)&=D_y(25)
\\
D_y(x^2)+D_y(y^2)&=0
\\
D_x(x^2)D_y(x)+2y&=0
\\
2xD_y(x)+2y&=0
\\
2xD_y(x)&=-2y
\\
D_y(x)=\frac{dx}{dy}&=-\frac{y}{x}
\end{align}$$
