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.

Solution 1:

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 $\frac{dx}{dy}$: $$\begin{align} D_y(x^2+y^2)&=D_y(25) \\ D_y(x^2)+D_y(y^2)&=0 \\ D_x(x^2)D_y(x)+2y&=0 \\ 2xD_y(x)+2y&=0 \\ 2xD_y(x)&=-2y \\ D_y(x)=\frac{dx}{dy}&=-\frac{y}{x} \end{align}$$

Solution 2:

Isaac and Ryan have already answered your question in words. Now, in symbols, the chain rule gives: $$\frac{d(y^2)}{dx} = \frac{d(y^2)}{dy}\frac{dy}{dx} = 2y\frac{dy}{dx}$$