# 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.

I was just wondering where the y'/(dy/dx) in implicit differentiation comes from
${x}^{2}+{y}^{2}=25$
$\left(d/dx\right){x}^{2}+\left(d/dy\right){y}^{2}\ast \ast \left(dy/dx\right)\ast \ast =25\left(d/dx\right)$
$2x+2y\left(dy/dx\right)=0$
$\left(dy/dx\right)=-x/y$
Where does the bold part come from?
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Simeon Hester
When you implicitly differentiate ${x}^{2}+{y}^{2}=25$, you are differentiating with respect to a particular variable—in this case, x, so:
$\begin{array}{rl}\frac{d}{dx}\left({x}^{2}+{y}^{2}\right)& =\frac{d}{dx}25\\ \frac{d}{dx}\left({x}^{2}\right)+\frac{d}{dx}\left({y}^{2}\right)& =0\\ 2x+2y\frac{dy}{dx}& =0\\ 2y\frac{dy}{dx}& =-2x\\ \frac{dy}{dx}& =-\frac{x}{y}\end{array}$
From the 3rd line to the 4th line, $\frac{d}{dx}\left({y}^{2}\right)$ 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: 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: ${D}_{x}\left({x}^{2}\right)=2x$ is "the derivative with respect to x of ${x}^{2}$ is $2x$."

Now, rewriting the work above in this notation:
$\begin{array}{rl}{D}_{x}\left({x}^{2}+{y}^{2}\right)& ={D}_{x}\left(25\right)\\ {D}_{x}\left({x}^{2}\right)+{D}_{x}\left({y}^{2}\right)& =0\\ 2x+{D}_{y}\left({y}^{2}\right){D}_{x}\left(y\right)& =0\\ 2x+2y{D}_{x}\left(y\right)& =0\\ 2y{D}_{x}\left(y\right)& =-2x\\ {D}_{x}\left(y\right)=\frac{dy}{dx}& =-\frac{x}{y}\end{array}$
And, to your question of finding $\frac{dx}{dy}$:
$\begin{array}{rl}{D}_{y}\left({x}^{2}+{y}^{2}\right)& ={D}_{y}\left(25\right)\\ {D}_{y}\left({x}^{2}\right)+{D}_{y}\left({y}^{2}\right)& =0\\ {D}_{x}\left({x}^{2}\right){D}_{y}\left(x\right)+2y& =0\\ 2x{D}_{y}\left(x\right)+2y& =0\\ 2x{D}_{y}\left(x\right)& =-2y\\ {D}_{y}\left(x\right)=\frac{dx}{dy}& =-\frac{y}{x}\end{array}$
###### Did you like this example?
saucletbh
In symbols, the chain rule gives:
$\frac{d\left({y}^{2}\right)}{dx}=\frac{d\left({y}^{2}\right)}{dy}\frac{dy}{dx}=2y\frac{dy}{dx}$