# I have seen how people implicitly differentiate the equation x^2+y^2=c. d/dx(x^2)+d/dx(y^2)=d/dx(c) treating "y" as "f(x)" and using the chainrule we get 2x+2y(y′)=0 and solving for y′ y′=−2x/2y The problem is that I just don´t understand implicit differentiation, I do know the rules but they don´t make any sense to me. The fact that it is valid to differentiate both "x" and "y" on the same side of the equation is what´s bothering me and even if I see "y" as a function of "x" I just end up imagining x^2+(−x^2+c)=c which doesn´t help me. I also don´t know very much about partial derivatives but I´m willing to learn about them if that helps me understand implicit differentiation.

I have seen how people implicitly differentiate the equation ${x}^{2}+{y}^{2}=c$.
$d/dx\left({x}^{2}\right)+d/dx\left({y}^{2}\right)=d/dx\left(c\right)$
treating "$y$" as "$f\left(x\right)$" and using the chainrule we get
$2x+2y\left({y}^{\prime }\right)=0$
and solving for ${y}^{\prime }$
${y}^{\prime }=-2x/2y$
The problem is that I just don´t understand implicit differentiation, I do know the rules but they don´t make any sense to me. The fact that it is valid to differentiate both "$x$" and "$y$" on the same side of the equation is what´s bothering me and even if I see "$y$" as a function of "$x$" I just end up imagining
${x}^{2}+\left(-{x}^{2}+c\right)=c$
which doesn´t help me. I also don´t know very much about partial derivatives but I´m willing to learn about them if that helps me understand implicit differentiation.
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Tiberlaue
Maybe this helps. Take the function $f\left(x\right)={x}^{2}$ and some other functions $g\left(x\right)$ and $h\left(x\right)$.
Let's differentiate the expression
$f\left(x\right)+\left(g\left(x\right){\right)}^{2}=h\left(x\right)$
with respect to $x$ (not: "differentiate $x$").

We use the fact that you can differentiate each summand individually, and the chain rule for $\left(g\left(x\right){\right)}^{2}$, to get
${f}^{\prime }\left(x\right)+2g\left(x\right){g}^{\prime }\left(x\right)={h}^{\prime }\left(x\right).$
In your case, $f\left(x\right)={x}^{2}$, $g\left(x\right)=y\left(x\right)$, and $h\left(x\right)=c$, i.e.:
$2x+2y\left(x\right){y}^{\prime }\left(x\right)=0.$
Sometimes, the argument $\left(x\right)$ is omitted -- that's what you have there; but that's just notation (no deeper meaning, afaik).
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Aarav Atkins
We are just using the chain rule
If I said:
$f\left(x\right)=\left(2x+1\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{df}{dx}=2\phantom{\rule{0ex}{0ex}}g\left(x\right)={x}^{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{dg}{dx}=2x\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(g\circ f\right)\left(x\right)=\frac{dg}{df}\frac{df}{dx}=2\left(2x+1\right)\left(2\right)$
Do have a problem with that?
So, $y$ is a function of $x$. ${y}^{2}$ is a function of $y$
$\frac{d}{dx}{y}^{2}=\frac{d}{dy}\left({y}^{2}\right)\frac{dy}{dx}=2y\frac{dy}{dx}$
And now for the last little bit:
$\frac{d}{dx}\left({y}^{2}=c-{x}^{2}\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}2y\frac{dy}{dx}=-2x$
Is that so different from
$\frac{d}{dx}\left({x}^{2}+{y}^{2}=c\right)\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left({x}^{2}\right)+\frac{d}{dx}\left({y}^{2}\right)=\frac{d}{dx}c\phantom{\rule{0ex}{0ex}}2x+2y\frac{dy}{dx}=0\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=-\frac{x}{y}$
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