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Why, after differentiating $y$ on one side of the equation, is $dy/dx$ added?
As clarification an example I will provide an example:

Implicitly differentiate $y^2 = x$.

You get $$2y\frac {dy}{dx} = 1$$ as one of the first steps in differentiation. Why is the $dy/dx$ added after $y^2$ is differentiated?

Thanks!

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    $\begingroup$ because of the chain rule, $y$ is a function of $x$ and so the derivative of $y$ is $1$ multiplied by its derivative, which is $y'$ $\endgroup$ – imranfat Apr 9 '14 at 19:49
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    $\begingroup$ @Ethan $y^2=x$ is abuse of notation for $\left(y(x)\right)^2=x$. Just apply the chain rule. $\endgroup$ – Git Gud Apr 9 '14 at 19:50
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$$y^2=(y(x))^2=x\\ \dfrac{d(y^2)}{dx}=1\\ \implies \dfrac{d(y^2)}{dx}=\dfrac{d(y^2)}{dy}\dfrac{dy}{dx}\text{ (chain rule )}=1\\ \implies 2y\dfrac{dy}{dx}=1$$

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  • $\begingroup$ When you get $\frac{d(y^2)}{dy} \frac{dy}{dx}$, wouldn't the $dy$ cancel leaving $\frac{d(y^2)}{dx} = 2y$ ? $\endgroup$ – OpieDopee Apr 9 '14 at 20:21
  • $\begingroup$ @Ethan $$\dfrac{d(y^2)}{dy}\dfrac{dy}{dx}=2y\boxed{\dfrac{dy}{dx}}$$Of course, you can cancel out the $dy$s, but what do you get? You get $\dfrac{d(y^2)}{dx}$, which is what you began with. $\endgroup$ – user122283 Apr 9 '14 at 20:25
  • $\begingroup$ So because there are no $x$s in $\frac{d(y^2)}{dx}$, you can differentiate with respect to $y$ giving you $2y$ ? $\endgroup$ – OpieDopee Apr 9 '14 at 20:29
  • $\begingroup$ @Ethan Let me give an example. Let $y=x+1$. Then, $y^2=x^2+1+2x$. Then, $$\dfrac{d(y^2)}{dx}=\dfrac{d(x^2+1+2x)}{dx}=2x+2\\ \dfrac{d(y^2)}{dx}=\dfrac{d(y^2)}{dy}\dfrac{dy}{dx}=2y\dfrac{dy}{dx}=2(x+1)(1)=2x+2$$See that both are equal. $\endgroup$ – user122283 Apr 9 '14 at 20:34
  • $\begingroup$ In the second half of the example when you get $2y \frac{dy}{dx}$ how did $\frac{dy}{dx}$ simplifying to $1$? $\endgroup$ – OpieDopee Apr 9 '14 at 20:42
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The idea behind implicit differentiation is that while $y$ isn't (necessarily) a function of $x$ we treat it as it it was.

So say that $y = f(x)$. Then you want to find $f'(x)$ in the following $$f(x)^2 = x.$$ So you take the derivative on both sides and use the chain rule to find the derivative of $f(x)^2$: $$\begin{align} \frac{d}{dx} f(x)^2 &= \frac{d}{dx} x & \Rightarrow \\ 2f(x)f'(x) &= 1 \end{align} $$ Now then we just replace $f(x)$ by $y$ and get $$ 2y\frac{dy}{dx} = 1. $$ So to answer you question about where the $\frac{dy}{dx}$ comes from, you can think of it as the derivative of the inner function $y$.

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