



 
Hi, $x^2 + y^2 = 2$ does not define a function, in fact if you solve this equation for $y$ you get \[y = \pm \sqrt{2  x^2}\] which you can write as two functions \[y = \sqrt{2  x^2} \mbox{ and } y =  \sqrt{2  x^2}.\] I can differentiate each of these functions. If $y = \sqrt{2  x^2}.$ \[\frac{dy}{dx} = \frac12 \left(2  x^2\right)^{1/2} (2x) = \frac{x}{\left(2  x^2\right)^{1/2}}\]
\[\frac{dy}{dx} = \frac{x}{y}.\] If $y = \sqrt{2  x^2}.$ \[\frac{dy}{dx} = \frac12 \left(2  x^2\right)^{1/2} (2x) = \frac{x}{\left(2  x^2\right)^{1/2}}\]
\[\frac{dy}{dx} = \frac{x}{y}.\] Notice that in differentiating each of the two functions I used the chain rule. Implicit differentiation allows us to show that \[\frac{dy}{dx} = \frac{x}{y}\] without solving for $y.$ What we can do is think of $y$ as a function of $x,$ even though we don't know an explicit expression for $y.$ Differentiate both sides of $x^2 + y^2 = 2$ with respect to $x.$ The right side is a constant so its derivative is zero. Since $y$ is some function of $x$ I will have to use the chain rule to differentiate it. Thus I get \[2x + 2 y \frac{dy}{dx} = 0\] and simplifying I get \[\frac{dy}{dx} = \frac{x}{y}.\] The real power of implicit differentiation is seen if you can't solve for $y$ as we did above. For example if \[x \; y^7  \frac{x}{y^2} = x^3 y\] then implicit differentiation allows you to find an expression for $\large \frac{dy}{dx}$ in terms of $x$ and $y.$ I hope this helps,  


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