# Few questions on implicit function [duplicate]

When we trying to differentiate a function $y=f(x)$, we are actually finding the rate of change of $y$. But what do we exactly mean by differentiating both sides of the equation, say $x^2+y^2=1$, with respect to $x$?

Another question is that we know that some functions are not differentiable, is it true that there are also some equations that cannot be differentiated on both sides with respect to $x$? If yes, what are the conditions? If no, why not?

## marked as duplicate by Git Gud, Community♦Nov 24 '15 at 14:59

What we mean is that we assume that $y=f(x)$ and that $y$ satisfies $$x^2+y^2=1,\tag{\heartsuit}$$ then proceed accordingly. Differentiating on both sides (and applying the Chain Rule) then gives us $$2x+2y\frac{dy}{dx}=0.\tag{\star}$$ Note that if $y=0,$ then $(\star)$ implies that $x=0,$ as well, which is impossible by $(\heartsuit).$ This is because $y$ is not differentiable when $y=0,$ which you should be able to see by looking at a graph. Hence, since $y\ne 0,$ then we solve $(\star)$ to get $$\frac{dy}{dx}=-\frac{x}{y}.$$ This gives us yet another way to see that $y$ isn't differentiable when $y=0.$
We are trying to find the unit vector emanating from $(x, y)$ along the direction in which $x^2+y^2-1$ remains constant or the direction of a unit vector in $R^2$ which is such that the rate of change of the function $f(x, y): = x^2+y^2-1$ is equal to zero.
This interpretation is a result of the implicit function theorem according to which if $\phi$ is a function such that $f(x, \phi (x)) = c$ where c is some constant, then $${dy\over dx} = {d\phi(x)\over dx} -{f_x(x, y)\over f_y(x, y)}$$ and ${f_x(x, y) \ i + f_y(x, y) \ j\over {{[(f_x(x, y))^2+(f_x(x, y))^2]}^{1/2}}} = {{d\phi (x)\over dx} \ i- j\over {{({d\phi (x)\over dx}^2+1)}^{1/2}}}$ will give you the unit normal vector to $y = \phi (x)$ .Computing ${dy\over dx}$ may even help you to find the direction of the unit tangent vector to the curve $y = \phi (x)$
In other words, for any point $(x, y) \ \epsilon \ R^2$, ${dy\over dx}$ at that point gives you the slope of the line passing through the point $(x, y)$ along which the rate of change of $x^2+y^2-1$ is maximum.